Exam Fitter https://examfitter.com/ Fitting You to Excel Tue, 10 Nov 2026 20:05:38 +0000 en-US hourly 1 https://wordpress.org/?v=7.1.4 https://examfitter.com/wp-content/uploads/2026/09/ef-fav.png Exam Fitter https://examfitter.com/ 32 32 Topic 6.8: Enthalpy of Formation Worksheet https://examfitter.com/ap-chemistry/worksheet/enthalpy-formation-problems/ https://examfitter.com/ap-chemistry/worksheet/enthalpy-formation-problems/#respond Tue, 10 Nov 2026 20:05:37 +0000 https://examfitter.com/?p=2930 How to Use This Worksheet Work in order. Section A checks the foundations: the definition, the zero rule, formation equations, and why states and signs behave as they do. Section B practises the full calculation on six reactions, including one back-calculation of an unknown formation value. Section C is exam style: a multi-part ammonia problem, ... Read more

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Work in order. Section A checks the foundations: the definition, the zero rule, formation equations, and why states and signs behave as they do. Section B practises the full calculation on six reactions, including one back-calculation of an unknown formation value. Section C is exam style: a multi-part ammonia problem, six claims to judge, and a limestone problem that ends in grams and kilojoules.

Reference data. Standard enthalpies of formation, ΔH°f in kJ mol⁻¹ at 298 K, the same table as the revision notes: CO₂(g) −393.5, H₂O(l) −285.8, H₂O(g) −241.8, CH₄(g) −74.8, CO(g) −110.5, NH₃(g) −46.1, NO(g) +90.3, NO₂(g) +33.2, O₃(g) +142.7, SO₂(g) −296.8, SO₃(g) −395.7, CaCO₃(s) −1206.9, CaO(s) −635.1, NaCl(s) −411.2, HCl(g) −92.3, C₂H₅OH(l) −277.7, C₆H₁₂O₆(s) −1273.3. Every element in its standard state has ΔH°f = 0.

Three habits protect every answer. Write the two sums separately, products first, with each value multiplied by its coefficient, before subtracting anything. Match every value to the state printed in the equation, especially water. Finish by checking the sign against the chemistry: combustions must come out negative, and a decomposition of a very stable compound should come out positive.

Do not look at the answer key until you have attempted every question.

Section A: Recall and Concepts

A1. (a) Define the standard enthalpy of formation of a compound, giving the conditions. (b) State the amount of compound formed in a formation reaction. (c) State what the reactants in a formation reaction must be. (d) Explain why fractional coefficients, such as ½O₂, appear in formation equations.

A2. (a) State the ΔH°f of O₂(g), Fe(s) and C(s, graphite), and give the rule that sets them. (b) Diamond is a form of carbon. State whether its ΔH°f is zero, and explain why. (c) Ozone, O₃(g), has ΔH°f = +142.7 kJ mol⁻¹. Explain what the positive sign tells you about ozone relative to O₂. (d) A compound has ΔH°f = −1206.9 kJ mol⁻¹. State what a large negative value indicates about the compound.

A3. Write the formation equation for each compound, with states, and state its ΔH°f from the reference table: (a) H₂O(l); (b) CO₂(g); (c) NH₃(g); (d) CaCO₃(s). (e) Explain why 2H₂(g) + O₂(g) → 2H₂O(l) is not a formation equation, even though only elements appear on the left.

A4. (a) Write the formula for ΔH°rxn in terms of formation enthalpies, including what the coefficient does. (b) State the order of subtraction used with bond enthalpies in Topic 6.7, and explain why the two methods run in opposite orders. (c) Explain why the ΔH°f of O₂ never changes the result of a calculation, whatever its coefficient. (d) A reaction has ΔH°rxn = +178.3 kJ. State, in terms of the two sums, why the answer is positive.

A5. (a) ΔH°f of H₂O(l) is −285.8 kJ mol⁻¹ and of H₂O(g) is −241.8 kJ mol⁻¹. Calculate the difference and name the process it prices. (b) Explain why a calculation must use the value matching the state in the equation. (c) Every ΔH°f in a reaction is negative, yet ΔH°rxn is positive. Explain how that is possible. (d) State the temperature and pressure at which standard enthalpies of formation are tabulated.

Section B: Core Practice

B1. Ammonia synthesis: N₂(g) + 3H₂(g) → 2NH₃(g). (a) Write the products sum and the reactants sum separately. (b) Calculate ΔH°rxn. (c) State ΔH per mole of ammonia formed, and connect it to ΔH°f of NH₃.

B2. Limestone decomposition: CaCO₃(s) → CaO(s) + CO₂(g). (a) Calculate ΔH°rxn. (b) State whether the reaction is exothermic or endothermic, and what that means for a lime kiln. (c) Calculate the heat that must be supplied to decompose 0.50 mol of CaCO₃.

B3. Ethanol combustion: C₂H₅OH(l) + 3O₂(g) → 2CO₂(g) + 3H₂O(l). (a) Write the products sum and the reactants sum separately. (b) Calculate ΔH°rxn. (c) Explain why the 3O₂ contributes nothing to the reactants sum.

B4. Nitric oxide formation: N₂(g) + O₂(g) → 2NO(g). (a) Calculate ΔH°rxn. (b) Calculate ΔH per mole of NO formed. (c) Explain why this reaction’s ΔH°rxn and the ΔH°f of NO carry the same number per mole, when most reactions’ do not.

B5. Finding an unknown formation value. Burning glucose, C₆H₁₂O₆(s) + 6O₂(g) → 6CO₂(g) + 6H₂O(l), has ΔH°rxn = −2802.5 kJ. (a) Calculate the products sum. (b) Set up the formation formula with ΔH°f of glucose as the unknown x. (c) Solve for x. (d) State what the sign and size of x say about glucose relative to its elements.

B6. The state trap. Methane burns two ways on paper: CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(l) and CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(g). (a) Calculate ΔH°rxn for the liquid-water version. (b) Calculate ΔH°rxn for the gas-water version. (c) Calculate the difference between the two answers. (d) Explain the difference using the gap between the two water values and the coefficient of water.

Section C: Exam Style

C1. Ammonia is made by N₂(g) + 3H₂(g) → 2NH₃(g), ΔH°f of NH₃(g) = −46.1 kJ mol⁻¹. (a) Calculate ΔH°rxn, showing both sums. (b) The plant runs the reaction in reverse, decomposing ammonia. State ΔH for 2NH₃(g) → N₂(g) + 3H₂(g). (c) Calculate the heat released when 0.40 mol of NH₃ is formed. (d) A student calculates ΔH°rxn as reactants minus products and gets +92.2 kJ. Identify the error and state what the wrong sign would claim about the reaction.

C2. Judge each claim, supporting each verdict with the rule it depends on. (a) “With formation enthalpies, ΔH°rxn is reactants minus products.” (b) “Diamond has ΔH°f = 0 because it is a pure element.” (c) “If every ΔH°f in a reaction is negative, the reaction must be exothermic.” (d) “Coefficients can be ignored, because ΔH°f is already per mole.” (e) “H₂O(g) and H₂O(l) have the same ΔH°f, since both are water.” (f) “Ozone cannot exist, because its ΔH°f is positive.”

C3. A lime kiln decomposes calcium carbonate: CaCO₃(s) → CaO(s) + CO₂(g). Use ΔH°f values CaCO₃(s) −1206.9, CaO(s) −635.1, CO₂(g) −393.5 kJ mol⁻¹, and take the molar mass of CaCO₃ as 100.09 g mol⁻¹. (a) Calculate ΔH°rxn, showing both sums. (b) Write the formation equation for CaCO₃(s). (c) Calculate the number of moles in 250 g of CaCO₃. (d) Calculate the heat that must be supplied to decompose 250 g of CaCO₃. (e) Explain why the value in (a) is positive even though CaCO₃ has the most negative ΔH°f in the equation.

Answer Key

Show the answer key and full working (only after you have attempted every question)

A1. (a) The standard enthalpy of formation is the enthalpy change when one mole of a compound forms from its elements in their standard states, at 298 K and 1 atm (1 bar). (b) Exactly one mole. (c) The elements that make up the compound, each in its standard state; no compounds may appear on the reactant side. (d) One mole of product fixes the amounts of the elements, and those amounts are often halves or three-halves of a mole: making one mole of water takes half a mole of O₂. The fraction is part of the definition, not an unbalanced equation.

A2. (a) All three are 0 kJ mol⁻¹. An element in its standard state has ΔH°f = 0 by definition, because forming it from itself involves no change. (b) No. Graphite is the most stable form of carbon at standard conditions, so graphite is the reference; diamond sits slightly higher, at about +1.9 kJ mol⁻¹. (c) Forming ozone from O₂ absorbs 142.7 kJ per mole, so ozone holds more enthalpy than the standard form of oxygen and is less stable than O₂. (d) The compound sits far lower in enthalpy than its elements, so it is stable relative to them: a large amount of energy was released when it formed.

A3. (a) H₂(g) + ½O₂(g) → H₂O(l); ΔH°f = −285.8 kJ mol⁻¹. (b) C(s, graphite) + O₂(g) → CO₂(g); ΔH°f = −393.5 kJ mol⁻¹. (c) ½N₂(g) + 3/2H₂(g) → NH₃(g); ΔH°f = −46.1 kJ mol⁻¹. (d) Ca(s) + C(s, graphite) + 3/2O₂(g) → CaCO₃(s); ΔH°f = −1206.9 kJ mol⁻¹. (e) It forms two moles of water. A formation equation forms exactly one mole of product, so this equation carries twice ΔH°f, −571.6 kJ, and must be halved to give the formation value.

A4. (a) ΔH°rxn = Σ nΔH°f(products) − Σ nΔH°f(reactants), where each ΔH°f is multiplied by its coefficient n in the balanced equation. (b) Bond enthalpies use broken minus formed: the reactants’ bonds are broken first, so the reactant side is priced first. Formation enthalpies price where each species sits relative to the elements, so the products’ total is compared against the reactants’ total, products first. The orders differ because the two methods price different journeys. (c) Its value is 0, and any coefficient times 0 is still 0, so the term vanishes from the reactants sum however much O₂ the equation uses. (d) The products sum is higher than the reactants sum by 178.3 kJ: the products hold more enthalpy relative to the elements than the reactants do, so the difference must be absorbed from the surroundings.

A5. (a) −241.8 − (−285.8) = 44.0 kJ mol⁻¹. It is the enthalpy of vaporization of water at 298 K: the energy needed to convert one mole of liquid water into gas. (b) ΔH°f is a property of a specific state, because the state itself holds a different enthalpy. Using the gas value where the equation makes liquid water understates the release by 44.0 kJ for every mole of water. (c) ΔH°rxn is a difference between two sums. If the reactants’ total is more negative than the products’ total, subtracting it gives a positive result: the reactants sit lower relative to the elements than the products do, so forming the products from the reactants climbs in enthalpy. (d) 298 K and 1 atm (1 bar).

B1. (a) Products sum: 2(−46.1) = −92.2 kJ. Reactants sum: 0 + 3(0) = 0 kJ, because both reactants are elements in their standard states. (b) ΔH°rxn = −92.2 − 0 = −92.2 kJ. (c) The equation forms 2 mol of NH₃, so per mole: −92.2 ÷ 2 = −46.1 kJ mol⁻¹, which is exactly ΔH°f of NH₃, as it must be: the synthesis equation is the formation equation doubled.

B2. (a) Products sum: −635.1 + (−393.5) = −1028.6 kJ. Reactants sum: −1206.9 kJ. ΔH°rxn = −1028.6 − (−1206.9) = +178.3 kJ. (b) Endothermic. The kiln must be heated continuously: every mole of limestone decomposed absorbs 178.3 kJ from the fuel burning around it. (c) 0.50 × 178.3 = 89.15 kJ, about 89.2 kJ, must be supplied.

B3. (a) Products sum: 2(−393.5) + 3(−285.8) = −787.0 + (−857.4) = −1644.4 kJ. Reactants sum: −277.7 + 3(0) = −277.7 kJ. (b) ΔH°rxn = −1644.4 − (−277.7) = −1366.7 kJ. (c) O₂ is an element in its standard state, so its ΔH°f is 0, and 3 × 0 adds nothing to the sum. The coefficient matters for balancing, not for the total.

B4. (a) Products sum: 2(+90.3) = +180.6 kJ. Reactants sum: 0 + 0 = 0 kJ. ΔH°rxn = +180.6 − 0 = +180.6 kJ. (b) The equation forms 2 mol of NO, so per mole: +180.6 ÷ 2 = +90.3 kJ mol⁻¹. (c) Both reactants are elements with ΔH°f = 0, so the reactants sum vanishes and ΔH°rxn is just the products sum: coefficient times ΔH°f of the one product. Dividing by the coefficient then returns ΔH°f itself. In most reactions the reactants are compounds with nonzero values, so the subtraction changes the number and no such shortcut exists.

B5. (a) Products sum: 6(−393.5) + 6(−285.8) = −2361.0 + (−1714.8) = −4075.8 kJ. (b) ΔH°rxn = products − reactants: −2802.5 = −4075.8 − (x + 6(0)). (c) x = −4075.8 + 2802.5 = −1273.3 kJ mol⁻¹. (d) The large negative value means glucose sits far below its elements in enthalpy: forming it from carbon, hydrogen and oxygen releases a great deal of energy, so it is stable relative to its elements.

B6. (a) Products sum: −393.5 + 2(−285.8) = −965.1 kJ. Reactants sum: −74.8 kJ. ΔH°rxn = −965.1 − (−74.8) = −890.3 kJ. (b) Products sum: −393.5 + 2(−241.8) = −877.1 kJ. Reactants sum: −74.8 kJ. ΔH°rxn = −877.1 − (−74.8) = −802.3 kJ. (c) −802.3 − (−890.3) = 88.0 kJ: the gas version releases 88.0 kJ less. (d) Each mole of water made as gas instead of liquid keeps 44.0 kJ of the vaporization enthalpy, and the equation makes 2 mol of water, so 2 × 44.0 = 88.0 kJ less is released. The products hold that energy as separated gas molecules rather than giving it up on condensing.

C1. (a) Products sum: 2(−46.1) = −92.2 kJ. Reactants sum: 0 + 3(0) = 0 kJ. ΔH°rxn = −92.2 − 0 = −92.2 kJ. (b) Reversing a reaction reverses the sign at the same size: +92.2 kJ. (c) Forming NH₃ releases 46.1 kJ per mole, so 0.40 × 46.1 = 18.44 kJ is released. (d) The student used the bond enthalpy order on formation data. The wrong sign would claim the synthesis is endothermic, absorbing heat, when forming ammonia from its elements actually releases it.

C2. (a) False. Formation enthalpies are subtracted as products minus reactants. Reactants minus products is the shape of the bond enthalpy formula, and it flips the sign of every formation calculation. (b) False. Only the standard state of an element is zero, and carbon’s standard state is graphite. Diamond is a different, slightly higher-enthalpy form, at about +1.9 kJ mol⁻¹. (c) False. ΔH°rxn is the difference between the products sum and the reactants sum. Limestone decomposition has all-negative values and ΔH = +178.3 kJ, because the reactant’s value is more negative than the products’ total. (d) False. ΔH°f prices one mole, so each value must be multiplied by its coefficient: the 2 in front of SO₃ doubles its contribution. Ignoring coefficients undercounts every species used more than once. (e) False. They differ by 44.0 kJ mol⁻¹, the enthalpy of vaporization: the gas holds more enthalpy than the liquid, so its formation value is less negative. (f) False. A positive ΔH°f means the compound is less stable than its elements and absorbs energy when it forms; it says nothing about whether the compound can exist. Ozone forms in the upper atmosphere and is detectable there, despite sitting 142.7 kJ mol⁻¹ above O₂.

C3. (a) Products sum: −635.1 + (−393.5) = −1028.6 kJ. Reactants sum: −1206.9 kJ. ΔH°rxn = −1028.6 − (−1206.9) = +178.3 kJ. (b) Ca(s) + C(s, graphite) + 3/2O₂(g) → CaCO₃(s). (c) 250 ÷ 100.09 = 2.498 mol, about 2.50 mol. (d) Each mole decomposed absorbs 178.3 kJ, so 2.498 × 178.3 = 445.3 kJ, about 445 kJ, must be supplied. (e) ΔH°rxn compares sums, not single values. CaCO₃’s −1206.9 kJ mol⁻¹ puts the reactant very low, but the products’ total, −1028.6 kJ, is higher than that: the products together sit above the carbonate, so reaching them from the carbonate climbs in enthalpy and absorbs heat. A very negative reactant value makes a positive ΔH more likely, not less.

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]]> https://examfitter.com/ap-chemistry/worksheet/enthalpy-formation-problems/feed/ 0 Topic 6.5: Energy of Phase Changes Worksheet https://examfitter.com/ap-chemistry/worksheet/phase-changes-problems/ https://examfitter.com/ap-chemistry/worksheet/phase-changes-problems/#comments Tue, 10 Nov 2026 20:05:19 +0000 https://examfitter.com/?p=2914 How to Use This Worksheet Work in order. Section A checks the language of the heating curve: its five segments, the plateau idea, the six changes of state and their signs, and the sizes of the numbers. Section B practises stage-by-stage pricing: single plateaus, two-stage heating paths, cooling paths with negative totals, and a reverse ... Read more

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Work in order. Section A checks the language of the heating curve: its five segments, the plateau idea, the six changes of state and their signs, and the sizes of the numbers. Section B practises stage-by-stage pricing: single plateaus, two-stage heating paths, cooling paths with negative totals, and a reverse problem that starts from the heat and finds the mass. Section C is exam style: a full five-stage path, five claims to judge, and a heater problem that connects plateau length to enthalpy.

Reference facts, all for water. Specific heat capacities: ice 2.09, liquid water 4.18, steam 2.01 J g⁻¹ °C⁻¹. ΔHfus = 6.01 kJ mol⁻¹. ΔHvap = 40.7 kJ mol⁻¹. Molar mass 18.02 g mol⁻¹. Warming stages are priced by q = mcΔT with the mass in grams; plateau stages are priced by q = nΔH, with n = mass ÷ molar mass. Signs are taken from the sample’s point of view: heat absorbed is positive, heat released is negative.

Three habits protect every answer. Split the path at every phase boundary before touching a number, and price each stage separately. Convert grams to moles for the plateau stages, and bring joules and kilojoules to one unit before adding anything. Finish by checking the total: its sign must match the direction of the journey, and on any path that boils water the vaporization term should be the largest single term.

Do not look at the answer key until you have attempted every question.

Section A: Recall and Concepts

A1. (a) List the five segments of the heating curve of water, in order, from ice below 0 °C to steam above 100 °C. (b) State which segments are sloped and which are flat. (c) State the equation that prices each sloped segment, and name what each symbol stands for. (d) State the equation that prices each flat segment. (e) State the temperature of each flat segment for water.

A2. (a) Define the enthalpy of fusion, ΔHfus, and give its value for water. (b) Define the enthalpy of vaporization, ΔHvap, and give its value for water. (c) Write down the enthalpy change for freezing one mole of water, and for condensing one mole of steam. (d) Explain, at the particle level, why ΔHvap is so much larger than ΔHfus.

A3. (a) Ice at 0 °C is heated steadily, yet its temperature does not rise while it melts. State where the added heat goes. (b) State what temperature measures, and use that to explain why the temperature cannot rise during the melting. (c) State what starts to happen to the temperature the moment the last crystal has melted. (d) A student claims the plateau means no heat is being absorbed there. Correct the claim.

A4. (a) Name the six changes of state between solid, liquid and gas. (b) Mark each one as endothermic or exothermic. (c) Pair each change with its exact reverse. (d) Write ΔHsub for water in terms of ΔHfus and ΔHvap, and calculate its value. (e) State the enthalpy change when one mole of water vapour deposits directly as ice.

A5. Four processes are proposed for exactly 1.00 mol of water: warming the ice by 10 °C; melting the ice; warming the steam by 10 °C; vaporizing the liquid. (a) Calculate the heat absorbed in each process. (b) Rank the four from smallest to largest. (c) State what the ranking shows about where the energy sits in a full heating path. (d) Using the same numbers, explain why the vaporization plateau lasts far longer than the melting plateau when the heat is supplied at a steady rate.

Section B: Core Practice

B1. (a) Calculate the number of moles in 54.1 g of ice. (b) Calculate the heat needed to melt all of it at 0 °C. (c) State why q = mcΔT cannot price this step.

B2. 45.0 g of liquid water at 0 °C freezes solid, and the ice then cools to −15.0 °C. (a) Calculate the number of moles of water. (b) Calculate q for the freezing stage. (c) Calculate q for the cooling stage. (d) Calculate the total q for the journey, and state what its sign tells you.

B3. 20.0 g of ice at −20.0 °C is heated until it is liquid water at 0 °C. (a) Name the two stages of the journey. (b) Calculate q for the first stage. (c) Calculate q for the second stage, showing the mole conversion. (d) Calculate the total heat absorbed, in kJ.

B4. 15.0 g of water at 75.0 °C is heated until it is entirely steam at 100 °C. (a) Name the two stages of the journey. (b) Calculate q for the first stage. (c) Calculate q for the second stage, showing the mole conversion. (d) Calculate the total heat absorbed, in kJ. (e) State which stage dominates the total, and explain why.

B5. 30.0 g of steam at 100 °C is cooled until it is ice at 0 °C. (a) Name the three stages of the journey. (b) Calculate q for each stage. (c) Calculate the total q, in kJ. (d) State the heat needed for the exact reverse journey, ice at 0 °C to steam at 100 °C for the same sample, and justify the answer without recalculating.

B6. (a) How many moles of ice at 0 °C can 25.0 kJ of heat melt? (b) What mass of ice is that? (c) Calculate the heat absorbed when 0.500 mol of ice sublimes. (d) Calculate q for the deposition of 0.500 mol of water vapour as ice, and state whether the surroundings warm or cool.

Section C: Exam Style

C1. 40.0 g of ice at −15.0 °C is heated steadily until it is steam at 120.0 °C. (a) List the five stages of the journey in order. (b) Calculate q for each stage, showing the mole conversion where one is needed. (c) Calculate the total heat absorbed, in kJ. (d) State which stage takes the largest share of the total, and connect that fact to the shape of the heating curve. (e) A student prices the whole journey as q = mcΔT, using c = 4.18 J g⁻¹ °C⁻¹ and ΔT = 135 °C, and gets 22.6 kJ. Identify the two distinct errors in that working.

C2. Judge each claim, supporting each verdict with the rule it depends on. (a) “While ice melts, the temperature stays constant because the heater’s energy stops entering the sample.” (b) “A single q = mcΔT calculation can price a whole heating curve, provided the overall temperature change is used.” (c) “Freezing one mole of water releases exactly the amount of heat that melting one mole of ice absorbs.” (d) “ΔHvap is larger than ΔHfus mainly because vaporization happens at a higher temperature.” (e) “On a cooling curve, the condensation plateau sits at a lower temperature than the boiling plateau on a heating curve.”

C3. A 500 W heater supplies 500 J every second, and all of it reaches the sample. It is used on 100 g of ice at 0 °C. (a) Calculate the heat needed to melt all of the ice, and hence the time the melting plateau lasts. (b) Calculate the time the resulting water takes to warm from 0 °C to 100 °C. (c) Calculate the heat needed to vaporize all of the water at 100 °C, and hence the time the boiling plateau lasts. (d) Explain why the two plateau times differ so much, using the values of ΔHfus and ΔHvap. (e) Calculate the total time from ice at 0 °C to steam at 100 °C.

Show the answer key and full working (only after you have attempted every question)

A1. (a) Ice warming; melting; liquid water warming; vaporization; steam warming. (b) Sloped: the three warmings. Flat: melting and vaporization. (c) q = mcΔT, where m is the mass of the sample in grams, c is the specific heat capacity of the phase being warmed, and ΔT is the temperature change. (d) q = nΔH, with ΔHfus at the melting plateau and ΔHvap at the boiling plateau. (e) Melting at 0 °C; vaporization at 100 °C.

A2. (a) The heat absorbed to melt one mole of a solid at its melting point; 6.01 kJ mol⁻¹ for water. (b) The heat absorbed to vaporize one mole of a liquid at its boiling point; 40.7 kJ mol⁻¹ for water. (c) Freezing: −6.01 kJ mol⁻¹. Condensing: −40.7 kJ mol⁻¹. Each reverse change keeps the magnitude and flips the sign. (d) Melting only loosens the ordered solid: the particles stay in contact and many attractions survive into the liquid. Vaporization pulls the particles completely apart, so every attraction between them must be overcome, and that complete separation costs far more energy per mole.

A3. (a) Into overcoming the intermolecular attractions that hold the particles in the solid: it raises the particles’ potential energy. (b) Temperature measures the average kinetic energy of the particles. While the melt lasts, none of the added heat becomes kinetic energy, so the average, and with it the temperature, cannot move. (c) It starts to rise: with no solid left to melt, further heat has nowhere to go except into kinetic energy, and the liquid warms. (d) Heat is being absorbed all through the plateau; that absorption is precisely what drives the change of state. The plateau shows heat being absorbed without a temperature rise, not heat failing to enter.

A4. (a) Melting, freezing, vaporization, condensation, sublimation, deposition. (b) Endothermic: melting, vaporization, sublimation. Exothermic: freezing, condensation, deposition. (c) Melting pairs with freezing; vaporization with condensation; sublimation with deposition. (d) ΔHsub = ΔHfus + ΔHvap = 6.01 + 40.7 = 46.7 kJ mol⁻¹. (e) −46.7 kJ mol⁻¹: deposition is the reverse of sublimation, endothermic forward, so the reverse is exothermic with the same magnitude.

A5. (a) Warming the ice: per mole of ice, 18.02 × 2.09 × 10 = 377 J, about 0.38 kJ. Melting: 6.01 kJ. Warming the steam: 18.02 × 2.01 × 10 = 362 J, about 0.36 kJ. Vaporizing: 40.7 kJ. (b) Warming the steam (0.36 kJ), then warming the ice (0.38 kJ), then melting (6.01 kJ), then vaporizing (40.7 kJ). (c) The phase changes hold nearly all the energy, and vaporization holds most of that; the warming stages are small corrections either side of the plateaus. (d) At a steady supply rate, a plateau’s length in time is set by the energy it must deliver per mole. Vaporization needs 40.7 kJ for every mole against melting’s 6.01 kJ, about 6.8 times as much, so its plateau lasts about 6.8 times longer.

B1. (a) n = 54.1 ÷ 18.02 = 3.00 mol. (b) q = nΔHfus = 3.00 × 6.01 = 18.0 kJ, absorbed. (c) During melting the temperature does not change, so ΔT = 0 and q = mcΔT would return zero; the energy goes into potential energy against the intermolecular forces, and only a per-mole price, nΔH, captures it.

B2. (a) n = 45.0 ÷ 18.02 = 2.497 mol. (b) Freezing: q = −nΔHfus = −2.497 × 6.01 = −15.0 kJ, that is −15,008 J. (c) Cooling the ice: q = mcΔT = 45.0 × 2.09 × (−15.0) = −1,411 J, that is −1.41 kJ. (d) Total: −15,008 + (−1,411) = −16,419 J, about −16.4 kJ. The negative sign says the sample releases heat overall: the journey is exothermic from the sample’s point of view, and the surroundings receive 16.4 kJ.

B3. (a) Warm the ice from −20.0 °C to 0 °C, then melt it at 0 °C. (b) q = mcΔT = 20.0 × 2.09 × 20.0 = 836 J. (c) n = 20.0 ÷ 18.02 = 1.110 mol, so q = nΔHfus = 1.110 × 6.01 = 6,670 J. (d) Total: 836 + 6,670 = 7,506 J = 7.51 kJ absorbed.

B4. (a) Warm the liquid from 75.0 °C to 100 °C, then vaporize it at 100 °C. (b) q = mcΔT = 15.0 × 4.18 × 25.0 = 1,568 J. (c) n = 15.0 ÷ 18.02 = 0.832 mol, so q = nΔHvap = 0.832 × 40.7 = 33,879 J, that is 33.9 kJ. (d) Total: 1,568 + 33,879 = 35,447 J, about 35.4 kJ absorbed. (e) Vaporization, by a wide margin: 33.9 kJ of the 35.4 kJ. Warming moves particles faster within the liquid, but vaporizing has to separate every particle from all of its neighbours, and complete separation is priced per mole at 40.7 kJ.

B5. (a) Condense the steam at 100 °C; cool the liquid from 100 °C to 0 °C; freeze it at 0 °C. (b) n = 30.0 ÷ 18.02 = 1.665 mol. Condensing: q = −nΔHvap = −1.665 × 40.7 = −67,758 J. Cooling the liquid: q = mcΔT = 30.0 × 4.18 × (0 − 100) = −12,540 J. Freezing: q = −nΔHfus = −1.665 × 6.01 = −10,006 J. (c) Total: −67,758 − 12,540 − 10,006 = −90,304 J, about −90.3 kJ. (d) +90.3 kJ must be absorbed. Every stage of the reverse journey is the exact reverse of a stage here, and a reverse change keeps its magnitude and flips its sign, so the totals are equal in size and opposite in sign.

B6. (a) n = 25.0 ÷ 6.01 = 4.16 mol. (b) Mass = 4.16 × 18.02 = 75.0 g. (c) ΔHsub = 6.01 + 40.7 = 46.71 kJ mol⁻¹, so q = 0.500 × 46.71 = 23.4 kJ absorbed. (d) q = −23.4 kJ: deposition releases the heat that sublimation absorbs. Released heat flows into the surroundings, so the surroundings warm.

C1. (a) Warm the ice, −15.0 °C to 0 °C; melt at 0 °C; warm the water, 0 °C to 100 °C; vaporize at 100 °C; warm the steam, 100 °C to 120.0 °C. (b) The sample is n = 40.0 ÷ 18.02 = 2.220 mol throughout. Warm the ice: q = 40.0 × 2.09 × 15.0 = 1,254 J. Melt: q = 2.220 × 6.01 = 13,341 J. Warm the water: q = 40.0 × 4.18 × 100.0 = 16,720 J. Vaporize: q = 2.220 × 40.7 = 90,344 J. Warm the steam: q = 40.0 × 2.01 × 20.0 = 1,608 J. (c) Total: 1,254 + 13,341 + 16,720 + 90,344 + 1,608 = 123,267 J, about 123 kJ absorbed. (d) Vaporization: 90,344 J of the 123,267 J, roughly 73 percent. On the heating curve that is the long flat stretch: the boiling plateau stretches far beyond the melting plateau because separating the particles completely costs the most energy per mole. (e) First, a single mcΔT term prices only warming and captures none of the energy absorbed at the two plateaus, where most of the true total sits. Second, it applies the liquid’s specific heat capacity to all five stages, but the ice stage needs 2.09 and the steam stage 2.01 J g⁻¹ °C⁻¹. Both errors shrink the answer, which is why 22.6 kJ lands far below the true 123 kJ.

C2. (a) False. The heater’s energy enters the sample continuously through the plateau; it is spent raising potential energy against the intermolecular forces rather than kinetic energy, which is why the temperature, the measure of average kinetic energy, stays fixed. (b) False. The plateau stages absorb large amounts of heat while ΔT is zero, and they are priced per mole by q = nΔH. A single warming term can never include them, whatever overall ΔT is inserted. (c) True. Freezing is the exact reverse of melting: same magnitude, opposite sign, −6.01 kJ released against +6.01 kJ absorbed per mole for water. (d) False. The difference is the separation achieved: melting loosens the particles while they stay in contact, vaporization separates them completely. The boiling point temperature is where the change happens, not the reason it costs more. (e) False. Condensation and boiling are the same phase boundary crossed in opposite directions, so both plateaus sit at 100 °C for water at the same pressure; the cooling curve mirrors the heating curve.

C3. (a) n = 100 ÷ 18.02 = 5.549 mol, so melting needs q = 5.549 × 6.01 = 33,352 J. At 500 J per second, the plateau lasts 33,352 ÷ 500 = 66.7 s. (b) Warming the water needs q = mcΔT = 100 × 4.18 × 100 = 41,800 J, taking 41,800 ÷ 500 = 83.6 s. (c) Vaporizing needs q = 5.549 × 40.7 = 225,860 J, taking 225,860 ÷ 500 = 451.7 s. (d) Both plateaus serve the same 5.549 mol at the same supply rate, so each plateau’s time is proportional to its per-mole price. Vaporization separates the particles completely at 40.7 kJ per mole, while melting only loosens them at 6.01 kJ per mole, so the boiling plateau lasts about 6.8 times longer than the melting plateau. (e) Total: 66.7 + 83.6 + 451.7 = 602 s, about ten minutes, most of it spent on the boiling plateau.

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]]> https://examfitter.com/ap-chemistry/worksheet/phase-changes-problems/feed/ 1 Topic 6.6: Introduction to Enthalpy of Reaction Worksheet https://examfitter.com/ap-chemistry/worksheet/enthalpy-reaction-problems/ https://examfitter.com/ap-chemistry/worksheet/enthalpy-reaction-problems/#respond Tue, 10 Nov 2026 20:05:18 +0000 https://examfitter.com/?p=2929 How to Use This Worksheet Work in order. Section A checks the definitions and rules: what enthalpy is, what a state function means, what a thermochemical equation must carry, and the standard conditions behind ΔH°. Section B practises the two working rules, reversing and scaling, on six different reactions, including the per-mole pricing that every ... Read more

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Work in order. Section A checks the definitions and rules: what enthalpy is, what a state function means, what a thermochemical equation must carry, and the standard conditions behind ΔH°. Section B practises the two working rules, reversing and scaling, on six different reactions, including the per-mole pricing that every heat question reduces to. Section C is exam style: a multi-part peroxide problem, six claims to judge, and a Haber process calculation that ends with heat per gram of product.

Reference data. Thermochemical equations used throughout: 2H₂(g) + O₂(g) → 2H₂O(l), ΔH = −572 kJ. N₂(g) + 3H₂(g) → 2NH₃(g), ΔH = −92 kJ. CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(l), ΔH = −890 kJ. CaCO₃(s) → CaO(s) + CO₂(g), ΔH = +178 kJ. 2CO(g) + O₂(g) → 2CO₂(g), ΔH = −566 kJ. H₂(g) + Cl₂(g) → 2HCl(g), ΔH = −184 kJ. 2SO₂(g) + O₂(g) → 2SO₃(g), ΔH = −198 kJ. 2H₂O₂(l) → 2H₂O(l) + O₂(g), ΔH = −196 kJ. Molar masses: H₂O₂ 34.02 g mol⁻¹, NH₃ 17.03 g mol⁻¹, N₂ 28.02 g mol⁻¹. Signs are taken from the system’s point of view: heat released by the system is negative, heat absorbed is positive.

Three habits protect every answer. Read the ΔH as pricing the equation exactly as written, never as a per-mole value, until you have divided by a coefficient yourself. When you rewrite an equation, carry the ΔH with it: reverse it and flip the sign, scale it and scale the value by the same factor. Finish by checking the sign against the equation’s own ΔH: formation and combustion reactions here release heat, so their answers must come out negative; the limestone decomposition absorbs heat, so its answer comes out positive, but the peroxide decomposition in C1 releases heat, so its answer comes out negative. Never guess the sign from the reaction type alone.

Do not look at the answer key until you have attempted every question.

Section A: Recall and Concepts

A1. (a) Define enthalpy and write the equation that defines it, naming each symbol. (b) At constant pressure, the heat exchanged by a reaction equals which quantity? (c) Use that fact to explain why enthalpy is more useful than internal energy for reactions run in open flasks. (d) State the sign of ΔH for an exothermic reaction and for an endothermic reaction, and state what each sign says about the direction of heat flow.

A2. (a) State what it means to call enthalpy a state function. (b) The same reactants reach the same products by two different routes, one using a catalyst and one not. Compare the ΔH of the two routes and justify the comparison. (c) Write ΔH in terms of the enthalpies of the reactants and the products. (d) A reaction runs in five small steps instead of one. State, with a reason, whether its overall ΔH changes.

A3. (a) List the three things a thermochemical equation must include. (b) Explain why the physical states belong in it. (c) For 2H₂(g) + O₂(g) → 2H₂O(l), ΔH = −572 kJ, state exactly which amounts of which substances the −572 kJ refers to. (d) A card shows only “H₂ + O₂ → H₂O”. Name three faults in it as a thermochemical equation.

A4. (a) Write the symbol for a standard enthalpy change and state what the symbol’s extra mark means. (b) State the standard pressure for a gas. (c) State the temperature at which thermochemical data are usually tabulated. (d) State the standard concentration for a substance in solution. (e) State the standard enthalpy of formation of O₂(g), and explain why it has that value.

A5. Start from 2H₂(g) + O₂(g) → 2H₂O(l), ΔH = −572 kJ. Write the ΔH for each rewritten version: (a) the equation halved; (b) the equation tripled; (c) the equation reversed; (d) the amount of H₂ named in the original equation replaced by 1 mol of H₂ reacting, coefficients adjusted to match; (e) the heat released per mole of O₂ consumed.

Section B: Core Practice

B1. Ammonia synthesis: N₂(g) + 3H₂(g) → 2NH₃(g), ΔH = −92 kJ. (a) Calculate the heat released per mole of NH₃ formed. (b) Write the thermochemical equation for the formation of 1 mol of NH₃, fractional coefficients allowed, and state its ΔH. (c) Calculate the heat released when 6.0 mol of H₂ reacts completely. (d) Calculate the heat released when 0.25 mol of N₂ reacts completely.

B2. Methane combustion: CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(l), ΔH = −890 kJ. (a) Calculate the heat released when 2.5 mol of CH₄ burns completely. (b) Calculate the heat released per mole of H₂O formed. (c) Write the thermochemical equation for the combustion of 2 mol of CH₄ and state its ΔH. (d) State whether the ΔH in part (c) describes a more or less exothermic reaction than the original, and explain the answer in one sentence.

B3. Limestone decomposition: CaCO₃(s) → CaO(s) + CO₂(g), ΔH = +178 kJ. (a) Calculate the heat absorbed when 0.50 mol of CaCO₃ decomposes. (b) Write the thermochemical equation for the reverse process, CaO reacting with CO₂, and state its ΔH. (c) Calculate the heat released when 2.0 mol of CO₂ is captured by CaO in that reverse process. (d) State, with a reason, whether the decomposition is endothermic or exothermic.

B4. Carbon monoxide combustion: 2CO(g) + O₂(g) → 2CO₂(g), ΔH = −566 kJ. (a) Calculate the heat released per mole of CO burned. (b) Calculate the heat released per mole of CO₂ formed, and comment on how it compares with part (a). (c) Calculate the heat released when 0.40 mol of CO burns. (d) Write the thermochemical equation rewritten per mole of CO, and state its ΔH.

B5. Hydrogen chloride formation: H₂(g) + Cl₂(g) → 2HCl(g), ΔH = −184 kJ. (a) Calculate the heat released when 5.0 mol of HCl is formed. (b) State the ΔH for 2H₂(g) + 2Cl₂(g) → 4HCl(g). (c) Calculate the heat absorbed when 1.0 mol of HCl is decomposed back into its elements. (d) Calculate the heat released per mole of Cl₂ consumed.

B6. Sulfur trioxide formation: 2SO₂(g) + O₂(g) → 2SO₃(g), ΔH = −198 kJ. (a) Calculate the heat released per mole of SO₃ formed. (b) Calculate the heat released when 10 mol of SO₃ is formed. (c) This ΔH was measured with the reaction open to the air. State which quantity the measured heat equals, and the condition that makes the equality hold. (d) A student claims a negative ΔH means this reaction must be fast. Correct the claim, using the state function idea.

Section C: Exam Style

C1. Hydrogen peroxide decomposes: 2H₂O₂(l) → 2H₂O(l) + O₂(g), ΔH = −196 kJ. (a) Calculate the heat released per mole of H₂O₂ decomposed. (b) A first-aid solution decomposes completely, and the peroxide in it amounts to 3.4 g. Calculate the heat released. (c) Write the thermochemical equation for the reverse process and state its ΔH. (d) A manganese(IV) oxide catalyst is added and the same sample decomposes in seconds instead of days. State the ΔH of the catalysed decomposition and justify the answer. (e) Calculate the heat released when 0.50 mol of H₂O₂ decomposes.

C2. Judge each claim, supporting each verdict with the rule it depends on. (a) “Doubling every coefficient in a thermochemical equation leaves ΔH unchanged, because it is still the same reaction.” (b) “The reverse reaction has the same ΔH as the forward reaction, sign included.” (c) “For a reaction run in an open beaker, the heat exchanged equals ΔH.” (d) “ΔH = −572 kJ for 2H₂ + O₂ → 2H₂O means each mole of H₂ releases 572 kJ.” (e) “The standard enthalpy of formation of O₂(g) is zero by definition.” (f) “A positive ΔH means the reaction gives out heat to the surroundings.”

C3. The Haber process runs N₂(g) + 3H₂(g) → 2NH₃(g), ΔH = −92 kJ. (a) A batch uses 28.02 g of N₂ with excess H₂ and converts it completely. Calculate the number of moles of N₂ and the heat released. (b) Calculate the mass of NH₃ the batch produces. (c) Calculate the heat released per gram of NH₃ produced. (d) A second batch uses half as much of each reactant. State the heat it releases and the mass of NH₃ it produces. (e) The plant then doubles its hourly throughput at the same recipe. State what happens to (i) the total heat released per hour and (ii) the ΔH of the equation as written, and explain why the two answers differ.

Show the answer key and full working (only after you have attempted every question)

A1. (a) Enthalpy is the internal energy of a system plus the product of its pressure and volume: H = E + PV, where E is the internal energy, P the pressure and V the volume. (b) The enthalpy change, ΔH: at constant pressure, q_p = ΔH. (c) Open flasks sit at the constant pressure of the atmosphere, so the heat a bench reaction exchanges is ΔH directly. Internal energy misses the expansion work done against the atmosphere when gases are produced, and enthalpy includes it through the PV term. (d) Exothermic: ΔH is negative, because the system loses enthalpy as heat flows out to the surroundings. Endothermic: ΔH is positive, because the system gains enthalpy as heat flows in from the surroundings.

A2. (a) Its value depends only on the current state of the system, the substances, amounts, states, temperature and pressure, and not on the path by which that state was reached. (b) They are equal. ΔH is fixed by the initial and final states alone, and both routes share the same two states; a catalyst changes the path and the speed, neither of which a state function records. (c) ΔH = H(products) − H(reactants). (d) It does not change. The overall change is the difference between the same two endpoint states, however many steps connect them.

A3. (a) A balanced chemical equation; the physical state of every reactant and product; and the ΔH value for the equation as written. (b) Different physical states are different states with different enthalpies: forming liquid water and forming steam from the same reactants give different ΔH values, so an equation without states does not fix which change its ΔH prices. (c) It refers to 2 mol of H₂ reacting with 1 mol of O₂ to form 2 mol of liquid water, releasing 572 kJ in total for those amounts. (d) It is not balanced (one O atom on the right against two on the left); it shows no physical states; and it carries no ΔH value, so no heat is priced at all.

A4. (a) ΔH°. The small circle marks standard conditions: the value applies when every substance is in its standard state. (b) 1 atm (1 bar in the newer convention). (c) 298 K. (d) 1 mol L⁻¹. (e) Zero. By convention the standard enthalpy of formation of an element in its standard state is defined as zero, making the elements the reference from which all formation values are measured; O₂(g) is oxygen’s standard state, so it sits on the reference line itself.

A5. (a) Halving scales ΔH by one half: −572 ÷ 2 = −286 kJ. (b) Tripling scales it by three: 3 × (−572) = −1,716 kJ. (c) Reversing keeps the size and flips the sign: +572 kJ. (d) One mole of H₂ is half the equation’s 2 mol, so the release is 572 ÷ 2 = 286 kJ, and ΔH = −286 kJ for H₂(g) + ½O₂(g) → H₂O(l). (e) The equation consumes 1 mol of O₂ for the full 572 kJ, so the heat released per mole of O₂ consumed is 572 kJ.

B1. (a) The equation forms 2 mol of NH₃ per 92 kJ released, so per mole: 92 ÷ 2 = 46 kJ released. (b) ½N₂(g) + ³⁄₂H₂(g) → NH₃(g), ΔH = −46 kJ: halving the equation halves ΔH. (c) The equation consumes 3 mol of H₂ per 92 kJ released. 6.0 mol of H₂ is two equation-loads, so the heat released is 2 × 92 = 184 kJ. (d) The equation consumes 1 mol of N₂ per 92 kJ released, so 0.25 mol releases 0.25 × 92 = 23 kJ.

B2. (a) The equation burns 1 mol of CH₄ per 890 kJ released, so 2.5 mol releases 2.5 × 890 = 2,225 kJ. (b) The equation forms 2 mol of H₂O per 890 kJ released, so per mole of water: 890 ÷ 2 = 445 kJ released. (c) 2CH₄(g) + 4O₂(g) → 2CO₂(g) + 4H₂O(l), ΔH = 2 × (−890) = −1,780 kJ. (d) Neither more nor less exothermic per mole of methane: doubling the equation doubles the material burned, so the heat doubles with it, while the release per mole of CH₄ is unchanged at 890 kJ.

B3. (a) The equation decomposes 1 mol of CaCO₃ per 178 kJ absorbed, so 0.50 mol absorbs 0.50 × 178 = 89 kJ. (b) CaO(s) + CO₂(g) → CaCO₃(s), ΔH = −178 kJ: reversal keeps the magnitude and flips the sign. (c) The reverse equation captures 1 mol of CO₂ per 178 kJ released, so 2.0 mol of CO₂ releases 2.0 × 178 = 356 kJ. (d) Endothermic: its ΔH is positive, meaning the system absorbs heat from the surroundings, which a kiln must supply continuously to keep limestone decomposing.

B4. (a) The equation burns 2 mol of CO per 566 kJ released, so per mole: 566 ÷ 2 = 283 kJ released. (b) The equation also forms 2 mol of CO₂ per 566 kJ, so per mole of CO₂: 566 ÷ 2 = 283 kJ released, exactly the same as part (a), because CO and CO₂ carry the same coefficient in this equation. (c) 0.40 mol of CO at 283 kJ per mole releases 0.40 × 283 = 113 kJ. (d) CO(g) + ½O₂(g) → CO₂(g), ΔH = −283 kJ: halving the equation halves ΔH.

B5. (a) The equation forms 2 mol of HCl per 184 kJ released, so per mole: 92 kJ. Forming 5.0 mol releases 5.0 × 92 = 460 kJ. (b) Doubling the equation doubles ΔH: 2 × (−184) = −368 kJ. (c) Decomposition is the reverse process: forming 1 mol of HCl releases 92 kJ, so decomposing 1.0 mol of HCl absorbs 92 kJ. (d) The equation consumes 1 mol of Cl₂ per 184 kJ released, so the heat released per mole of Cl₂ consumed is 184 kJ.

B6. (a) The equation forms 2 mol of SO₃ per 198 kJ released, so per mole: 198 ÷ 2 = 99 kJ released. (b) 10 mol × 99 kJ = 990 kJ released. (c) The measured heat equals ΔH, and the equality holds because the reaction runs at constant pressure, open to the air: q_p = ΔH. (d) The claim confuses thermodynamics with kinetics. ΔH compares only the enthalpies of the reactants and products, the two endpoint states; it records nothing about the path between them or how fast the reaction travels it, so a very exothermic reaction can still be very slow.

C1. (a) The equation decomposes 2 mol of H₂O₂ per 196 kJ released, so per mole: 196 ÷ 2 = 98 kJ released. (b) n = 3.4 ÷ 34.02 = 0.10 mol of H₂O₂. At 98 kJ released per mole, the heat released is 0.10 × 98 = 9.8 kJ. (c) 2H₂O(l) + O₂(g) → 2H₂O₂(l), ΔH = +196 kJ: the reverse process keeps the magnitude and flips the sign. (d) ΔH is unchanged at −196 kJ for the equation as written. Enthalpy is a state function, so ΔH depends only on the peroxide at the start and the water and oxygen at the end; the catalyst changes the route and the speed, never the endpoints. (e) 0.50 mol at 98 kJ released per mole gives 0.50 × 98 = 49 kJ released.

C2. (a) False. ΔH is extensive: it prices the amounts in the equation, so doubling the amounts doubles the heat and ΔH doubles with the coefficients. (b) False. Reversing swaps reactants and products in ΔH = H(products) − H(reactants), so the value keeps its magnitude and flips its sign. (c) True. An open beaker sits at the constant pressure of the atmosphere, and at constant pressure the heat exchanged is exactly ΔH. (d) False. The 572 kJ prices the whole equation as written, in which 2 mol of H₂ reacts; per mole of H₂ the release is 572 ÷ 2 = 286 kJ. (e) True. An element in its standard state is the reference of the formation scale, and its standard enthalpy of formation is defined as zero. (f) False. Positive ΔH means the system absorbs heat from the surroundings: the process is endothermic. Heat given out carries a negative ΔH.

C3. (a) n = 28.02 ÷ 28.02 = 1.00 mol of N₂. The equation consumes 1 mol of N₂ per 92 kJ released, so the batch releases 92 kJ. (b) The equation forms 2 mol of NH₃ per mole of N₂, so the batch forms 2.00 mol of NH₃, a mass of 2.00 × 17.03 = 34.06 g, about 34.1 g. (c) Heat per gram of NH₃: 92 ÷ 34.06 = 2.70 kJ per gram. (d) Half the material follows the same equation at half the scale: the heat released is 46 kJ, and the NH₃ produced is 1.00 mol, a mass of 17.03 g. (e) (i) The total heat released per hour doubles, because twice as much reaction happens per hour and each equation-load of reaction releases its 92 kJ. (ii) ΔH for the equation as written is unchanged at −92 kJ. The two answers differ because ΔH is a price per equation-load, a fixed property of the reaction’s endpoints, while the total heat is that price multiplied by how many loads run: throughput scales the total, never the price.

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]]> https://examfitter.com/ap-chemistry/worksheet/enthalpy-reaction-problems/feed/ 0 Topic 6.9: Hess’s Law Worksheet https://examfitter.com/ap-chemistry/worksheet/hess-law-problems/ https://examfitter.com/ap-chemistry/worksheet/hess-law-problems/#respond Tue, 10 Nov 2026 20:03:58 +0000 https://examfitter.com/?p=2928 How to Use This Worksheet Work in order. Section A checks the foundations: the law, the three manipulations, cancellation and states. Section B practises building target equations on six cycles, including one unknown step and one cycle that needs a state change step. Section C is exam style: a graphite and diamond problem, six claims ... Read more

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Work in order. Section A checks the foundations: the law, the three manipulations, cancellation and states. Section B practises building target equations on six cycles, including one unknown step and one cycle that needs a state change step. Section C is exam style: a graphite and diamond problem, six claims to judge, and a propane problem that recovers a formation enthalpy.

Every step value is given inside its question. States are printed in every equation because states decide what may cancel.

Three habits protect every answer. Write the target first and keep it in view. Apply each manipulation to the equation and its ΔH together, never to one without the other. Before adding ΔH values, add the manipulated equations and check the sum reproduces the target exactly.

Do not look at the answer key until you have attempted every question.

Section A: Recall and Concepts

A1. (a) State Hess’s law. (b) Define a state function, and name two other state functions besides enthalpy. (c) Explain why Hess’s law follows from enthalpy being a state function. (d) State the main practical use of the law.

A2. A step P → Q has ΔH = −120 kJ. State the equation and ΔH after each manipulation: (a) the step is reversed; (b) the step is multiplied by 2; (c) the step is halved; (d) the step is reversed and then multiplied by 3.

A3. (a) State the three conditions a species must meet before it cancels when equations are added. (b) One step produces H₂O(l) and another step consumes H₂O(g). State whether they cancel, and explain why, giving the enthalpy difference between the two states. (c) A summed equation has 3H₂O(l) on the product side and 2H₂O(l) on the reactant side. State what remains after cancellation. (d) State whether C(s, graphite) can cancel C(s, diamond) in a summed equation, and explain why.

A4. (a) Describe what a Hess cycle diagram shows. (b) A cycle has a direct route with ΔH(direct) and an indirect route through two steps with ΔH values ΔH(1) and ΔH(2). Write the equation connecting them. (c) Explain why the enthalpy change for C(s, graphite) → C(s, diamond) is found from a cycle rather than measured directly. (d) A catalyst is added to the direct route of a cycle. State whether ΔH(direct) changes, and explain your answer.

A5. (a) State the first thing to write down when tackling a Hess calculation. (b) Explain which given step is usually manipulated first, and what decides whether it is reversed or scaled. (c) A species needed on the product side of the target appears only on the reactant side of its step. State the manipulation required. (d) State the check that must pass before the ΔH values are added, and what a failed check means.

Section B: Core Practice

B1. Given (1) N₂(g) + O₂(g) → 2NO(g), ΔH = +180.6 kJ; (2) 2NO(g) + O₂(g) → 2NO₂(g), ΔH = −114.2 kJ. Target: N₂(g) + 2O₂(g) → 2NO₂(g). (a) State which species cancels when the steps are added, and whether either step needs reversing or scaling. (b) Calculate ΔH for the target. (c) Calculate ΔH per mole of NO₂ formed. (d) The target is endothermic even though one step is exothermic. Explain why.

B2. Given (1) C(s) + O₂(g) → CO₂(g), ΔH = −393.5 kJ; (2) CO(g) + ½O₂(g) → CO₂(g), ΔH = −283.0 kJ. Target: C(s) + ½O₂(g) → CO(g). (a) State which step must be reversed, and write the reversed step with its new ΔH. (b) Show which species cancel fully and which cancels only in part. (c) Calculate ΔH for the target. (d) The target is a formation equation. State which tabulated quantity the answer equals.

B3. Given 2H₂(g) + O₂(g) → 2H₂O(l), ΔH = −571.6 kJ. (a) Calculate ΔH for H₂(g) + ½O₂(g) → H₂O(l). (b) State the manipulation, or pair of manipulations, that converts the given step into 3H₂O(l) → 3H₂(g) + 3/2O₂(g), and calculate its ΔH. (c) Calculate ΔH for 4H₂(g) + 2O₂(g) → 4H₂O(l). (d) Explain why halving an equation in the middle of a cycle is legitimate, even though half a molecule cannot react.

B4. Given (1) S(s) + O₂(g) → SO₂(g), ΔH = −296.8 kJ; (2) 2SO₂(g) + O₂(g) → 2SO₃(g), ΔH = −197.8 kJ. Target: S(s) + 3/2O₂(g) → SO₃(g). (a) Write the halved second step, with its ΔH. (b) State which species cancels when the steps are added. (c) Calculate ΔH for the target. (d) State which tabulated quantity the answer equals, and justify the claim from the target equation.

B5. Finding an unknown step. The overall reaction 2NO₂(g) → N₂(g) + 2O₂(g) has ΔH = −66.4 kJ. It is the sum of step 1, 2NO₂(g) → 2NO(g) + O₂(g) with ΔH = +114.2 kJ, and step 2, 2NO(g) → N₂(g) + O₂(g), whose ΔH is unknown. (a) Write the Hess sum as an equation with the unknown as x. (b) Solve for x. (c) State which forward reaction step 2 is the reverse of, using the steps in B1. (d) State what the sign of x says about step 2.

B6. The state conversion. Given (1) CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(l), ΔH = −890.3 kJ; (2) H₂O(l) → H₂O(g), ΔH = +44.0 kJ. Target: CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(g). (a) State the factor by which step 2 must be multiplied, and why that factor is forced by the target. (b) Calculate ΔH for the target. (c) Calculate how much less heat the target reaction releases than step 1. (d) Explain why burning methane to water vapour releases less heat than burning it to liquid water.

Section C: Exam Style

C1. Carbon has two solid forms. Given C(s, graphite) + O₂(g) → CO₂(g), ΔH = −393.5 kJ, and C(s, diamond) + O₂(g) → CO₂(g), ΔH = −395.4 kJ. (a) State the manipulations that build the target C(s, graphite) → C(s, diamond) from these steps. (b) Calculate ΔH for the target. (c) Calculate the heat absorbed when 24.02 g of graphite is converted to diamond. Take the molar mass of carbon as 12.01 g mol⁻¹. (d) Explain why this ΔH cannot be measured by a direct experiment. (e) A student argues: “Diamond burns more exothermically, so diamond must be the more stable form.” Judge the argument.

C2. Judge each claim, supporting each verdict with the rule it depends on. (a) “Reversing an equation doubles its ΔH.” (b) “The ΔH of the target is the average of the step ΔH values.” (c) “H₂O(l) produced in one step cancels H₂O(g) consumed in another, because both are water.” (d) “Doubling an equation leaves ΔH unchanged, because the ratio of products to reactants is the same.” (e) “A catalyst lowers ΔH, so a catalysed reaction has a different ΔH from the same reaction uncatalysed.” (f) “If the steps add to the target, their ΔH values may be added as given, whatever manipulations were applied to the equations.”

C3. Propane’s formation enthalpy cannot be measured directly, but its combustion can. Given (1) C(s) + O₂(g) → CO₂(g), ΔH = −393.5 kJ; (2) H₂(g) + ½O₂(g) → H₂O(l), ΔH = −285.8 kJ; (3) C₃H₈(g) + 5O₂(g) → 3CO₂(g) + 4H₂O(l), ΔH = −2219.9 kJ. (a) Write the target formation equation for propane, C₃H₈(g), with states. (b) State the manipulation each of the three given steps needs: the factor for steps 1 and 2, and what must happen to step 3. (c) Calculate ΔH for the target, showing each manipulated value. (d) Calculate the heat released when 0.50 mol of propane burns completely to liquid water. (e) Explain why reacting carbon and hydrogen together gives no clean measurement of the target in (a).

Answer Key

Show the answer key and full working (only after you have attempted every question)

A1. (a) The enthalpy change for a reaction is the same whichever route is taken from reactants to products, provided the initial and final states are the same; equivalently, ΔH for a target reaction equals the sum of the ΔH values of steps that add to it. (b) A state function is a property whose value depends only on the current state of the system, not on the path taken to reach it. Pressure, volume and temperature are state functions, as are internal energy and enthalpy itself. (c) Because enthalpy is fixed by the state alone, the enthalpy gap between given reactants and products is fixed, so the step values along every route must sum to that same gap. (d) To find ΔH for reactions that cannot be measured directly: too slow, too dangerous, or spoiled by side reactions.

A2. (a) Q → P, ΔH = +120 kJ: reversing flips the sign and keeps the size. (b) 2P → 2Q, ΔH = −240 kJ: scaling by 2 doubles ΔH. (c) ½P → ½Q, ΔH = −60 kJ: halving halves ΔH. (d) 3Q → 3P, ΔH = +360 kJ: the reversal flips −120 to +120, and multiplying by 3 gives +360 kJ.

A3. (a) The same chemical formula, in the same physical state, with the same total coefficient, appearing on opposite sides of the summed equation. (b) No. H₂O(l) and H₂O(g) hold different enthalpies, differing by 44.0 kJ mol⁻¹, the enthalpy of vaporization, so they are different species for cancellation. They connect only through H₂O(l) → H₂O(g), ΔH = +44.0 kJ. (c) 1H₂O(l) remains on the product side: 3 mol produced minus 2 mol consumed leaves a net 1 mol. Partial cancellation of this kind is normal. (d) No. Graphite and diamond are different allotropes with different enthalpies, differing by 1.9 kJ mol⁻¹, so neither cancels the other; converting one into the other is itself a reaction with its own ΔH.

A4. (a) A diagram with the reactants and products at two corners and at least two routes drawn between them: a direct route, and an indirect route passing through intermediate states, each route labelled with its ΔH. (b) ΔH(direct) = ΔH(1) + ΔH(2): going round the cycle and returning to the start gives a net enthalpy change of zero, so the two routes carry equal totals. (c) Graphite converts to diamond immeasurably slowly under ordinary conditions, so no calorimeter can capture the change directly. Both forms burn readily, so the cycle through their common combustion product, CO₂, prices the conversion from two measurable steps. (d) No. A catalyst changes the route and the rate, but the reactants and products are unchanged, so the enthalpy gap between them is unchanged; only the path and the activation energy along it differ.

A5. (a) The target equation, written in full with states. Every later decision is a comparison against it. (b) The step containing a species that appears in only one step, because it has no partner to share the species with. Match it to the target: reverse it if the species sits on the wrong side, scale it if the coefficient is wrong, or both. (c) Reverse the step, which moves the species to the product side and flips the sign of its ΔH. (d) Adding the manipulated equations must reproduce the target exactly: same species, same coefficients, same states, nothing left over. A failed check means a manipulation is wrong, and the ΔH sum built on it cannot be trusted.

B1. (a) 2NO cancels: step 1 produces it and step 2 consumes it, with the same coefficient and state. Neither step needs reversing or scaling; both are used as written. (b) ΔH = +180.6 + (−114.2) = +66.4 kJ. The summed equation is N₂(g) + 2O₂(g) → 2NO₂(g), exactly the target. (c) The target forms 2 mol of NO₂, so per mole: +66.4 ÷ 2 = +33.2 kJ mol⁻¹. (d) Forming NO costs +180.6 kJ, and oxidising it returns only 114.2 kJ, so the products sit 66.4 kJ higher in enthalpy than the reactants, which is what a positive ΔH records.

B2. (a) Step 2 must be reversed, because CO₂ is a product in both given steps but appears nowhere in the target. Reversed: CO₂(g) → CO(g) + ½O₂(g), ΔH = +283.0 kJ. (b) CO₂ cancels fully: 1 mol produced by step 1, 1 mol consumed by the reversed step 2. O₂ cancels in part: step 1 consumes 1 mol, the reversed step 2 produces ½ mol, leaving ½ mol consumed, as the target requires. (c) ΔH = −393.5 + 283.0 = −110.5 kJ. (d) It equals the standard enthalpy of formation of CO(g), ΔH°f = −110.5 kJ mol⁻¹, because the target forms one mole of CO from its elements in their standard states.

B3. (a) Halving the equation halves ΔH: −571.6 ÷ 2 = −285.8 kJ. (b) Reverse the step and multiply by 3/2, in either order. Reversing alone gives 2H₂O(l) → 2H₂(g) + O₂(g), ΔH = +571.6 kJ; scaling by 3/2 gives 3H₂O(l) → 3H₂(g) + 3/2O₂(g), ΔH = +571.6 × 3/2 = +857.4 kJ. (c) Doubling the given step doubles ΔH: 2 × (−571.6) = −1143.2 kJ. (d) A cycle step is a bookkeeping entry, not a claim about a single molecular event. Scaling scales every amount together, so the ratios and the chemistry are unchanged; only the batch size changes, and ΔH, being extensive, scales with it.

B4. (a) SO₂(g) + ½O₂(g) → SO₃(g), ΔH = −197.8 ÷ 2 = −98.9 kJ. (b) SO₂ cancels: step 1 produces 1 mol and the halved step 2 consumes 1 mol, same state. The O₂ adds: 1 + ½ = 3/2 mol consumed, as the target requires. (c) ΔH = −296.8 + (−98.9) = −395.7 kJ. (d) It equals ΔH°f of SO₃(g), −395.7 kJ mol⁻¹, because the target forms exactly one mole of SO₃ from its elements in their standard states.

B5. (a) −66.4 = +114.2 + x. (b) x = −66.4 − 114.2 = −180.6 kJ. Check: +114.2 + (−180.6) = −66.4 kJ, the overall value. (c) Step 2, 2NO(g) → N₂(g) + O₂(g), is the reverse of step 1 in question B1, N₂(g) + O₂(g) → 2NO(g), and its ΔH is the negative of that step’s value: −(+180.6) = −180.6 kJ, agreeing with the solved x by a different route. (d) The negative sign means step 2 is exothermic: breaking 2NO down to its elements releases 180.6 kJ.

B6. (a) By 2. The target makes 2 mol of H₂O(g), and each use of step 2 converts only 1 mol, so step 2 is taken twice: 2H₂O(l) → 2H₂O(g), ΔH = 2 × (+44.0) = +88.0 kJ. (b) Add step 1 and the doubled step 2: the 2H₂O(l) produced by step 1 is consumed by the doubled step 2 and cancels, leaving CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(g). ΔH = −890.3 + 88.0 = −802.3 kJ. (c) −802.3 releases 88.0 kJ less than −890.3: exactly the cost of vaporising the 2 mol of water. (d) Liquid water’s molecules sit in a hydrogen bonded network at lower enthalpy. Making the vapour pulls that network apart, so 44.0 kJ per mole stays held in the products.

C1. (a) Keep the graphite combustion as written. Reverse the diamond combustion, giving CO₂(g) → C(s, diamond) + O₂(g), ΔH = +395.4 kJ, then add. (b) The CO₂ and the O₂ both cancel, leaving C(s, graphite) → C(s, diamond). ΔH = −393.5 + 395.4 = +1.9 kJ. (c) Moles of carbon: 24.02 ÷ 12.01 = 2.00 mol. Heat absorbed: 2.00 × 1.9 = 3.8 kJ. (d) The conversion is immeasurably slow at ordinary conditions, so a calorimeter sees nothing happen. The cycle works because both forms burn quickly and completely to the same product, CO₂. (e) The argument reads the cycle backwards. Burning diamond releases more heat precisely because diamond starts higher in enthalpy: it has further to fall to reach CO₂. The ΔH of +1.9 kJ confirms graphite sits lower and is the more stable form.

C2. (a) False. Reversing changes the sign of ΔH and keeps its size: a step of −40 kJ becomes +40 kJ, not −80 kJ. Doubling happens only when the equation is multiplied by 2. (b) False. ΔH values add, after manipulation. An average would shrink as more steps describe the same reaction, while the true ΔH is fixed by the reactants and products. (c) False. Cancellation needs the same formula, state and coefficient. H₂O(l) and H₂O(g) differ by 44.0 kJ mol⁻¹, so treating them as identical drops a real enthalpy change from the sum. (d) False. ΔH is extensive, priced for the equation as written, so doubling every amount doubles the heat absorbed or released. (e) False. A catalyst changes the route and lowers the activation energy, but reactants and products are the same states, so ΔH is unchanged. The claim would make Hess’s law fail on a change of path, the opposite of what the law says. (f) False. Each ΔH must carry the manipulation applied to its equation: reversed steps enter with flipped signs, scaled steps with scaled values. Adding the original values prices different equations, not the ones that sum to the target.

C3. (a) 3C(s) + 4H₂(g) → C₃H₈(g). (b) Step 1 multiplied by 3: 3C(s) + 3O₂(g) → 3CO₂(g), ΔH = 3 × (−393.5) = −1180.5 kJ. Step 2 multiplied by 4: 4H₂(g) + 2O₂(g) → 4H₂O(l), ΔH = 4 × (−285.8) = −1143.2 kJ. Step 3 reversed, so propane becomes a product and CO₂ and H₂O become reactants that cancel: 3CO₂(g) + 4H₂O(l) → C₃H₈(g) + 5O₂(g), ΔH = +2219.9 kJ. (c) Adding the manipulated steps cancels 3CO₂, 4H₂O(l) and all 5O₂, leaving the target. ΔH = −1180.5 + (−1143.2) + 2219.9 = −103.8 kJ, which is ΔH°f of C₃H₈(g), −103.8 kJ mol⁻¹. (d) Burning releases 2219.9 kJ per mole, from step 3 as written, so 0.50 mol releases 0.50 × 2219.9 = 1109.95 kJ, about 1110 kJ. (e) Carbon and hydrogen heated together do not react cleanly to propane alone: the mixture gives a spread of hydrocarbons and unreacted material, too slow and tangled to price in a calorimeter. Because the direct route cannot be isolated, its ΔH is recovered from steps that can be.

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]]> https://examfitter.com/ap-chemistry/worksheet/hess-law-problems/feed/ 0 Topic 6.3: Heat Transfer and Thermal Equilibrium Worksheet https://examfitter.com/ap-chemistry/worksheet/heat-transfer-problems/ https://examfitter.com/ap-chemistry/worksheet/heat-transfer-problems/#respond Tue, 10 Nov 2026 20:03:53 +0000 https://examfitter.com/?p=2907 How to Use This Worksheet Work in order. Section A checks the definitions and the rules: heat against temperature, the heat equation and its signs, specific heat capacity, and thermal equilibrium. Section B practises the calculations: warming and cooling with q = mcΔT, finding an unknown specific heat, and mixing samples to an equilibrium temperature. ... Read more

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]]> How to Use This Worksheet

Work in order. Section A checks the definitions and the rules: heat against temperature, the heat equation and its signs, specific heat capacity, and thermal equilibrium. Section B practises the calculations: warming and cooling with q = mcΔT, finding an unknown specific heat, and mixing samples to an equilibrium temperature. Section C is exam style: a full metal-in-water equilibrium, judging claims, and a timed heater problem.

Reference facts. Specific heat capacities, in J g⁻¹ °C⁻¹: water 4.18, aluminium 0.897, iron 0.450, copper 0.385. q = mcΔT, with ΔT = T(final) − T(initial), and every sign is taken from the system’s point of view: q is positive when the system absorbs heat, negative when it releases heat. In an insulated mixture, heat lost by the hot sample equals heat gained by the cold sample.

Three habits protect every answer. Always work out ΔT first, as final minus initial, before touching the equation. Keep the sign that the arithmetic gives you, and say in words which way the heat went. In mixing problems, compare the mc products before calculating: they tell you which starting temperature the answer must sit near.

Do not look at the answer key until you have attempted every question.

Section A: Recall and Concepts

A1. (a) Define heat, and state its units. (b) State what temperature measures, and state its units. (c) State the direction in which heat flows spontaneously between two objects at different temperatures. (d) State two properties of the objects that do not decide that direction.

A2. (a) Write the equation connecting heat, mass, specific heat capacity and temperature change, and define each symbol with its units. (b) Write ΔT in terms of the initial and final temperatures. (c) State the condition that must hold for this equation to capture the whole heat transfer. (d) State when q is positive and when q is negative, from the system’s point of view.

A3. (a) Define specific heat capacity, and state its units. (b) State the specific heat capacity of liquid water. (c) Equal masses of a high-c substance and a low-c substance receive equal heat. State which one warms more, and explain why in one sentence. (d) State what the product mc represents for a sample, with its units.

A4. For each event, state the sign of q for the named system. (a) The water in a beaker warming from 20 °C to 35 °C; system: the water. (b) A metal block cooling from 80 °C to 60 °C; system: the block. (c) Soup cooling on a counter; system: the soup. (d) Hands warming while wrapped around a hot mug; system: the hands. (e) A sample held for an hour at exactly the temperature of its surroundings; state the net q.

A5. (a) Name the route by which heat is transferred through direct contact. (b) Name the other two routes, and state briefly how each carries energy. (c) Explain at particle level why the far end of a metal rod warms when only one end is heated. (d) State the two conditions that define thermal equilibrium between two objects.

Section B: Core Practice

B1. 75 g of water warms from 15 °C to 45 °C. (a) Calculate ΔT. (b) Calculate q, showing the substitution. (c) State the sign of q and what it means. (d) Give q in kilojoules. (e) A heater supplying 50 J per second did the warming. Calculate how long it ran.

B2. 120 g of water cools from 85 °C to 30 °C. (a) Calculate ΔT. (b) Calculate q. (c) State how much heat was released, in kilojoules. (d) All of that released heat is captured by 100 g of water starting at 20 °C. Calculate the temperature rise of that water, and its final temperature.

B3. A heater supplies 385 J to a 50 g sample of copper. (a) Calculate the temperature rise of the copper. (b) State its final temperature if it started at 25 °C. (c) Calculate the temperature rise if the same 385 J were supplied to 50 g of water instead. (d) Explain the difference between the two rises in one sentence.

B4. A 60 g sample of an unknown metal absorbs 540 J and warms from 25 °C to 45 °C. (a) Calculate ΔT. (b) Calculate the specific heat capacity of the metal, showing the rearrangement you use. (c) Identify the metal from the reference values. (d) Calculate the heat needed to warm 600 g of this metal by the same temperature change.

B5. Water is mixed with water in an insulated jug. (a) 200 g at 90 °C is mixed with 200 g at 10 °C. State the final temperature, and the rule that gives it at once. (b) 150 g at 60 °C is mixed with 50 g at 20 °C. Calculate the final temperature, setting heat lost equal to heat gained. (c) What mass of water at 10 °C must be added to 200 g at 90 °C for the mixture to land at 50 °C?

B6. A small heater delivers 300 J to 30 g of water and, separately, 300 J to 30 g of copper. (a) Calculate the temperature rise of each sample. (b) Calculate the ratio of the copper rise to the water rise. (c) Show that this ratio equals the inverse ratio of the two specific heat capacities. (d) A student says the copper warms more because it is denser. State the property that actually decides the rises.

Section C: Exam Style

C1. A 50 g block of copper at 100 °C is dropped into 200 g of water at 25 °C in an insulated cup. (a) Write the heat balance equation for this mixture in symbols, using T for the final temperature. (b) Solve it for the final temperature. (c) Calculate the heat gained by the water. (d) Check your answer by calculating the heat lost by the copper. (e) Explain why the final temperature sits near 25 °C rather than near 62.5 °C, the average of the two starting temperatures.

C2. Judge each claim, supporting each verdict with the rule it depends on. (a) “At thermal equilibrium, the two objects contain the same amount of heat.” (b) “Heat always flows from the object with more internal energy to the object with less.” (c) “Doubling the mass doubles the heat needed for the same temperature rise.” (d) “Substance X warmed twice as much as substance Y when equal masses received equal heat, so X has twice the specific heat capacity of Y.” (e) “At thermal equilibrium, heat stops moving between the two objects.”

C3. A small immersion heater supplies heat at 25 J per second. (a) Calculate the heat needed to raise 80 g of water by 10 °C. (b) Calculate how long the heater takes to do this. (c) The heater is then used on 80 g of cooking oil, with c = 1.90 J g⁻¹ °C⁻¹, for the same 10 °C rise. Calculate the heat needed and the time taken. (d) If the oil were left on the heater for the full time found in part (b), calculate its temperature rise. (e) State whether running the heater for longer changes the specific heat capacity of the oil, and give your reason.

Show the answer key and full working (only after you have attempted every question)

A1. (a) Heat is energy transferred between objects because of a temperature difference; it is measured in joules. (b) Temperature measures the average kinetic energy of the particles in a substance; measured in degrees Celsius or kelvin. (c) From the object at the higher temperature to the object at the lower temperature. (d) Any two of: mass, density, total internal energy (amount of stored energy).

A2. (a) q = mcΔT: q is the heat transferred, in joules; m is the mass, in grams; c is the specific heat capacity, in J g⁻¹ °C⁻¹; ΔT is the temperature change, in °C or K. (b) ΔT = T(final) − T(initial). (c) The substance must stay in a single phase throughout: no melting, boiling, freezing or condensing during the change. (d) q is positive when the system absorbs heat (warming); q is negative when the system releases heat (cooling).

A3. (a) The heat needed to raise the temperature of one gram of a substance by one degree Celsius; units J g⁻¹ °C⁻¹. (b) 4.18 J g⁻¹ °C⁻¹. (c) The low-c substance warms more: with q and m fixed, ΔT = q ÷ (mc), so the smaller c gives the larger temperature change. (d) mc is the heat capacity of the sample itself: the heat needed to raise that whole sample by one degree, in J °C⁻¹.

A4. (a) Positive: the water warms, so it absorbs heat. (b) Negative: the block cools, so it releases heat. (c) Negative: the soup cools, releasing heat to the room. (d) Positive: the hands warm, absorbing heat from the mug. (e) Net q = 0: temperatures are equal, so there is no net transfer, which is thermal equilibrium.

A5. (a) Conduction. (b) Convection: energy carried by the bulk movement of a fluid. Radiation: energy carried by electromagnetic waves, with no material contact. (c) Particles at the heated end gain kinetic energy and collide with their slower neighbours, passing energy along the rod particle by particle until the far end’s average kinetic energy has risen. (d) The two objects have the same temperature, and the net heat flow between them is zero.

B1. (a) ΔT = 45 − 15 = +30 °C. (b) q = mcΔT = 75 × 4.18 × 30 = 9,405 J. (c) Positive: the water absorbed heat as it warmed. (d) 9,405 J ÷ 1,000 = 9.41 kJ. (e) Time = 9,405 ÷ 50 = 188.1 s, about 188 s: at a fixed supply rate, the time is simply the heat needed divided by the joules arriving each second.

B2. (a) ΔT = 30 − 85 = −55 °C. (b) q = mcΔT = 120 × 4.18 × (−55) = −27,588 J. (c) The negative sign means release: 27,588 J, which is 27.6 kJ, was released. (d) ΔT = 27,588 ÷ (100 × 4.18) = 27,588 ÷ 418 = 66.0 °C. Final temperature = 20 + 66.0 = 86 °C.

B3. (a) ΔT = q ÷ (mc) = 385 ÷ (50 × 0.385) = 385 ÷ 19.25 = 20.0 °C. (b) 25 + 20.0 = 45 °C. (c) ΔT = 385 ÷ (50 × 4.18) = 385 ÷ 209 = 1.84 °C. (d) The same heat is spread over the same mass in both cases, but water’s specific heat capacity is nearly eleven times copper’s, so its temperature rise is nearly eleven times smaller.

B4. (a) ΔT = 45 − 25 = +20 °C. (b) c = q ÷ (mΔT) = 540 ÷ (60 × 20) = 540 ÷ 1,200 = 0.450 J g⁻¹ °C⁻¹. (c) Iron, whose reference value is 0.450 J g⁻¹ °C⁻¹. (d) q = mcΔT = 600 × 0.450 × 20 = 5,400 J, or 5.40 kJ: ten times the mass needs ten times the heat for the same rise.

B5. (a) 50 °C. Equal masses of the same substance have equal mc products, so the final temperature is the average of the two starts: (90 + 10) ÷ 2 = 50 °C. (b) Heat lost by the hot portion = heat gained by the cold portion: 150 × (60 − T) = 50 × (T − 20), the c values cancelling because both portions are water. Expanding: 9,000 − 150T = 50T − 1,000, so 10,000 = 200T and T = 50 °C. (c) Heat lost by the 200 g must equal heat gained by the added mass m: 200 × (90 − 50) = m × (50 − 10), so 8,000 = 40m and m = 200 g.

B6. (a) Water: ΔT = 300 ÷ (30 × 4.18) = 300 ÷ 125.4 = 2.39 °C. Copper: ΔT = 300 ÷ (30 × 0.385) = 300 ÷ 11.55 = 26.0 °C. (b) Ratio = 26.0 ÷ 2.39 = 10.9. (c) Inverse ratio of specific heats = 4.18 ÷ 0.385 = 10.9. The ratios match: with q and m equal, ΔT is inversely proportional to c, so the ratio of rises is the ratio of the c values turned upside down. (d) Specific heat capacity. Density never enters q = mcΔT; copper’s low c, not its density, is why its temperature climbs further on the same heat.

C1. (a) Heat lost by copper = heat gained by water: 50 × 0.385 × (100 − T) = 200 × 4.18 × (T − 25), that is, 19.25 × (100 − T) = 836 × (T − 25). (b) Expanding: 1,925 − 19.25T = 836T − 20,900. Collecting: 22,825 = 855.25T, so T = 26.7 °C. (c) Heat gained by the water = mcΔT = 836 × (26.688 − 25) = 836 × 1.688 = 1,411 J, about 1.41 kJ. (d) Heat lost by the copper = 19.25 × (100 − 26.688) = 19.25 × 73.312 = 1,411 J, about 1.41 kJ. The two routes agree at 1,411 J, as conservation requires: using the rounded 26.7 °C in part (c) would give 1,421 J, which is a rounding artefact, not a real gap. (e) Averaging the starts would be right only if the two mc products were equal. Here the water’s mc is 836 J °C⁻¹ against the copper’s 19.25 J °C⁻¹, over forty times larger, so the final temperature is dragged to within 1.7 °C of the water’s start.

C2. (a) False. Objects do not contain heat: heat is energy in transit, not a stored quantity. Equilibrium means the two objects share a temperature and the net flow between them is zero; their internal energies can be very different. (b) False. Direction is set by temperature alone. A large lukewarm object can hold more internal energy than a small hot one, yet heat flows from the small hot object into it, as with the nail and the bath in the notes. (c) True. In q = mcΔT, with c and ΔT unchanged, q is proportional to m, so double the mass needs double the heat. (d) False, and backwards. With q and m equal, ΔT is inversely proportional to c, so warming twice as much means X has half the specific heat capacity of Y, not twice. (e) False at particle level. Energy is still exchanged in both directions across the boundary; the flows are equal, so the net flow is zero and the temperatures hold steady.

C3. (a) q = mcΔT = 80 × 4.18 × 10 = 3,344 J. (b) Time = 3,344 ÷ 25 = 133.76 s, about 134 s. (c) q = 80 × 1.90 × 10 = 1,520 J. Time = 1,520 ÷ 25 = 60.8 s. (d) In 133.76 s the heater supplies the same 3,344 J, so ΔT = 3,344 ÷ (80 × 1.90) = 3,344 ÷ 152 = 22.0 °C: the oil rises 22.0 °C, more than double the water’s rise, because its specific heat capacity is less than half of water’s. (e) No. Specific heat capacity is a property of the substance; a longer run delivers more heat and produces a larger ΔT, but c stays at 1.90 J g⁻¹ °C⁻¹ throughout.

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]]> https://examfitter.com/ap-chemistry/worksheet/heat-transfer-problems/feed/ 0 Topic 6.7: Bond Enthalpies Worksheet https://examfitter.com/ap-chemistry/worksheet/bond-enthalpy-problems/ https://examfitter.com/ap-chemistry/worksheet/bond-enthalpy-problems/#respond Tue, 10 Nov 2026 20:03:40 +0000 https://examfitter.com/?p=2923 How to Use This Worksheet Work in order. Section A checks the foundations: what a bond enthalpy is, how bonds are counted from Lewis structures, what controls a bond’s size, and why the method only estimates. Section B practises the full calculation on six reactions, ending with a back-calculation of an unknown bond enthalpy. Section ... Read more

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Work in order. Section A checks the foundations: what a bond enthalpy is, how bonds are counted from Lewis structures, what controls a bond’s size, and why the method only estimates. Section B practises the full calculation on six reactions, ending with a back-calculation of an unknown bond enthalpy. Section C is exam style: a full combustion inventory, six claims to judge, and a hydrazine problem built around a structure to price bond by bond.

Reference data. Average bond enthalpies in kJ mol⁻¹, the same table as the revision notes: H−H 436, H−F 567, H−Cl 431, H−Br 366, H−I 299, Cl−Cl 242, Br−Br 193, O=O 498, N≡N 945, N−N 163, N−H 391, O−H 467, C−H 413, C−C 347, C=C 614, C≡C 839, C−O 358, C=O 799, C−Cl 328. Every reaction in this worksheet is treated as taking place entirely in the gas phase.

Three habits protect every answer. Build the inventory first: list every bond broken and every bond formed before reaching for a calculator. Keep the formula the right way round: broken minus formed, because breaking is the cost and forming is the payback. Finish by checking the sign against the chemistry: combustions must come out negative, and a positive result there means the sums were almost certainly swapped.

Do not look at the answer key until you have attempted every question.

Section A: Recall and Concepts

A1. (a) Define bond enthalpy, including the phase it is defined for. (b) State the sign of the energy change when a bond breaks, and when the same bond forms. (c) The H−H bond enthalpy is 436 kJ mol⁻¹. State, in words, the two energy changes this one number describes. (d) Explain why the unit is quoted per mole of bonds rather than per mole of molecules.

A2. Ammonia synthesis: N₂(g) + 3H₂(g) → 2NH₃(g). (a) List every bond broken, with its count. (b) List every bond formed, with its count. (c) Explain why the nitrogen bond is counted once, at 945 kJ mol⁻¹, and not as three separate bonds. (d) Each NH₃ molecule carries a lone pair on nitrogen. State whether it enters either list, and why.

A3. (a) Arrange C−C, C=C and C≡C in order of increasing bond enthalpy, giving each value. (b) Explain the order in terms of bond order. (c) Arrange H−F, H−Cl, H−Br and H−I in order of decreasing bond enthalpy. (d) Explain that order in terms of atomic size and bond length. (e) A triple bond is the strongest of the three carbon bonds. State whether its enthalpy is three times the single bond’s, and support the answer with the numbers.

A4. (a) Write the formula that estimates ΔH from bond enthalpies. (b) Explain why the formed sum is subtracted rather than added. (c) State the sign of ΔH, and whether the reaction is exothermic or endothermic, when the formed total is bigger than the broken total. (d) State ΔH for a reaction in which the two totals happen to be exactly equal, and what that equality means physically.

A5. (a) Explain why bond enthalpies are average values, using C−H in methane and in ethane as the example. (b) State the phase assumption built into every bond enthalpy calculation. (c) A calculation for burning methane gives about −818 kJ, but the measured ΔH for methane burning to liquid water is −890 kJ. Give two reasons for the gap. (d) State one job for which a bond enthalpy estimate is good enough, and one for which it is not.

Section B: Core Practice

B1. Hydrogen bromide formation: H₂(g) + Br₂(g) → 2HBr(g). (a) List the bonds broken and formed, with counts. (b) Calculate each total. (c) Calculate ΔH and state whether the reaction is exothermic or endothermic.

B2. Steam formation: 2H₂(g) + O₂(g) → 2H₂O(g). (a) List the bonds broken and formed, with counts. (b) Calculate ΔH for the equation as written. (c) Calculate ΔH per mole of water formed. (d) A student counts only 2 O−H bonds as formed. State the ΔH the student gets and identify the error.

B3. Methanol combustion: 2CH₃OH(g) + 3O₂(g) → 2CO₂(g) + 4H₂O(g). Each methanol molecule contains 3 C−H bonds, 1 C−O bond and 1 O−H bond. (a) List the bonds broken, with counts for the equation as written. (b) List the bonds formed, with counts. (c) Calculate ΔH for the equation as written. (d) Calculate ΔH per mole of methanol burned.

B4. Ethyne hydrogenation: C₂H₂(g) + 2H₂(g) → C₂H₆(g). Ethyne contains one C≡C bond and two C−H bonds; ethane contains one C−C bond and six C−H bonds. (a) List every bond broken and every bond formed, counting the full inventories. (b) Calculate ΔH. (c) Repeat the calculation pricing only the bonds that change, and confirm the same answer. (d) State which C−H bonds cancel, and how many.

B5. Chloromethane formation: CH₄(g) + Cl₂(g) → CH₃Cl(g) + HCl(g). (a) Using cancellation, list only the bonds that change. (b) Calculate ΔH. (c) State the three C−H bonds that survive, and explain why omitting them from both lists leaves ΔH unchanged. (d) A student prices the C−Cl bond as a second H−Cl bond. State the student’s ΔH and the size of the error.

B6. Finding an unknown bond enthalpy. For H₂(g) + F₂(g) → 2HF(g), the measured ΔH is −537 kJ. (a) Write the bond enthalpy equation for this reaction, with F−F as the unknown x. (b) Solve for x. (c) Explain why the F−F value comes out so much smaller than Cl−Cl (242 kJ mol⁻¹) would lead you to expect. (Think about the small fluorine atoms and their lone pairs.) (d) State why the answer is still only an estimate of the true F−F bond enthalpy.

Section C: Exam Style

C1. Ethane combustion: 2C₂H₆(g) + 7O₂(g) → 4CO₂(g) + 6H₂O(g). Each ethane molecule contains one C−C bond and six C−H bonds. (a) List the bonds broken, with counts and the total. (b) List the bonds formed, with counts and the total. (c) Calculate ΔH for the equation as written. (d) Calculate ΔH per mole of ethane burned. (e) Calculate ΔH per mole of CO₂ formed, and explain why it differs from the answer to (d).

C2. Judge each claim, supporting each verdict with the rule it depends on. (a) “Combustion is exothermic because breaking the bonds in the fuel releases energy.” (b) “ΔH is calculated as bonds formed minus bonds broken.” (c) “Each C=O bond in CO₂ should be priced as two C−O bonds, 2 × 358 = 716 kJ.” (d) “An estimate of −818 kJ against a measured −890 kJ for methane combustion proves the arithmetic was wrong.” (e) “The lone pairs on the oxygen atoms in water count toward the formed total.” (f) “Because C≡C is a triple bond, its enthalpy is exactly three times that of C−C.”

C3. Hydrazine, N₂H₄, burns in oxygen: N₂H₄(g) + O₂(g) → N₂(g) + 2H₂O(g). The hydrazine molecule has the structure H₂N−NH₂: four N−H bonds and one N−N bond. (a) List the bonds broken, with counts and the total. (b) List the bonds formed, with counts and the total. (c) Calculate ΔH, and state the heat released per mole of hydrazine burned. (d) The reaction breaks an N−N single bond yet is strongly exothermic. Explain, using the totals, what pays for the breaking. (e) State two reasons the calculated value will not match a measured ΔH exactly.

Answer Key

Show the answer key and full working (only after you have attempted every question)

A1. (a) Bond enthalpy is the energy required to break one mole of a particular bond in gaseous molecules, separating the atoms completely; it is defined for the gas phase. (b) Breaking absorbs energy, so its enthalpy change is positive. Forming releases the same amount, so its enthalpy change is negative. (c) Breaking one mole of H−H bonds in gaseous H₂ into separate atoms requires 436 kJ, and forming one mole of H₂ from separated gaseous hydrogen atoms releases 436 kJ. (d) The table prices bonds, not molecules: a mole of CH₄ contains four moles of C−H bonds, and each mole of bonds carries the same price, so counting is done in moles of bonds.

A2. (a) Broken: 1 N≡N and 3 H−H. (b) Formed: 6 N−H, because each of the 2 NH₃ molecules holds 3 N−H bonds. (c) A triple bond is one bond of order three, a single entry in the structure, and the table prices it as one unit at 945 kJ mol⁻¹; it is not three separate single bonds, and three N−N values (3 × 163 = 489) would badly underprice it. (d) It enters neither list. A lone pair is a non-bonding pair of electrons, not a bond, and bond enthalpy tables price only bonds.

A3. (a) C−C 347, then C=C 614, then C≡C 839 kJ mol⁻¹. (b) Higher bond order means more shared electron pairs between the same two atoms, a stronger attraction, and more energy needed to separate them, so the enthalpy rises from single to double to triple. (c) H−F 567, then H−Cl 431, then H−Br 366, then H−I 299 kJ mol⁻¹. (d) Down the group the halogen atom grows, so the shared pair sits further from both nuclei, the bond is longer, and a longer bond is weaker and cheaper to break. (e) No. Three single bonds would cost 3 × 347 = 1,041 kJ mol⁻¹, well above the triple bond’s 839 kJ mol⁻¹. Each added shared pair strengthens the bond by less than the one before.

A4. (a) ΔH ≈ Σ(bond enthalpies of bonds broken) − Σ(bond enthalpies of bonds formed). (b) Forming releases energy, which is a negative contribution to the system’s enthalpy change. Subtracting the formed total is how that release enters the sum: cost of breaking, minus energy paid back by forming. (c) ΔH is negative and the reaction is exothermic: the products hold less enthalpy than the reactants, and the difference leaves as heat. (d) ΔH = 0. The energy absorbed in breaking is exactly repaid by the energy released in forming, so there is no net heat exchange with the surroundings.

A5. (a) The environment of a bond changes slightly from molecule to molecule: a C−H bond in methane sits beside three other C−H bonds, while in ethane it sits beside a C−C bond, and the two differ slightly in strength. Tables publish one average gathered over many compounds. (b) All reactants and products are gases. (c) First, the table values are averages, so the methane C−H and water O−H prices are not exact for these molecules. Second, the method prices gaseous water, while the measured value forms liquid water; the condensation of steam releases extra energy that bond enthalpies never include. (d) Good enough: predicting whether a proposed reaction is exothermic and roughly by how much, or comparing two fuels. Not good enough: any purpose needing a precise ΔH, such as engineering design.

B1. (a) Broken: 1 H−H and 1 Br−Br. Formed: 2 H−Br. (b) Broken total: 436 + 193 = 629 kJ. Formed total: 2 × 366 = 732 kJ. (c) ΔH = 629 − 732 = −103 kJ. Negative, so the reaction is exothermic.

B2. (a) Broken: 2 H−H and 1 O=O. Formed: 4 O−H, because each of the 2 water molecules holds 2 O−H bonds. (b) Broken total: 2 × 436 + 498 = 1,370 kJ. Formed total: 4 × 467 = 1,868 kJ. ΔH = 1,370 − 1,868 = −498 kJ. (c) The equation forms 2 mol of water, so per mole: −498 ÷ 2 = −249 kJ mol⁻¹. (d) The student forms only 2 × 467 = 934 kJ, giving ΔH = 1,370 − 934 = +436 kJ. The error is forgetting the coefficient: the 2 in front of H₂O doubles the O−H count to 4, and missing it even flips the sign of the answer.

B3. (a) Broken: 6 C−H, 2 C−O, 2 O−H and 3 O=O, doubling the methanol inventory for the coefficient 2. (b) Formed: 4 C=O (two in each CO₂) and 8 O−H (two in each of the 4 water molecules). (c) Broken total: 6 × 413 + 2 × 358 + 2 × 467 + 3 × 498 = 2,478 + 716 + 934 + 1,494 = 5,622 kJ. Formed total: 4 × 799 + 8 × 467 = 3,196 + 3,736 = 6,932 kJ. ΔH = 5,622 − 6,932 = −1,310 kJ. (d) The equation burns 2 mol of methanol, so per mole: −1,310 ÷ 2 = −655 kJ mol⁻¹.

B4. (a) Broken: 1 C≡C, 2 C−H and 2 H−H. Formed: 1 C−C and 6 C−H. (b) Broken total: 839 + 2 × 413 + 2 × 436 = 839 + 826 + 872 = 2,537 kJ. Formed total: 347 + 6 × 413 = 347 + 2,478 = 2,825 kJ. ΔH = 2,537 − 2,825 = −288 kJ. (c) Pricing only the changes: broken 1 C≡C + 2 H−H = 839 + 872 = 1,711 kJ; formed 1 C−C + 4 new C−H = 347 + 1,652 = 1,999 kJ; ΔH = 1,711 − 1,999 = −288 kJ, the same answer. (d) Two C−H bonds cancel: ethyne’s two C−H bonds are still present in ethane, so they are broken on paper and re-formed at the same price, contributing nothing to the difference.

B5. (a) Only one C−H bond and the Cl−Cl bond break; one C−Cl bond and one H−Cl bond form. The other three C−H bonds are untouched. (b) Broken: 413 + 242 = 655 kJ. Formed: 328 + 431 = 759 kJ. ΔH = 655 − 759 = −104 kJ. (c) The three surviving C−H bonds would add 3 × 413 = 1,239 kJ to both totals if they were listed, and equal additions to both sides of a subtraction cancel, so ΔH is unchanged. (d) The student forms 2 × 431 = 862 kJ in place of 759 kJ, giving ΔH = 655 − 862 = −207 kJ. The error is 103 kJ: the reaction looks far more exothermic than it is because C−Cl (328) was priced at the stronger H−Cl value (431).

B6. (a) Broken: 1 H−H and 1 F−F, total 436 + x. Formed: 2 H−F, total 2 × 567 = 1,134 kJ. ΔH = (436 + x) − 1,134 = −537 kJ. (b) 436 + x = 1,134 − 537 = 597, so x = 597 − 436 = 161 kJ mol⁻¹. (c) Fluorine atoms are so small that the lone pairs on the two bonded atoms sit very close and repel each other, weakening the F−F bond well below the trend set by the other halogens. (d) The measured ΔH is exact for the reaction, but the calculation leans on the H−F table value, which is an average; any averaging error in H−F is transferred into x, so x is an estimate of the F−F bond enthalpy, not a direct measurement.

C1. (a) Broken: 2 C−C, 12 C−H and 7 O=O. Total: 2 × 347 + 12 × 413 + 7 × 498 = 694 + 4,956 + 3,486 = 9,136 kJ. (b) Formed: 8 C=O (two in each of the 4 CO₂ molecules) and 12 O−H (two in each of the 6 water molecules). Total: 8 × 799 + 12 × 467 = 6,392 + 5,604 = 11,996 kJ. (c) ΔH = 9,136 − 11,996 = −2,860 kJ. (d) The equation burns 2 mol of ethane, so per mole: −2,860 ÷ 2 = −1,430 kJ mol⁻¹. (e) Per mole of CO₂: −2,860 ÷ 4 = −715 kJ. It differs because the two prices divide the same total by different amounts: the equation names 2 mol of ethane but 4 mol of carbon dioxide, so the per-mole price for CO₂ is half that for ethane.

C2. (a) False. Breaking bonds always absorbs energy. Combustion is exothermic because the bonds formed in CO₂ and H₂O release more energy than the breaking absorbs; the release comes from forming, never from breaking. (b) False. The formula is broken minus formed. Forming is the payback and is subtracted from the cost of breaking; the claim reverses the sums and flips every sign. (c) False. A double bond is one bond priced at its own enthalpy, 799 kJ mol⁻¹. Two C−O prices give 716 kJ mol⁻¹, which is both the wrong structure and the wrong value: a double bond is not two single bonds. (d) False. A gap is expected: the table values are averages, and the estimate prices steam while the measured value forms liquid water, whose condensation releases extra energy. Neither gap implies an arithmetic slip. (e) False. Lone pairs are non-bonding electron pairs, not bonds, and bond enthalpy tables price bonds only. They enter neither the broken nor the formed total. (f) False. C≡C is 839 kJ mol⁻¹ and 3 × 347 = 1,041 kJ mol⁻¹. Bond enthalpy rises with bond order, but each extra shared pair adds less than the previous one.

C3. (a) Broken: 4 N−H and 1 N−N, plus 1 O=O. Total: 4 × 391 + 163 + 498 = 1,564 + 163 + 498 = 2,225 kJ. (b) Formed: 1 N≡N and 4 O−H (two in each water molecule). Total: 945 + 4 × 467 = 945 + 1,868 = 2,813 kJ. (c) ΔH = 2,225 − 2,813 = −588 kJ for the equation as written, which burns 1 mol of hydrazine, so the heat released per mole of hydrazine is 588 kJ. (d) The formed total is what pays. Forming the very strong N≡N triple bond (945 kJ) and four O−H bonds releases 2,813 kJ, comfortably more than the 2,225 kJ spent breaking every bond in hydrazine and oxygen; the weak N−N bond is a small part of the cost. (e) The table values are averages, so the N−H price gathered from other molecules is not exact for hydrazine; and the method assumes gases throughout, so any real measurement involving liquid hydrazine or liquid water includes intermolecular effects the table never prices.

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]]> https://examfitter.com/ap-chemistry/worksheet/bond-enthalpy-problems/feed/ 0 Topic 6.4: Heat Capacity and Calorimetry Worksheet https://examfitter.com/ap-chemistry/worksheet/calorimetry-problems/ https://examfitter.com/ap-chemistry/worksheet/calorimetry-problems/#respond Tue, 10 Nov 2026 20:02:59 +0000 https://examfitter.com/?p=2918 How to Use This Worksheet Work in order. Section A checks the vocabulary and the two instruments: heat capacity in its three sizes, the calorimeter constant, the energy ledger and its signs, and the assumptions that limit accuracy. Section B practises the calculations: a bomb run, a dissolving salt, a hot metal in a calorimeter, ... Read more

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Work in order. Section A checks the vocabulary and the two instruments: heat capacity in its three sizes, the calorimeter constant, the energy ledger and its signs, and the assumptions that limit accuracy. Section B practises the calculations: a bomb run, a dissolving salt, a hot metal in a calorimeter, a calibration, and an endothermic dissolving with the hardware counted. Section C is exam style: a full experiment with error analysis, five claims to judge, and a combustion in a bomb calorimeter.

Reference facts. Specific heat capacity of water and of dilute aqueous solutions: 4.18 J g⁻¹ °C⁻¹. Take the density of a dilute solution as 1.00 g mL⁻¹, so a volume in millilitres gives the same number as a mass in grams. q = mcΔT for a mass of solution warming or cooling; q = CΔT for a whole object or a calorimeter; q_reaction = −(q_water + q_calorimeter). Signs are taken from the system’s point of view: a reaction that warms its surroundings is exothermic, with q_reaction and ΔH negative.

Three habits protect every answer. Work out ΔT first, as final minus initial, and keep the sign it gives you. Count every place the heat went: the water and the hardware both. Finish by dividing by the moles of the limiting reactant, and check whether the instrument you used reports ΔH or ΔE.

Do not look at the answer key until you have attempted every question.

Section A: Recall and Concepts

A1. (a) State what a coffee-cup calorimeter holds constant, and the quantity its heat measurement gives directly. (b) State what a bomb calorimeter holds constant, and the quantity it gives. (c) Explain why the coffee cup is open to the air. (d) State two reasons a bomb calorimeter, rather than a coffee cup, is used for combustions.

A2. (a) Define the heat capacity of an object, and state its units and its equation. (b) Define specific heat capacity, and state its units. (c) Define molar heat capacity, and state its units. (d) Calculate the heat capacity of 250 g of water. (e) State how the heat capacity of a pure sample is obtained from its specific heat capacity.

A3. (a) State what the calorimeter constant, C_cal, accounts for. (b) State its units, and write the equation for the heat the hardware absorbs. (c) Describe how C_cal is determined experimentally. (d) State which temperature change is used in the hardware equation, and why it is that one.

A4. (a) Write q_reaction in terms of q_water and q_calorimeter, and state the assumption the equation rests on. (b) A reaction in a coffee-cup calorimeter makes the solution temperature rise. State the signs of q_water, q_reaction and ΔH. (c) Repeat part (b) for a reaction that makes the temperature fall. (d) A hot metal is placed in a calorimeter of water. Write the energy balance connecting the metal, the water and the hardware.

A5. (a) State the main source of error in coffee-cup calorimetry. (b) State the direction in which that error pushes the calculated ΔH. (c) A student leaves the calorimeter constant out of the calculation. State what happens to the magnitude of the calculated ΔH, and what happens to its sign. (d) State what would happen to temperature changes, and to the precision of the measurements, if the specific heat capacity of water were much lower. (e) State two further assumptions made when a coffee-cup result is calculated.

Section B: Core Practice

B1. A sample burns in a bomb calorimeter with C_cal = 7.50 kJ °C⁻¹, raising the temperature by 2.80 °C. (a) Calculate the heat absorbed by the calorimeter. (b) State the heat released by the combustion, with its sign from the reaction’s point of view. (c) The sample mass was 0.600 g and its molar mass is 60.0 g mol⁻¹. Calculate the moles burned. (d) Calculate ΔE for the combustion, in kJ mol⁻¹. (e) State why the answer is ΔE and not ΔH.

B2. 5.55 g of CaCl₂ (molar mass 111 g mol⁻¹) dissolves in water in a coffee-cup calorimeter: CaCl₂(s) → Ca²⁺(aq) + 2Cl⁻(aq). The total solution mass is 100 g, and the temperature rises from 24.0 °C to 33.7 °C. (a) Calculate ΔT. (b) Calculate q for the solution, showing the substitution. (c) State q_reaction, and whether the dissolving is exothermic or endothermic. (d) Calculate the moles of CaCl₂ dissolved. (e) Calculate the enthalpy of solution, ΔH, in kJ mol⁻¹.

B3. A 50.0 g block of metal at 100.0 °C is placed in a calorimeter holding 100 g of water at 25.0 °C. The calorimeter constant is 30.0 J °C⁻¹, and everything settles at 28.0 °C. (a) Calculate the heat gained by the water. (b) Calculate the heat gained by the calorimeter hardware. (c) State the heat lost by the metal. (d) Calculate the specific heat capacity of the metal. (e) State the assumption that lets part (c) follow from parts (a) and (b).

B4. A student calibrates a calorimeter by pouring 60.0 g of water at 55.0 °C into 60.0 g of water at 25.0 °C already inside it. The mixture settles at 38.5 °C. (a) Calculate the heat lost by the hot water. (b) Calculate the heat gained by the cold water. (c) Calculate the heat absorbed by the calorimeter hardware. (d) Calculate C_cal. (e) State the temperature change you used for the hardware in part (d), and why it is that value.

B5. 10.7 g of NH₄Cl (molar mass 53.5 g mol⁻¹) dissolves in water in a coffee-cup calorimeter with C_cal = 50 J °C⁻¹: NH₄Cl(s) → NH₄⁺(aq) + Cl⁻(aq). The total solution mass is 200 g, and the temperature falls from 22.0 °C to 18.2 °C. (a) Calculate ΔT, with its sign. (b) Calculate q for the solution. (c) Calculate q for the calorimeter hardware. (d) Calculate q_reaction, and state whether the dissolving is exothermic or endothermic. (e) Calculate ΔH of solution, in kJ mol⁻¹.

B6. (a) Calculate the heat capacity of 400 g of water. (b) A calorimeter assembly with C_cal = 85 J °C⁻¹ warms by 4.0 °C during a run. Calculate the heat it absorbed. (c) Using the molar mass of water, 18.0 g mol⁻¹, calculate the molar heat capacity of water. (d) A student doubles the mass of water in a coffee-cup calorimeter and repeats exactly the same reaction. State what happens to ΔT and to the calculated ΔH, and explain both answers.

Section C: Exam Style

C1. A student reacts 0.243 g of magnesium ribbon (molar mass 24.3 g mol⁻¹) with excess hydrochloric acid in a coffee-cup calorimeter: Mg + 2HCl → MgCl₂ + H₂. The solution has mass 50.0 g, the calorimeter constant is 20 J °C⁻¹, and the temperature rises from 20.0 °C to 40.4 °C. (a) Calculate the heat absorbed by the solution. (b) Calculate the heat absorbed by the calorimeter hardware. (c) Calculate q_reaction, and hence ΔH for the reaction in kJ per mole of magnesium, stating the sign evidence as you go. (d) A second student ignores the calorimeter constant. Calculate the ΔH she obtains, and state how it compares with the part (c) value. (e) The accepted ΔH is more negative than even the part (c) value. Give the reason, naming the assumption that fails.

C2. Judge each claim, supporting each verdict with the rule it depends on. (a) “A bomb calorimeter measures ΔH, because the combustion heat ends up in the water around the bomb.” (b) “The temperature in a coffee-cup calorimeter fell during a reaction, so ΔH for the reaction is negative.” (c) “Doubling the mass of solution in the cup doubles the temperature rise produced by the same reaction heat.” (d) “Neglecting the calorimeter constant makes the calculated ΔH too large in magnitude.” (e) “Adding a catalyst changes the ΔH that a calorimeter measures for a reaction.”

C3. 0.920 g of ethanol (molar mass 46.0 g mol⁻¹) burns completely in a bomb calorimeter with C_cal = 10.5 kJ °C⁻¹: C₂H₅OH + 3O₂ → 2CO₂ + 3H₂O. The temperature rises by 2.60 °C. (a) Calculate the heat absorbed by the calorimeter. (b) Calculate the moles of ethanol burned. (c) Calculate ΔE for the combustion of ethanol, in kJ mol⁻¹. (d) State what the water bath around the bomb vessel does in this measurement. (e) State two reasons the same combustion is never run in an open coffee cup instead.

Show the answer key and full working (only after you have attempted every question)

A1. (a) Pressure: the open cup stays at atmospheric pressure, so the measured heat is q_p, which gives ΔH directly. (b) Volume: the rigid, sealed vessel cannot change volume, so the measured heat gives ΔE. (c) Open to the air, the contents stay at atmospheric pressure even as the temperature changes and gases form; that constant pressure is what makes the heat equal to ΔH. (d) Any two: the sealed steel vessel contains the hot, pressurised gases of a combustion safely; the fixed volume means no gas escapes carrying heat away; the vessel withstands the pressure a combustion generates.

A2. (a) The heat needed to raise the temperature of the whole object by one degree Celsius; units J °C⁻¹; q = CΔT. (b) The heat needed to raise one gram of a substance by one degree Celsius; units J g⁻¹ °C⁻¹. (c) The heat needed to raise one mole of a substance by one degree Celsius; units J mol⁻¹ °C⁻¹. (d) C = mc = 250 × 4.18 = 1,045 J °C⁻¹. (e) Multiply the specific heat capacity by the mass of the sample: C = mc.

A3. (a) The heat absorbed by the calorimeter hardware itself: the cup, lid, stirrer and thermometer, rather than by the water. (b) J °C⁻¹; q_cal = C_cal × ΔT. (c) Mix known masses of hot and cold water inside the calorimeter, record the temperatures, and take the difference between the heat lost by the hot water and the heat gained by the cold water as the heat the hardware absorbed; divide by the hardware’s temperature change. (d) The same temperature change as the water: hardware and contents start at one temperature and settle at the same final temperature, so they share ΔT.

A4. (a) q_reaction = −(q_water + q_calorimeter). It rests on the calorimeter being perfectly insulated, so no heat crosses between the apparatus and the room. (b) q_water is positive, because the solution warmed; q_reaction is negative; ΔH is negative: the reaction is exothermic. (c) q_water is negative, because the solution cooled; q_reaction is positive; ΔH is positive: the reaction is endothermic. (d) Heat lost by the metal = heat gained by the water + heat gained by the calorimeter hardware.

A5. (a) Heat exchange with the surroundings: the foam cup is not a perfect insulator, so heat leaks to or from the room air. (b) It makes the measured temperature change smaller than the reaction alone would produce, so the calculated ΔH is too small in magnitude. (c) The magnitude comes out too small, because real heat absorbed by the hardware goes uncounted; the sign is unchanged. (d) The same heat would produce larger temperature changes, and larger changes are read with less relative error, so the measurements would be more precise. (e) Any two: the solution has the density and specific heat capacity of pure water; the reaction goes to completion; all of the limiting reactant reacts; the final temperature is read at its true extreme before cooling or warming towards room temperature sets in.

B1. (a) q = C_cal × ΔT = 7.50 × 2.80 = 21.0 kJ absorbed by the calorimeter. (b) The combustion released that heat, so from the reaction’s point of view q_reaction = −21.0 kJ. (c) n = 0.600 ÷ 60.0 = 0.0100 mol. (d) ΔE = q_reaction ÷ n = −21.0 ÷ 0.0100 = −2,100 kJ mol⁻¹. (e) The bomb’s rigid walls hold the volume constant, and heat measured at constant volume is ΔE; ΔH is the heat at constant pressure.

B2. (a) ΔT = 33.7 − 24.0 = +9.7 °C. (b) q = mcΔT = 100 × 4.18 × 9.7 = +4,054.6 J, about +4.05 kJ. (c) q_reaction = −4,054.6 J. The solution gained heat, so the dissolving released it: exothermic. (d) n = 5.55 ÷ 111 = 0.0500 mol. (e) ΔH = −4,054.6 ÷ 0.0500 = −81,092 J mol⁻¹ = −81.1 kJ mol⁻¹.

B3. (a) The water warmed by 28.0 − 25.0 = 3.0 °C: q_water = 100 × 4.18 × 3.0 = +1,254 J. (b) The hardware shared that same 3.0 °C rise: q_cal = 30.0 × 3.0 = +90 J. (c) Heat lost by the metal = 1,254 + 90 = 1,344 J. (d) The metal cooled by 100.0 − 28.0 = 72.0 °C, so c = 1,344 ÷ (50.0 × 72.0) = 1,344 ÷ 3,600 = 0.373 J g⁻¹ °C⁻¹. (e) That the calorimeter is insulated from the room: every joule the metal lost stayed inside, split between the water and the hardware.

B4. (a) The hot water cooled by 55.0 − 38.5 = 16.5 °C: heat lost = 60.0 × 4.18 × 16.5 = 4,138.2 J. (b) The cold water warmed by 38.5 − 25.0 = 13.5 °C: heat gained = 60.0 × 4.18 × 13.5 = 3,385.8 J. (c) Heat absorbed by the hardware = 4,138.2 − 3,385.8 = 752.4 J. (d) C_cal = 752.4 ÷ 13.5 = 55.7 J °C⁻¹. (e) 13.5 °C: the hardware held the cold water, so it started at 25.0 °C and finished at the mixture’s 38.5 °C, the same rise as the cold water’s.

B5. (a) ΔT = 18.2 − 22.0 = −3.8 °C. (b) q_solution = mcΔT = 200 × 4.18 × (−3.8) = −3,176.8 J. (c) q_cal = C_cal × ΔT = 50 × (−3.8) = −190 J. (d) q_reaction = −(−3,176.8 + (−190)) = +3,366.8 J. Positive: the dissolving absorbed heat from the solution and the hardware, so it is endothermic. (e) n = 10.7 ÷ 53.5 = 0.200 mol, so ΔH = 3,366.8 ÷ 0.200 = 16,834 J mol⁻¹ = +16.8 kJ mol⁻¹.

B6. (a) C = mc = 400 × 4.18 = 1,672 J °C⁻¹. (b) q = CΔT = 85 × 4.0 = 340 J. (c) Molar heat capacity = 4.18 × 18.0 = 75.24, about 75.2 J mol⁻¹ °C⁻¹. (d) ΔT halves: the reaction releases the same heat, but q = mcΔT now spreads it over twice the mc product. The calculated ΔH does not change: q_reaction is the same heat and it is still divided by the same moles of limiting reactant; the temperature rise is only the measuring device, not the quantity reported.

C1. (a) ΔT = 40.4 − 20.0 = +20.4 °C, so q_solution = 50.0 × 4.18 × 20.4 = +4,263.6 J. (b) q_calorimeter = 20 × 20.4 = +408 J. (c) q_reaction = −(4,263.6 + 408) = −4,671.6 J. The temperature rose, so the surroundings gained heat and the reaction released it: negative q_reaction, exothermic. n(Mg) = 0.243 ÷ 24.3 = 0.0100 mol, the limiting reactant since the acid is in excess. ΔH = −4,671.6 ÷ 0.0100 = −467,160 J mol⁻¹, about −467 kJ mol⁻¹. (d) Ignoring the hardware: q_reaction = −4,263.6 J, so ΔH = −4,263.6 ÷ 0.0100 = −426 kJ mol⁻¹. It is too small in magnitude, less negative by about 41 kJ mol⁻¹, because 408 J of real reaction heat went uncounted. (e) Heat exchange with the surroundings: during the run the warm solution leaks heat to the room through and above the foam cup, so the measured rise is smaller than the reaction’s heat warrants. The assumption of perfect insulation fails.

C2. (a) False. The bomb’s rigid walls fix the volume, so its heat is ΔE whatever the heat ends up warming; ending up in water is only how the change is read, and ΔH belongs to constant pressure. (b) False. A falling temperature means the solution lost heat to the reaction: the reaction absorbed heat, so it is endothermic and ΔH is positive. (c) False, and backwards. Doubling the mass doubles mc in q = mcΔT, so the same heat is spread over twice the heat capacity and ΔT halves. (d) False. The hardware absorbs part of the reaction’s heat; leaving it out undercounts the total heat, so the calculated ΔH is too small in magnitude. (e) False. A catalyst changes the pathway and so the rate, but ΔH depends only on the reactants and the products, which a catalyst leaves unchanged; the calorimeter measures the same ΔH, just reached sooner.

C3. (a) q = C_cal × ΔT = 10.5 × 2.60 = 27.3 kJ absorbed by the calorimeter. (b) n = 0.920 ÷ 46.0 = 0.0200 mol. (c) ΔE = −27.3 ÷ 0.0200 = −1,365 kJ mol⁻¹. The sign is negative because the combustion released the heat that the calorimeter absorbed, and the per-mole quantity is ΔE because the bomb holds the volume constant. (d) It absorbs the heat the combustion releases and shows it as a temperature rise; together with the vessel it forms the surroundings whose total heat capacity is the C_cal used in part (a). (e) Any two: the open cup cannot contain the hot pressurised gases a combustion produces, so they would escape carrying heat away and the ledger could never close; the foam cup could not withstand a combustion safely; combustion in open air may be incomplete, so the heat per mole would not match complete burning in oxygen.

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]]> https://examfitter.com/ap-chemistry/worksheet/calorimetry-problems/feed/ 0 Topic 6.1: Endothermic and Exothermic Processes Worksheet https://examfitter.com/ap-chemistry/worksheet/endothermic-exothermic/ https://examfitter.com/ap-chemistry/worksheet/endothermic-exothermic/#respond Tue, 10 Nov 2026 20:01:47 +0000 https://examfitter.com/?p=2915 How to Use This Worksheet Work in order. Section A checks the definitions and the system and surroundings labels. Section B practises the skills: reading enthalpy diagrams, running an energy ledger between system and surroundings, and classifying phase changes with numbers. Section C is exam style: judging claims and two full scenarios. Reference facts. ΔH ... Read more

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]]> How to Use This Worksheet

Work in order. Section A checks the definitions and the system and surroundings labels. Section B practises the skills: reading enthalpy diagrams, running an energy ledger between system and surroundings, and classifying phase changes with numbers. Section C is exam style: judging claims and two full scenarios.

Reference facts. ΔH is always reported from the system’s point of view: positive for endothermic (system absorbs heat), negative for exothermic (system releases heat). Whatever the system gains, the surroundings lose, and the reverse: the two changes are equal in size and opposite in sign. Separating particles against their attractions (melting, evaporating, subliming) is endothermic; forming attractions (freezing, condensing, depositing) is exothermic. Useful values: melting 1 mol of ice absorbs 6.01 kJ; evaporating 1 mol of water absorbs 40.7 kJ; burning methane has ΔH = −890 kJ/mol; neutralisation of a strong acid with a strong base has ΔH ≈ −57 kJ/mol.

Three habits protect every answer. Name the system before you assign any sign. Check every energy ledger balances: gain on one side, equal loss on the other. And read what a hand or thermometer feels as a report about the surroundings, not the system.

Do not look at the answer key until you have attempted every question.

Section A: Recall and Concepts

A1. (a) Define the system in a chemistry experiment. (b) Define the surroundings. (c) In a coffee-cup calorimetry experiment, a student mixes two solutions and records the temperature change of the mixture. State what the system is, and state two things that belong to the surroundings. (d) State why the choice of boundary matters when a sign is assigned to ΔH.

A2. (a) State what heat is, and the condition that makes heat flow. (b) State the direction of heat flow in an endothermic process. (c) State the direction of heat flow in an exothermic process. (d) State what happens to the temperature of the surroundings in each case.

A3. (a) State the sign of ΔH for an endothermic process, from the system’s point of view, and explain the sign. (b) State the sign of ΔH for an exothermic process, from the system’s point of view, and explain the sign. (c) A process has ΔH = +45 kJ/mol. State whether it is endothermic or exothermic, and state how much heat is transferred, and in which direction, when 1 mol reacts. (d) A reaction gives off 120 kJ of heat per mole. Write its ΔH value with its sign and units.

A4. Classify each process as endothermic or exothermic, giving a one-line reason for each. (a) Ice melting in a drink. (b) Water freezing in a freezer tray. (c) Alcohol evaporating from skin. (d) Methane burning in oxygen: CH₄ + 2O₂ → CO₂ + 2H₂O. (e) Steam condensing on a bathroom mirror. (f) Dry ice subliming at room temperature.

A5. (a) State what temperature measures at the particle level. (b) State whether heat and temperature are the same quantity, and justify your answer in one sentence. (c) The same amount of heat is released into 50 g of water and into 500 g of water. State which water shows the larger temperature rise, and explain why in terms of particles. (d) A thermometer in a calorimeter shows a temperature rise. State which side of the boundary the thermometer is reporting, and state what that rise tells you about the reaction.

Section B: Core Practice

B1. An enthalpy diagram shows reactants at 95 kJ/mol and products at 140 kJ/mol. (a) Calculate ΔH, showing the subtraction you use. (b) Classify the reaction as endothermic or exothermic, using the sign and the relative levels. (c) State how much heat is transferred when 1 mol reacts, and in which direction. (d) Calculate ΔH for the reverse reaction and classify it. (e) State what the surroundings gain or lose when 1 mol of the forward reaction occurs.

B2. Methane burns with ΔH = −890 kJ/mol, from the system’s point of view. (a) State how much heat is released to the surroundings when 1 mol of methane burns. (b) Calculate the heat released when 0.50 mol of methane burns. (c) Write the energy ledger for that 0.50 mol burn: the change for the system and the change for the surroundings. (d) State what happens to the average kinetic energy of the particles in the surroundings, and what a thermometer placed there shows. (e) Explain, at the level of bonds, why this reaction is exothermic.

B3. A student dissolves 5.35 g of ammonium chloride (0.100 mol) in water and the temperature of the solution falls. The dissolving absorbs 1.5 kJ of heat. (a) Calculate ΔH for the dissolving, per mole, with its sign. (b) State the system and the surroundings in this experiment. (c) Explain why the beaker feels cold, in terms of energy crossing the boundary. (d) State what happens to the average kinetic energy of the water particles. (e) A classmate says the cold solution “contains more cold than before”. Correct the statement.

B4. Use the reference values: melting ice, ΔH = +6.01 kJ/mol; evaporating water, ΔH = +40.7 kJ/mol, both from the system’s point of view. (a) Calculate the heat absorbed when 0.25 mol of water evaporates. (b) Write ΔH for condensing steam, per mole, with its sign, and calculate the heat transferred when 2.0 mol of steam condenses, stating the direction. (c) Write ΔH for freezing water, per mole, with its sign, and calculate the heat transferred when 5.0 mol of water freezes, stating the direction. (d) Explain why evaporation cools the skin while condensation warms a mirror, using the same particle-level idea for both.

B5. A student mixes HCl solution with NaOH solution in a foam cup. The temperature of the mixture rises by several degrees: HCl + NaOH → NaCl + H₂O. (a) State the system and the surroundings. (b) State the direction of heat flow across the boundary. (c) Classify the neutralisation as endothermic or exothermic. (d) State the sign of ΔH, from the system’s point of view, and justify it from the temperature reading. (e) If 0.20 mol of water forms and ΔH = −57 kJ/mol, calculate the heat gained by the surroundings.

B6. Photosynthesis and respiration are reverse processes: photosynthesis, 6CO₂ + 6H₂O → C₆H₁₂O₆ + 6O₂; respiration, C₆H₁₂O₆ + 6O₂ → 6CO₂ + 6H₂O. (a) Photosynthesis is endothermic. State the sign of its ΔH and state where the absorbed energy ends up. (b) State the classification and the sign of ΔH for respiration. (c) The combustion of glucose has ΔH = −2,808 kJ/mol. State the value of ΔH for photosynthesis, per mole of glucose formed. (d) Explain why the two ΔH values must be equal in size and opposite in sign. (e) A leaf in sunlight stays cool while photosynthesis runs. Suggest why, in terms of energy flow.

Section C: Exam Style

C1. Judge each claim, supporting each verdict with the definition or the sign rule it depends on. (a) “The beaker felt cold during the reaction, so the reaction released heat.” (b) “ΔH = +26 kJ/mol means the surroundings gained 26 kJ of enthalpy per mole.” (c) “Energy is conserved in every reaction, so ΔH is always zero.” (d) “Condensation is endothermic, because the gas particles slow down as they become a liquid.” (e) “Doubling the amount of reactant doubles the heat transferred, but ΔH in kJ/mol stays the same.”

C2. A cold pack contains ammonium nitrate and water in separate compartments. Squeezing the pack mixes them and the pack turns cold: NH₄NO₃(s) → NH₄⁺(aq) + NO₃⁻(aq). In a test, dissolving 0.100 mol absorbs 2.6 kJ of heat. (a) Calculate ΔH for the dissolving, per mole, with its sign, from the system’s point of view. (b) Identify the system and the surroundings when the pack is held against an injured ankle. (c) Explain, in terms of heat flow and particle kinetic energy, why the ankle feels cold. (d) State the energy change of the surroundings for the 0.100 mol test, with its sign, and show that the total energy change of system plus surroundings is zero. (e) The pack is left until it returns to room temperature. State whether the dissolving process was endothermic or exothermic, and state what supplied the energy that the system gained.

C3. An athlete’s muscles respire glucose: C₆H₁₂O₆ + 6O₂ → 6CO₂ + 6H₂O, ΔH = −2,808 kJ/mol from the system’s point of view. (a) Calculate the heat released to the surroundings when 0.10 mol of glucose is respired. (b) State where that heat goes in the body, and what happens to the average kinetic energy of the particles it reaches. (c) The athlete sweats, and the sweat evaporates from the skin. Classify evaporation as endothermic or exothermic, state the direction of heat flow, and explain why it cools the athlete. (d) A spectator says: “Respiration releases heat, so the glucose must end up with more enthalpy than it started with.” Correct the statement, using the definition of ΔH. (e) State the classification and ΔH sign for the reverse of respiration, and name the process that carries it out.

Show the answer key and full working (only after you have attempted every question)

A1. (a) The part under study: the reactants and products of the process being investigated. (b) Everything outside the system that can exchange energy with it: solvent, container, thermometer, air, and anything touching the apparatus. (c) The system is the reacting chemicals (the two solutions as they react). The surroundings include the water the reaction sits in, the cup and lid, the thermometer and the stirrer (any two). (d) Because ΔH is defined from the system’s side of the boundary: the same heat flow is a loss for one side and a gain for the other, so the sign flips if the labels are swapped.

A2. (a) Heat is energy transferred across the boundary because of a temperature difference between system and surroundings. (b) From the surroundings into the system. (c) From the system into the surroundings. (d) Endothermic: the surroundings lose energy and their temperature falls. Exothermic: the surroundings gain energy and their temperature rises.

A3. (a) Positive. The system absorbs heat and gains enthalpy, so H(products) is greater than H(reactants) and the difference is positive. (b) Negative. The system releases heat and loses enthalpy, so H(products) is less than H(reactants) and the difference is negative. (c) Endothermic: 45 kJ of heat is absorbed by the system from the surroundings for each mole that reacts. (d) ΔH = −120 kJ/mol: the heat is given off, so the system loses enthalpy and the sign is negative.

A4. (a) Endothermic: melting separates particles against their attractions, which costs energy drawn from the drink. (b) Exothermic: freezing forms attractions as the liquid becomes solid, releasing heat to the surroundings. (c) Endothermic: evaporation separates the liquid’s particles, drawing heat from the skin. (d) Exothermic: combustion forms strong new bonds in CO₂ and H₂O, releasing more energy than breaking the reactant bonds cost. (e) Exothermic: condensing brings vapour particles together and forms attractions, releasing heat to the mirror. (f) Endothermic: subliming separates the solid’s particles completely, absorbing heat from the room.

A5. (a) Temperature measures the average kinetic energy of the particles. (b) No: heat is an amount of energy transferred, while temperature is an average per particle, so the same heat can produce different temperatures in different amounts of material. (c) The 50 g of water: the same energy is shared among fewer particles, so the average kinetic energy per particle, and therefore the temperature, rises more. (d) The surroundings: the thermometer sits outside the reacting system, in the water and cup. A rise there means heat arrived in the surroundings, so the system released it and the reaction is exothermic.

B1. (a) ΔH = H(products) − H(reactants) = 140 − 95 = +45 kJ/mol. (b) Endothermic: ΔH is positive and the products sit higher in enthalpy than the reactants. (c) 45 kJ of heat is absorbed by the system from the surroundings per mole. (d) ΔH(reverse) = −45 kJ/mol, exothermic: the reverse reaction runs downhill from 140 to 95 kJ/mol and releases the 45 kJ. (e) The surroundings lose 45 kJ per mole: they supply exactly what the system gains, so their temperature falls.

B2. (a) 890 kJ per mole burned. (b) 0.50 × 890 = 445 kJ released. (c) System: −445 kJ. Surroundings: +445 kJ. The two cancel, as conservation requires. (d) It increases: the transferred energy becomes kinetic energy of the surroundings’ particles, and the thermometer shows a temperature rise. (e) The bonds formed in the CO₂ and H₂O products are strong, and forming them releases more energy than was needed to break the bonds in CH₄ and O₂. The surplus leaves the system as heat.

B3. (a) ΔH = 1.5 ÷ 0.100 = +15 kJ/mol. Positive, because the dissolving system absorbs the heat. (b) The system is the dissolving ammonium chloride (the solute as it dissolves). The surroundings are the water, the beaker and the air and hand touching it. (c) Heat flows from the surroundings into the system to supply the dissolving. The beaker is part of the surroundings, so it loses energy and its temperature drops; your hand detects that loss as cold. (d) It decreases: the water particles gave up energy to the system, so their average kinetic energy, and the temperature, fall. (e) There is no such substance as cold to contain. The solution feels cold because it holds less energy than before: heat left the solution and entered the dissolving system.

B4. (a) 0.25 × 40.7 = 10.2 kJ absorbed by the system from the surroundings. (b) Condensing is the reverse of evaporating, so ΔH = −40.7 kJ/mol. For 2.0 mol: 2.0 × 40.7 = 81.4 kJ released by the system to the surroundings. (c) Freezing is the reverse of melting, so ΔH = −6.01 kJ/mol. For 5.0 mol: 5.0 × 6.01 = 30.05 kJ released by the system to the surroundings. (d) Evaporation must pull particles apart against their attractions, and the energy for that separation is taken from the skin, whose particles lose kinetic energy and cool. Condensation brings particles together and forms those attractions, releasing energy that arrives in the mirror and warms it.

B5. (a) The system is the reacting HCl and NaOH (the neutralisation reaction). The surroundings are the solution the reaction sits in, the foam cup, the thermometer and the air. (b) From the system into the surroundings. (c) Exothermic. (d) Negative. The thermometer sits in the surroundings and shows their temperature rising, which means heat arrived there from the system; a system that loses enthalpy has ΔH negative from its own point of view. (e) Heat released by the system = 0.20 × 57 = 11.4 kJ, and all of it (in the ideal calorimetry picture) is gained by the surroundings, so the surroundings gain 11.4 kJ.

B6. (a) Positive: an endothermic process has ΔH greater than zero from the system’s point of view. The absorbed energy is stored as chemical potential energy in the glucose and oxygen formed. (b) Respiration is exothermic, with ΔH negative. (c) +2,808 kJ/mol: photosynthesis is the exact reverse of the combustion, so its ΔH has the same size and the opposite sign. (d) Reversing a process swaps the roles of reactants and products in ΔH = H(products) − H(reactants), which flips the sign while keeping the size, because the same two enthalpy levels are involved. (e) The sunlight energy the leaf absorbs is carried into stored chemical energy by photosynthesis instead of raising the leaf particles’ kinetic energy.

C1. (a) False. A cold beaker is losing heat to the process inside it; heat flowing into the system is an endothermic process, not a release. (b) False. ΔH is defined from the system’s point of view: +26 kJ/mol means the system gained 26 kJ per mole. The surroundings lost 26 kJ per mole. (c) False. Conservation applies to system and surroundings combined: their changes must cancel, but the system’s own ΔH can be any positive or negative value. (d) False. The particles do slow down, but the classification follows the heat flow: condensing forms attractions and releases heat to the surroundings, so it is exothermic. Slowing particles are the result of losing energy, not evidence of absorbing it. (e) True. Heat transferred is an extensive amount and scales with the quantity reacted; ΔH in kJ/mol is defined per mole, so it is unchanged when the amount doubles.

C2. (a) ΔH = 2.6 ÷ 0.100 = +26 kJ/mol. (b) The system is the dissolving ammonium nitrate and the mixture it forms inside the pack. The surroundings are the pack material, the ankle and skin it touches, and the air. (c) The dissolving absorbs heat, so heat flows from the skin into the pack. The skin’s particles lose kinetic energy, their average kinetic energy falls, and that fall is what is felt as cold. (d) The system gained 2.6 kJ, so the surroundings changed by −2.6 kJ. Total: +2.6 + (−2.6) = 0, as conservation requires. (e) Endothermic: it absorbed heat throughout. The energy came from the surroundings: the water and pack material and, while held in place, the ankle, which is why they cooled.

C3. (a) 0.10 × 2,808 = 280.8 kJ released to the surroundings. (b) It passes to the body’s tissues and blood, the surroundings of the reacting system, whose particles gain kinetic energy, which is part of how the body stays warm. (c) Endothermic. Heat flows from the skin into the evaporating sweat. Separating the water particles against their attractions takes energy, the skin supplies it and loses kinetic energy, so the skin cools. (d) The statement has the enthalpy change backwards. ΔH = H(products) − H(reactants) = −2,808 kJ/mol is negative, so the products (CO₂ and H₂O) have lower enthalpy than the glucose and oxygen started with; the difference is exactly the heat released. (e) Endothermic, with ΔH positive: photosynthesis, carried out by plants, reverses respiration and stores the energy again.

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]]> https://examfitter.com/ap-chemistry/worksheet/endothermic-exothermic/feed/ 0 Topic 6.2: Energy Diagrams Worksheet https://examfitter.com/ap-chemistry/worksheet/energy-diagram-problems/ https://examfitter.com/ap-chemistry/worksheet/energy-diagram-problems/#comments Tue, 10 Nov 2026 20:00:50 +0000 https://examfitter.com/?p=2913 How to Use This Worksheet Work in order. Section A checks the parts of the diagram and the two readings taken from it. Section B practises the skills: reading full diagrams, reverse barriers, catalysed pathways, two-peak diagrams, sketching from data, and comparing two reactions. Section C is exam style: judging claims, reconstructing a diagram from ... Read more

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]]> How to Use This Worksheet

Work in order. Section A checks the parts of the diagram and the two readings taken from it. Section B practises the skills: reading full diagrams, reverse barriers, catalysed pathways, two-peak diagrams, sketching from data, and comparing two reactions. Section C is exam style: judging claims, reconstructing a diagram from two values, and a full two-step analysis.

Reference facts. All values are in kJ/mol and every ΔH is reported from the system’s point of view. ΔH = product level − reactant level. Eₐ(forward) = peak − reactants. Eₐ(reverse) = peak − products = Eₐ(forward) − ΔH. A catalyst lowers the peak only: the reactant and product levels, and therefore ΔH, do not move.

Three habits protect every answer. Subtract levels, never read a value straight off the axis from zero. Measure every barrier from the level that step starts from. And check reverse barriers with the identity Eₐ(reverse) = Eₐ(forward) − ΔH before you finish.

Do not look at the answer key until you have attempted every question.

Section A: Recall and Concepts

A1. (a) State what the vertical axis of an energy diagram shows, with its units. (b) State what the horizontal axis shows, and state what it does not show. (c) State the three levels or points whose energies carry every value on a one-step diagram. (d) State where reactants and products sit on the diagram, left or right.

A2. (a) Write the formula for ΔH in terms of the reactant and product levels. (b) Write the formula for Eₐ(forward) in terms of the reactant level and the peak. (c) Write the formula for Eₐ(reverse), and the identity connecting Eₐ(reverse), Eₐ(forward) and ΔH. (d) State what the peak represents, and the name of the atomic arrangement at the peak. (e) State two properties of that arrangement.

A3. For each description, state whether the reaction is exothermic or endothermic, and state the sign of ΔH from the system’s point of view. (a) Products sit 60 kJ/mol below the reactants. (b) Products sit 25 kJ/mol above the reactants. (c) The curve finishes at exactly the level it started. (d) The system absorbs heat from the surroundings as it reacts. (e) The surroundings warm up as the reaction runs.

A4. (a) State what a catalyst does to the peak of an energy diagram. (b) State what a catalyst does to the reactant level, the product level and ΔH. (c) State what happens to Eₐ(forward) and Eₐ(reverse) when a catalyst lowers the peak by 30 kJ/mol. (d) State whether heating the reaction or doubling a concentration redraws the diagram, and explain your answer in one sentence each.

A5. (a) A diagram shows two peaks. State how many elementary steps the reaction has, and what the valley between the peaks holds. (b) State how the activation energy of the second step is measured. (c) State which step is the rate-determining step. (d) A student says the tallest peak on a two-peak diagram is always the rate-determining step. State whether this is always true, and explain in one sentence.

Section B: Core Practice

B1. Profile P: reactants at 80 kJ/mol, peak at 210 kJ/mol, products at 20 kJ/mol. (a) Calculate Eₐ(forward), showing the subtraction. (b) Calculate ΔH, showing the subtraction, and classify the reaction. (c) State how much heat is transferred per mole, and in which direction. (d) Calculate Eₐ(reverse) directly from the levels. (e) Check your part (d) value with the identity.

B2. Profile Q: reactants at 20 kJ/mol, peak at 140 kJ/mol, products at 95 kJ/mol. (a) Calculate Eₐ(forward). (b) Calculate ΔH and classify the reaction. (c) Apply the built-in endothermic check: state what Eₐ must exceed, and confirm it does. (d) Calculate Eₐ(reverse) by both routes. (e) State which direction, forward or reverse, is exothermic, and which crosses the smaller barrier.

B3. An uncatalysed reaction has reactants at 60 kJ/mol, a peak at 190 kJ/mol and products at 10 kJ/mol. A catalyst draws a new peak at 140 kJ/mol. (a) Calculate the uncatalysed Eₐ(forward) and ΔH. (b) Calculate the catalysed Eₐ(forward). (c) Calculate Eₐ(reverse) on both pathways. (d) State by how much each barrier fell, and relate that number to the peak. (e) State the catalysed ΔH, and explain why it has that value.

B4. Profile R, a two-step reaction: reactants at 30 kJ/mol, first peak at 110 kJ/mol, valley at 90 kJ/mol, second peak at 150 kJ/mol, products at 50 kJ/mol. (a) State how many transition states and how many intermediates the diagram shows. (b) Calculate the activation energy of each step, stating the level each is measured from. (c) Identify the rate-determining step, and explain why the taller second peak does not decide the answer. (d) Calculate the ΔH of each step. (e) Calculate the overall ΔH directly from the endpoints, and check it against the sum of the step values.

B5. Sketch data: a reaction has Eₐ(forward) = 90 kJ/mol and ΔH = +40 kJ/mol, with reactants at 50 kJ/mol. (a) Calculate the energy of the peak. (b) Calculate the energy of the products. (c) Describe the sketch in words: the order of the three levels from lowest to highest. (d) Classify the reaction. (e) Calculate Eₐ(reverse), and confirm the endothermic check from part (c) of B2 holds.

B6. Two reactions start from reactants at 100 kJ/mol. Reaction A: peak at 150 kJ/mol, products at 0 kJ/mol. Reaction B: peak at 180 kJ/mol, products at 80 kJ/mol. (a) Calculate Eₐ and ΔH for each reaction. (b) State which reaction is faster, and what decides it. (c) State which reaction releases more heat per mole, and what decides it. (d) Calculate Eₐ(reverse) for each reaction. (e) State what these four values together show about the link between rate and enthalpy change.

Section C: Exam Style

C1. Judge each claim, supporting each verdict with the rule it depends on. (a) “The activation energy is the height of the peak above zero.” (b) “Adding a catalyst makes a reaction more exothermic, because the pathway is lower.” (c) “The species at the peak can be collected, because it sits at a definite energy.” (d) “A reaction with a very negative ΔH must be fast.” (e) “On a two-peak diagram, the valley species is a transition state.”

C2. A reaction is known only from two values: Eₐ(reverse) = 140 kJ/mol and ΔH(forward) = −50 kJ/mol, with reactants at 70 kJ/mol. (a) Calculate Eₐ(forward), showing which rearrangement of the identity you use. (b) Calculate the energy of the peak. (c) Calculate the energy of the products. (d) Classify the forward reaction, and state the classification and ΔH of the reverse reaction. (e) Verify all three original values from your reconstructed levels.

C3. A two-step mechanism is proposed for an endothermic reaction, with this profile: reactants at 40 kJ/mol, first peak at 130 kJ/mol, valley at 70 kJ/mol, second peak at 170 kJ/mol, products at 90 kJ/mol. (a) State the number of steps, transition states and intermediates the profile shows, and state what a mechanism must have to match it. (b) Calculate the activation energy of each step. (c) Calculate the ΔH of each step, then the overall ΔH, and check the sum. (d) Identify the rate-determining step, with your reason. (e) Calculate the reverse barrier of step 2 directly from the levels, and check it with the identity for that step. (f) A catalyst lowers the second peak to 140 kJ/mol. State the new step 2 barrier, state whether the rate-determining step changes, and state the overall ΔH on the catalysed route.

Show the answer key and full working (only after you have attempted every question)

A1. (a) Potential energy, in kJ/mol. (b) Reaction progress, the reaction coordinate: how far the rearrangement from reactants to products has advanced. It does not show time. (c) The reactant level, the peak, and the product level. (d) Reactants on the left, products on the right.

A2. (a) ΔH = product level − reactant level. (b) Eₐ(forward) = peak − reactant level. (c) Eₐ(reverse) = peak − product level, and Eₐ(reverse) = Eₐ(forward) − ΔH. (d) The peak is the transition state; the arrangement there is the activated complex, with old bonds partly broken and new bonds partly formed. (e) Any two: it is the energy maximum of the pathway; it exists only for the instant of the rearrangement; it cannot be isolated or stored; it is the least stable arrangement on the route.

A3. (a) Exothermic, ΔH negative: products below reactants means the system lost enthalpy. (b) Endothermic, ΔH positive. (c) Neither in net terms: ΔH = 0, because the levels are equal. (d) Endothermic, ΔH positive: heat flowing into the system raises its enthalpy. (e) Exothermic, ΔH negative: the surroundings can only warm if the system released heat to them.

A4. (a) It provides an alternative pathway with a lower peak. (b) Nothing: the reactant level, product level and ΔH are all unchanged. (c) Both fall by exactly 30 kJ/mol, because both climbs end at a peak that is 30 kJ/mol lower while their starting levels stay fixed. (d) Neither redraws it. Heating changes how many particles can cross the unchanged barrier, not the barrier; concentration changes how often particles meet, not any energy level.

A5. (a) Two elementary steps; the valley holds an intermediate, a real species formed in step 1 and consumed in step 2. (b) From the valley level up to the second peak. (c) The step with the largest activation energy, measured from its own starting level. (d) Not always: a later peak can stand higher on the axis yet have the smaller barrier, because its barrier is measured from its valley, not from the reactant level.

B1. (a) Eₐ(forward) = 210 − 80 = 130 kJ/mol. (b) ΔH = 20 − 80 = −60 kJ/mol. Negative, so exothermic. (c) 60 kJ per mole is released by the system to the surroundings. (d) Eₐ(reverse) = 210 − 20 = 190 kJ/mol. (e) Identity: Eₐ(forward) − ΔH = 130 − (−60) = 190 kJ/mol. The two routes agree.

B2. (a) Eₐ(forward) = 140 − 20 = 120 kJ/mol. (b) ΔH = 95 − 20 = +75 kJ/mol. Positive, so endothermic. (c) Eₐ must exceed ΔH: 120 is greater than 75, so the check holds. (d) Directly: 140 − 95 = 45 kJ/mol. By the identity: 120 − 75 = 45 kJ/mol. (e) The reverse direction is exothermic, because it runs from 95 down to 20 kJ/mol, and it crosses the smaller barrier, 45 kJ/mol against 120 kJ/mol.

B3. (a) Eₐ(forward) = 190 − 60 = 130 kJ/mol. ΔH = 10 − 60 = −50 kJ/mol. (b) Catalysed Eₐ(forward) = 140 − 60 = 80 kJ/mol. (c) Uncatalysed Eₐ(reverse) = 190 − 10 = 180 kJ/mol. Catalysed Eₐ(reverse) = 140 − 10 = 130 kJ/mol. (d) Each barrier fell by 50 kJ/mol, which is exactly the peak drop, 190 − 140 = 50 kJ/mol. (e) ΔH is still −50 kJ/mol: the catalyst moved only the peak, and ΔH is read from the endpoints, which stayed at 60 and 10 kJ/mol.

B4. (a) Two transition states (the two peaks) and one intermediate (in the valley). (b) Step 1: 110 − 30 = 80 kJ/mol, measured from the reactant level. Step 2: 150 − 90 = 60 kJ/mol, measured from the valley. (c) Step 1: its barrier, 80 kJ/mol, is larger than step 2’s 60 kJ/mol. The second peak stands higher on the axis, but a barrier is a climb, not an altitude, and step 2 climbs only from its valley at 90 kJ/mol. (d) Step 1 ΔH = 90 − 30 = +60 kJ/mol. Step 2 ΔH = 50 − 90 = −40 kJ/mol. (e) Overall ΔH = 50 − 30 = +20 kJ/mol. Sum of steps: +60 + (−40) = +20 kJ/mol. They agree, and the overall reaction is endothermic.

B5. (a) Peak = reactants + Eₐ = 50 + 90 = 140 kJ/mol. (b) Products = reactants + ΔH = 50 + 40 = 90 kJ/mol. (c) Lowest: reactants at 50, then products at 90, then the peak at 140 kJ/mol. (d) Endothermic: ΔH is positive and the products finish above the reactants. (e) Eₐ(reverse) = 140 − 90 = 50 kJ/mol. Check: Eₐ(forward) − ΔH = 90 − 40 = 50 kJ/mol, and Eₐ(forward) = 90 exceeds ΔH = 40, as an endothermic diagram requires.

B6. (a) Reaction A: Eₐ = 150 − 100 = 50 kJ/mol; ΔH = 0 − 100 = −100 kJ/mol. Reaction B: Eₐ = 180 − 100 = 80 kJ/mol; ΔH = 80 − 100 = −20 kJ/mol. (b) Reaction A is faster: its barrier is lower, 50 against 80 kJ/mol, so more of its particles can cross per second. (c) Reaction A releases more heat: its ΔH is more negative, 100 against 20 kJ per mole. (d) A: Eₐ(reverse) = 150 − 0 = 150 kJ/mol. B: Eₐ(reverse) = 180 − 80 = 100 kJ/mol. (e) Here the faster reaction also releases more heat, but that is coincidence, not a rule: rate is decided by Eₐ alone and heat by ΔH alone, and either can be large while the other is small.

C1. (a) False. Eₐ is the climb from the reactant level to the peak, not the peak’s height above zero: reactants at 50 with a peak at 120 give Eₐ = 70, not 120 kJ/mol. (b) False. A catalyst lowers only the peak; the endpoints are unchanged, so ΔH is identical and the reaction is exactly as exothermic as before. It is faster, not more exothermic. (c) False. The peak species is the activated complex, which exists only at the energy maximum for an instant and cannot be isolated or collected. (d) False. ΔH measures the endpoint gap and says nothing about the barrier: a very exothermic reaction can sit behind a tall peak and run slowly until it is lit or catalysed. (e) False. The valley holds an intermediate, a real species at a local minimum. Transition states are the peaks.

C2. (a) From Eₐ(reverse) = Eₐ(forward) − ΔH, rearranged: Eₐ(forward) = Eₐ(reverse) + ΔH = 140 + (−50) = 90 kJ/mol. (b) Peak = reactants + Eₐ(forward) = 70 + 90 = 160 kJ/mol. (c) Products = reactants + ΔH = 70 + (−50) = 20 kJ/mol. (d) Forward: exothermic, ΔH = −50 kJ/mol, because the products sit below the reactants. Reverse: endothermic, ΔH = +50 kJ/mol. (e) From the levels: Eₐ(forward) = 160 − 70 = 90 kJ/mol; ΔH = 20 − 70 = −50 kJ/mol; Eₐ(reverse) = 160 − 20 = 140 kJ/mol. All three match the values given.

C3. (a) Two steps, two transition states (the peaks) and one intermediate (in the valley). A matching mechanism must have exactly two elementary steps and exactly one intermediate; any other count cannot produce this profile. (b) Step 1: 130 − 40 = 90 kJ/mol, from the reactant level. Step 2: 170 − 70 = 100 kJ/mol, from the valley. (c) Step 1 ΔH = 70 − 40 = +30 kJ/mol. Step 2 ΔH = 90 − 70 = +20 kJ/mol. Overall ΔH = 90 − 40 = +50 kJ/mol. Check: +30 + 20 = +50 kJ/mol. (d) Step 2: its barrier, 100 kJ/mol, is larger than step 1’s 90 kJ/mol, so it is crossed most slowly and limits the overall rate. (e) Directly: Eₐ(reverse, step 2) = 170 − 90 = 80 kJ/mol. By the identity for step 2: 100 − 20 = 80 kJ/mol. (f) New step 2 barrier = 140 − 70 = 70 kJ/mol. Step 1, at 90 kJ/mol, now has the largest barrier, so the rate-determining step changes to step 1. The overall ΔH is unchanged at +50 kJ/mol, because the catalyst moved a peak, not an endpoint.

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]]> https://examfitter.com/ap-chemistry/worksheet/energy-diagram-problems/feed/ 1 Topic 6.9: Hess’s Law Revision Notes https://examfitter.com/ap-chemistry/revision-notes/hess-law/ https://examfitter.com/ap-chemistry/revision-notes/hess-law/#respond Tue, 10 Nov 2026 19:54:26 +0000 https://examfitter.com/?p=2866 What Hess’s Law States Hess’s law states that the enthalpy change for a reaction is the same whichever route is taken from reactants to products, provided the starting and finishing states are the same. In practice this means one thing: if a reaction can be written as a sum of step reactions, then its ΔH ... Read more

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]]> What Hess’s Law States

Hess’s law states that the enthalpy change for a reaction is the same whichever route is taken from reactants to products, provided the starting and finishing states are the same.

In practice this means one thing: if a reaction can be written as a sum of step reactions, then its ΔH is the sum of the step ΔH values.

ΔH(target) = ΔH(step 1) + ΔH(step 2) + ΔH(step 3) + …

The steps need only add up, on paper, to the target equation exactly.

Why It Works: Enthalpy Is a State Function

Enthalpy is a state function. Its value depends only on the state the system is in now, not on how it got there.

Pressure, volume and temperature behave the same way. Heat and work do not: they are path functions, and their values do depend on the route.

Because enthalpy is a state function, the difference in enthalpy between reactants and products is fixed. Every route between the same two states crosses the same enthalpy gap, so the step values along any route must add to the same total.

That is why Hess’s law is exact, not approximate. It also explains its main use: a reaction too slow, too dangerous or too tangled with side reactions to measure directly can still be priced, by adding up steps that can be measured.

The Three Manipulations

Reverse. Turning an equation round swaps reactants and products, so ΔH changes sign and keeps its size. If A → B has ΔH = +50 kJ, then B → A has ΔH = −50 kJ.

Scale. Multiplying every coefficient in an equation by a factor multiplies ΔH by the same factor, because enthalpy is extensive: twice the reaction involves twice the material and twice the heat. Halving an equation halves ΔH, and fractions such as ½ are used freely.

Add. Adding equations together adds their ΔH values. Any species that appears on both sides of the summed equation cancels, exactly as in algebra.

The three moves combine. Reversing and then tripling an equation flips the sign and triples the size: a step with ΔH = −40 kJ becomes +120 kJ. Changing the temperature is not a fourth move; ΔH values belong to the conditions they were measured at.

Building the Target Equation

Start from the target, not from the steps. Write the target equation out in full, with states, and keep it in view while you work.

Find a species that appears in only one step. That step usually decides first: if the species sits on the wrong side of its step, reverse the step; if its coefficient is wrong, scale the step so the coefficient matches the target.

Work species by species until every reactant and product of the target is accounted for. Species that appear in no step and not in the target must cancel between steps; if one is left over at the end, a manipulation is wrong.

Finish with a full check. Add the manipulated steps on paper and confirm the sum reproduces the target exactly: same species, same coefficients, same states. Only then add the manipulated ΔH values, each with its new sign and scale applied.

States Must Match Before Anything Cancels

Cancellation is exact matching. A species cancels only when the same formula, in the same state, with the same total coefficient, appears on both sides.

H₂O(l) and H₂O(g) are different species for this purpose. A step that produces liquid water cannot cancel a step that consumes water vapour: the two waters differ in enthalpy by 44.0 kJ mol⁻¹, the enthalpy of vaporization.

The same discipline applies to allotropes. C(s, graphite) and C(s, diamond) are both carbon, but they are not the same species, and neither cancels the other.

When states do not match, a step for the state change itself, such as

H₂O(l) → H₂O(g) with ΔH = +44.0 kJ,

can be added to convert one into the other before cancelling.

Solving for an Unknown Step

The same addition runs backwards. If the overall ΔH and every step but one are known, the unknown step is whatever makes the sum balance.

Write the sum with the unknown as x, then rearrange:

x = ΔH(overall) − sum of the known steps,

each known step already reversed or scaled as the cycle requires.

This is how many tabulated values are actually obtained. A formation enthalpy that cannot be measured by reacting the elements directly is recovered from a combustion that can be measured, by solving the cycle for the missing step. The formation method of Topic 6.8 is exactly this idea carried out through the elements.

Worked Examples

Example 1: forming carbon monoxide. Given C(s) + O₂(g) → CO₂(g), ΔH = −393.5 kJ, and CO(g) + ½O₂(g) → CO₂(g), ΔH = −283.0 kJ. Target: C(s) + ½O₂(g) → CO(g). Reverse the second step: CO₂(g) → CO(g) + ½O₂(g), ΔH = +283.0 kJ. Add: the CO₂ cancels, and ½O₂ of the O₂ cancels, leaving the target. ΔH = −393.5 + 283.0 = −110.5 kJ.

Example 2: forming nitrogen dioxide. Given N₂(g) + O₂(g) → 2NO(g), ΔH = +180.6 kJ, and 2NO(g) + O₂(g) → 2NO₂(g), ΔH = −114.2 kJ. Target: N₂(g) + 2O₂(g) → 2NO₂(g). Neither step needs changing: add them and the 2NO cancels. ΔH = +180.6 + (−114.2) = +66.4 kJ. Endothermic overall, because the first step costs more than the second returns.

Example 3: graphite to diamond. Given C(s, graphite) + O₂(g) → CO₂(g), ΔH = −393.5 kJ, and C(s, diamond) + O₂(g) → CO₂(g), ΔH = −395.4 kJ. Target: C(s, graphite) → C(s, diamond). Keep the first step, reverse the second: CO₂(g) → C(s, diamond) + O₂(g), ΔH = +395.4 kJ. Adding cancels both CO₂ and O₂. ΔH = −393.5 + 395.4 = +1.9 kJ. The direct conversion is far too slow to measure, so the two combustions price it instead.

Example 4: hydrogenating ethene. Given C₂H₄(g) + 3O₂(g) → 2CO₂(g) + 2H₂O(l), ΔH = −1411 kJ; H₂(g) + ½O₂(g) → H₂O(l), ΔH = −285.8 kJ; C₂H₆(g) + 7/2O₂(g) → 2CO₂(g) + 3H₂O(l), ΔH = −1560 kJ. Target: C₂H₄(g) + H₂(g) → C₂H₆(g). Take the first step as written, add the second, and reverse the third. The 2CO₂ cancels, all 3H₂O cancels, and the O₂ cancels. ΔH = −1411 + (−285.8) + 1560 = −136.8 kJ.

Traps That Cost Marks in Topic 6.9

“Flip the equation, keep the ΔH.” Reversing a step without flipping its sign is the commonest error in the topic. The reversed step carries the negative of its original ΔH, every time.

“Scale the equation, keep the ΔH.” ΔH is priced for the equation as written. Double the coefficients and the ΔH doubles; halve them and it halves. A scaled step with an unscaled ΔH breaks the sum.

“Cancel anything with the same formula.” States and allotropes are part of the species. H₂O(l) does not cancel H₂O(g), and graphite does not cancel diamond.

“Average the step values.” ΔH values add. They are never averaged, and the answer is neither the largest step nor the smallest step.

“Apply the manipulation to ΔH only.” Reversing or scaling acts on the whole step, equation and ΔH together. Changing one without the other produces a sum that matches nothing.

“Add first, check later.” A leftover species at the end means the manipulations are wrong, however tidy the arithmetic looks. The summed equation must reproduce the target exactly before its ΔH is trusted.

Key Relationships

  • Hess’s law: ΔH for a reaction is independent of the route taken, because enthalpy is a state function
  • ΔH(target) = sum of the ΔH values of the steps that add to the target, after each step is manipulated
  • Reversing an equation changes the sign of its ΔH and keeps its size
  • Multiplying every coefficient by a factor multiplies ΔH by the same factor; halving an equation halves ΔH
  • A species cancels only when formula, state and coefficient all match on both sides of the summed equation
  • An unknown step follows from x = ΔH(overall) − sum of the known manipulated steps
  • H₂O(l) and H₂O(g) differ by 44.0 kJ mol⁻¹ and never cancel each other
  • A catalyst changes the route and the rate but not ΔH, which is itself a demonstration of path independence

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