How to Use This Worksheet
Work in order. Section A checks the decision rule: comparing the forward and reverse rates, from reactants only, from products only, and from mixtures. Section B works with one data set, CO(g) + 2H₂(g) ⇌ CH₃OH(g), approached from the product side: a table, rates, a second vessel, snapshots and a removal. Section C is exam style: an ammonia vessel charged with product only, a rate-time graph, and five claims to judge.
Reference facts. Net direction follows the larger rate: forward rate greater means reactants fall and products rise; equal rates mean no net change. An absent side has a rate of zero, so a one-sided start forces the opening direction. Concentration changes always follow the coefficients. No equilibrium constants are needed here. Useful values: the methanol system settles at [CO] = 0.42, [H₂] = 0.84 and [CH₃OH] = 0.68 mol L⁻¹.
Three habits protect every answer. Name the direction from the equation as written, never from habit. Compare the two rates, not the two head counts. And check every concentration change against the coefficients before trusting it.
Do not look at the answer key until you have attempted every question.
Section A: Recall and Concepts
A1. (a) State what is meant by the net direction of a reversible reaction. (b) State the rule that decides the net direction at any moment. (c) A system has equal forward and reverse rates but very different reactant and product concentrations. State its net direction, and explain.
A2. A reversible reaction is started with reactants only in a sealed vessel. (a) State the initial value of the reverse rate, and why. (b) State which concentrations rise and which fall at the start. (c) State why the forward rate is at its maximum at the start, and describe what happens to the gap between the two rates as the reaction proceeds.
A3. A reversible reaction is started with products only in a sealed vessel. (a) State the initial value of the forward rate, and why. (b) State the net direction at the start. (c) Describe how each rate changes as the system moves toward equilibrium, and why the net change slows down.
A4. In 2NO₂(g) ⇌ N₂O₄(g), readings show [NO₂] falling by 0.20 mol L⁻¹ over an interval while [N₂O₄] rises by 0.10 mol L⁻¹. (a) State the net direction over the interval, as the equation is written. (b) Show that the two changes are consistent with the equation. (c) A second report claims [NO₂] fell by 0.20 while [N₂O₄] rose by 0.20 mol L⁻¹ in the same vessel. State whether both reports can describe the same closed vessel, and why.
A5. Two students describe the same flask. One writes the reaction as N₂O₄(g) ⇌ 2NO₂(g) and says the net reaction is reverse; the other writes it as 2NO₂(g) ⇌ N₂O₄(g) and says the net reaction is forward. (a) State what is physically happening to [NO₂] in the flask. (b) Explain how both students can be right. (c) State what a direction claim always needs before it means anything.
Section B: Core Practice
CO(g) + 2H₂(g) ⇌ CH₃OH(g) is followed in a sealed vessel at constant temperature, charged with a product-rich mixture. Concentrations in mol L⁻¹:
| Time (min) | [CO] | [H₂] | [CH₃OH] | |—|—|—|—| | 0 | 0.20 | 0.40 | 0.90 | | 2 | 0.30 | 0.60 | 0.80 | | 4 | 0.36 | 0.72 | 0.74 | | 6 | 0.40 | 0.80 | 0.70 | | 8 | 0.42 | 0.84 | 0.68 | | 10 | 0.42 | 0.84 | 0.68 |
B1. (a) State the net direction over the first 8 minutes, and the evidence in the table. (b) State the time by which equilibrium is reached, and the evidence. (c) State the equilibrium concentration of each species. (d) Calculate the change in concentration of each species between 0 and 8 minutes.
B2. (a) Show that the changes found in B1(d) obey the stoichiometry of the equation. (b) Calculate the average rate of disappearance of CH₃OH between 0 and 2 minutes. (c) Calculate the average rate of formation of CO between 6 and 8 minutes. (d) Calculate the average net rate of change of [CH₃OH] between 8 and 10 minutes, and explain the value, given that CH₃OH is still reacting in both directions.
B3. (a) At time zero, state which rate, forward or reverse, is larger, and explain the answer in terms of the concentrations. (b) Explain why the forward rate in this vessel is small at time zero but not zero. (c) Describe how each rate changes during the first 8 minutes, and the condition reached at 8 minutes. (d) At equilibrium the forward and reverse rates are each 0.015 mol L⁻¹ min⁻¹, measured as the rate at which CH₃OH is formed and consumed. A student says [CH₃OH] must therefore be falling at 0.015 mol L⁻¹ min⁻¹. Correct the student.
B4. A second vessel at the same temperature is charged with [CO] = 1.10 mol L⁻¹ and [H₂] = 2.20 mol L⁻¹, and no CH₃OH. (a) State the net direction at the start, with a reason based on the rates. (b) Show that this vessel holds the same totals of carbon, hydrogen and oxygen per litre as the B1 vessel. (c) Predict the equilibrium concentrations in this vessel, and justify the prediction. (d) Calculate the change in [CO] and in [H₂] needed to reach that equilibrium, and check the two changes against the equation.
B5. Snapshots are taken of A₂(g) + B₂(g) ⇌ 2AB(g) in a sealed box. Snapshot 1 shows 6 A₂, 6 B₂ and 4 AB particles. Snapshot 2, taken later, shows 4 A₂, 4 B₂ and 8 AB. The equilibrium snapshot for this box shows 3 A₂, 3 B₂ and 10 AB. (a) State the net direction between the snapshots, with evidence. (b) Verify that snapshots 1 and 2 hold the same total number of A atoms and of B atoms. (c) State whether the box has reached equilibrium by snapshot 2, and describe what happens to the counts next.
B6. The B1 vessel is at equilibrium when half of the CH₃OH is pumped out, dropping [CH₃OH] instantly from 0.68 to 0.34 mol L⁻¹; [CO] and [H₂] are unchanged at that instant. (a) State what happens to the reverse rate immediately, and why. (b) State what happens to the forward rate at that instant, and why. (c) State the net direction that follows. (d) Describe the new equilibrium: which way each concentration moves after the removal, and why [CH₃OH] cannot climb back to 0.68 mol L⁻¹.
Section C: Exam Style
C1. Ammonia synthesis is written N₂(g) + 3H₂(g) ⇌ 2NH₃(g). A sealed vessel is charged with NH₃ only, at [NH₃] = 0.90 mol L⁻¹. It settles at [NH₃] = 0.50, [N₂] = 0.20 and [H₂] = 0.60 mol L⁻¹. (a) Explain, in terms of rates, why the initial net change is toward reactants, giving the initial value of the forward rate. (b) Calculate the amount of NH₃ per litre that has decomposed by equilibrium, and check it against the N₂ and H₂ formed. (c) A second vessel at the same temperature starts with [N₂] = 0.45 and [H₂] = 1.35 mol L⁻¹, and no NH₃. Predict its equilibrium concentrations, justifying your answer with the total amounts of nitrogen and hydrogen in each vessel, and state the changes that take the second vessel there. (d) A student argues: “[H₂] in the second vessel, 1.35 mol L⁻¹, is more than double the equilibrium value of 0.60 mol L⁻¹, so the second vessel must react in reverse.” Judge the argument.
C2. A rate-time graph is drawn for 2NO₂(g) ⇌ N₂O₄(g) in a sealed vessel charged with a mixture of both gases. Both rates are quoted as rates of change of [N₂O₄]: the forward reaction forms N₂O₄, the reverse reaction consumes it. The forward rate starts at 0.60 mol L⁻¹ min⁻¹ and falls steadily; the reverse rate starts at 0.30 mol L⁻¹ min⁻¹ and rises. The curves meet at 0.42 mol L⁻¹ min⁻¹ at t = 9 min and run level together after that. (a) State which rate is larger at t = 0, and what is happening to [NO₂] and to [N₂O₄] then. (b) State when equilibrium is reached, and the evidence. (c) Explain why the forward rate falls and the reverse rate rises during the first 9 minutes. (d) At t = 3 min the forward rate is 0.52 and the reverse rate is 0.36 mol L⁻¹ min⁻¹. Calculate the net rate of change of [N₂O₄] and of [NO₂] at that moment.
C3. Judge each claim, supporting each verdict with the idea it depends on. (a) “If the concentration of products is higher than the concentration of reactants, the net reaction must be reverse.” (b) “During a net forward reaction, the reverse reaction is not happening.” (c) “Removing some product from an equilibrium mixture starts a net reverse reaction to replace it.” (d) “A system approaching equilibrium overshoots it, swings back, and settles only after a few swings.” (e) “Equal forward and reverse rates mean equal reactant and product concentrations.”
Show the answer key and full working (only after you have attempted every question)
A1. (a) The direction of the overall change: which conversion, forward or reverse, is happening faster, and so which side is consumed net. (b) Compare the forward and reverse rates: the larger rate gives the net direction, and equal rates give no net change. (c) There is no net direction: the system is at equilibrium. Equal rates mean every forward conversion is balanced by a reverse one, so concentrations stay constant.
A2. (a) Zero: no product exists yet, so nothing can react back. (b) Reactant concentrations fall and product concentrations rise. (c) Reactant concentrations are highest at the start, so the forward rate starts at its maximum. As products accumulate, the reverse rate climbs from zero while the forward rate falls, narrowing the gap.
A3. (a) Zero. No reactants exist for the forward reaction to use. (b) Net reverse: products are converted back into reactants. (c) The reverse rate falls as products fall; the forward rate rises from zero as reactants accumulate. The gap narrows, so the net change slows toward zero.
A4. (a) Net forward, as written: NO₂ is the reactant here and is being consumed, while N₂O₄ is formed. (b) The equation consumes 2NO₂ per N₂O₄ formed, so a fall of 0.20 mol L⁻¹ in [NO₂] pairs with a rise of half as much, 0.10, in [N₂O₄]. (c) No: a rise of 0.20 mol L⁻¹ in [N₂O₄] needs a fall of 0.40 in [NO₂]. The pair 0.20 and 0.20 breaks the ratio.
A5. (a) [NO₂] is falling: NO₂ is consumed net and N₂O₄ is formed. (b) Forward and reverse are defined by the writing. With N₂O₄ on the left, forming N₂O₄ is right-to-left: reverse. With NO₂ on the left, the same change is left-to-right: forward. Each student describes the flask correctly, relative to their own writing. (c) The equation as written. A direction claim means nothing until the writing it refers to is fixed.
B1. (a) Net reverse: [CH₃OH] falls while [CO] and [H₂] rise, so product is consumed net and reactants formed. (b) By 8 minutes: the readings at 8 and 10 minutes are identical for all three species. (c) [CO] = 0.42 mol L⁻¹, [H₂] = 0.84 mol L⁻¹, [CH₃OH] = 0.68 mol L⁻¹. (d) CO: 0.42 − 0.20 = +0.22 mol L⁻¹. H₂: 0.84 − 0.40 = +0.44 mol L⁻¹. CH₃OH: 0.68 − 0.90 = −0.22 mol L⁻¹.
B2. (a) Read in reverse, the equation consumes CH₃OH and produces CO and H₂ in the ratio 1 : 2 : 1; the measured changes match exactly. (b) (0.90 − 0.80) ÷ (2 − 0) = 0.050 mol L⁻¹ min⁻¹. (c) (0.42 − 0.40) ÷ (8 − 6) = 0.010 mol L⁻¹ min⁻¹. (d) (0.68 − 0.68) ÷ (10 − 8) = 0 mol L⁻¹ min⁻¹. Zero: the concentration does not change, because at equilibrium CH₃OH is formed and consumed at equal rates. The net change is the difference of two equal rates.
B3. (a) The reverse rate is larger. The vessel starts product-rich, [CH₃OH] = 0.90 mol L⁻¹, so product particles react back often, while the reactants driving the forward rate are dilute. (b) The forward rate is not zero because both its reactants are present, CO at 0.20 and H₂ at 0.40 mol L⁻¹: it can run, just more slowly than the reverse. A rate is zero only when a species on its side is absent. (c) The reverse rate falls as [CH₃OH] falls; the forward rate rises as [CO] and [H₂] build up, until at 8 minutes the rates are equal. (d) The net rate of change is forward minus reverse: 0.015 − 0.015 = 0 mol L⁻¹ min⁻¹. The value 0.015 is the rate at which CH₃OH is formed and, separately, consumed; concentration changes only by the difference.
B4. (a) Net forward: with no CH₃OH present the reverse rate is zero, while CO and H₂ at high concentrations put the forward rate at its maximum. (b) Carbon: B1 holds 0.20 from CO plus 0.90 from CH₃OH, total 1.10 mol L⁻¹; this vessel holds 1.10 from CO. Oxygen: B1 holds 0.20 plus 0.90 = 1.10; this vessel holds 1.10. Hydrogen atoms: B1 holds 2 × 0.40 + 4 × 0.90 = 4.40; this vessel holds 2 × 2.20 = 4.40. All three totals match. (c) The same equilibrium as B1: [CO] = 0.42, [H₂] = 0.84, [CH₃OH] = 0.68 mol L⁻¹, because the same reaction at the same temperature with the same totals settles at the same composition from either side. (d) [CO]: 0.42 − 1.10 = −0.68 mol L⁻¹. [H₂]: 0.84 − 2.20 = −1.36 mol L⁻¹, exactly twice the CO change, and the CH₃OH formed, +0.68, matches the CO consumed.
B5. (a) Net forward: A₂ and B₂ counts fell while the AB count rose, so reactants are consumed net and product formed. (b) A atoms: snapshot 1, 2 × 6 + 4 = 16; snapshot 2, 2 × 4 + 8 = 16. B atoms: the same two totals, 16 and 16. Both conserved. (c) No. Equilibrium counts are 3, 3 and 10; snapshot 2 shows 4, 4 and 8. Net forward reaction continues until the counts reach 3, 3 and 10.
B6. (a) The reverse rate falls immediately, because it is driven by [CH₃OH], just halved. (b) The forward rate is unchanged at that instant: [CO] and [H₂], which drive it, were untouched. (c) Net forward: the forward rate is now larger. (d) [CO] and [H₂] fall and [CH₃OH] rises from 0.34 mol L⁻¹ until the rates are equal again. [CH₃OH] cannot regain 0.68 mol L⁻¹: the vessel now holds only 0.42 + 0.34 = 0.76 mol L⁻¹ of carbon, and some must remain as CO.
C1. (a) The vessel holds NH₃ only, so the forward rate starts at zero: there is no N₂ or H₂ to react. The reverse reaction has its reactant at full concentration, so its rate is at maximum. With the reverse rate larger, net change runs toward reactants until the forward rate catches up. (b) NH₃ decomposed: 0.90 − 0.50 = 0.40 mol L⁻¹. Check: read in reverse, the equation turns 2NH₃ into N₂ and 3H₂, so 0.40 mol L⁻¹ of NH₃ forms 0.20 of N₂ and 0.60 of H₂, matching the equilibrium values. (c) Nitrogen totals: vessel 1, 0.90 mol L⁻¹; vessel 2, 2 × 0.45 = 0.90. Hydrogen totals: vessel 1, 3 × 0.90 = 2.70; vessel 2, 2 × 1.35 = 2.70. Same totals, same temperature: vessel 2 reaches the same equilibrium, [NH₃] = 0.50, [N₂] = 0.20, [H₂] = 0.60 mol L⁻¹, by net forward reaction: [N₂] falls by 0.25, [H₂] by 0.75 and [NH₃] rises by 0.50 mol L⁻¹, in the 1 : 3 : 2 ratio. (d) Wrong. Direction is set by comparing the two rates, not by comparing one concentration with its equilibrium value. Vessel 2 contains no NH₃, so its reverse rate is zero: the only net direction is forward, and the excess H₂ is consumed as NH₃ forms.
C2. (a) The forward rate is larger, 0.60 against 0.30 mol L⁻¹ min⁻¹, so the net reaction is forward: [N₂O₄] rises, [NO₂] falls. (b) Equilibrium is reached at t = 9 min, where the curves meet; after that the rates stay equal and level. (c) The forward rate falls because [NO₂] falls as NO₂ is consumed, so its particles collide less often. The reverse rate rises because [N₂O₄] climbs, so more N₂O₄ is present to break down. (d) Net rate for N₂O₄: 0.52 − 0.36 = +0.16 mol L⁻¹ min⁻¹, so [N₂O₄] rises at 0.16 mol L⁻¹ min⁻¹. Each N₂O₄ formed consumes 2NO₂, so [NO₂] falls at 2 × 0.16 = 0.32 mol L⁻¹ min⁻¹.
C3. (a) False. Direction is decided by comparing the rates, equivalently by comparing the mixture with the equilibrium composition. [H₂] = [I₂] = 0.50 mol L⁻¹ with [HI] = 1.00 mol L⁻¹ is product-rich yet short of equilibrium, [HI] = 1.56 mol L⁻¹, and runs net forward. (b) False. The reverse reaction runs whenever any product is present; net forward means the forward rate is larger, not that the reverse rate is zero. (c) False. Removing product lowers the reverse rate, so the forward rate becomes the larger one and net forward reaction replaces the lost product. (d) False. Approaching equilibrium, the winning rate falls and the losing rate rises; they meet and net change stops. No overshoot, no swinging. (e) False. Equilibrium needs equal rates, not equal concentrations: the methanol equilibrium of Section B sits at [CO] = 0.42, [H₂] = 0.84 and [CH₃OH] = 0.68 mol L⁻¹, three different values.