Topic 7.2: Direction of Reversible Reactions Quiz

Master AP Chemistry Topic 7.2 with this free practice quiz. Test your skills on approaching equilibrium, shift to products, shift to reactants, and initial rates to prepare for your AP exam.

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Topic 7.2: Direction of Reversible Reactions Quiz

18 MCQs

Master the reversible reactions with these 18 MCQs on AP Chemistry Topic 7.2: Direction of Reversible Reactions.

Covered Subtopics: Reversible reactions, approaching equilibrium, direction of net change

1 / 18

1. The rate of the forward reaction is greatest at the moment:

the system has already reached equilibrium.

only reactants are present and no products have formed yet.

the reverse rate exceeds the forward rate.

products outnumber reactants by ten to one.

The forward rate depends on reactant concentration, which is at its maximum before any products form.

The forward rate depends on reactant concentration, which is at its maximum before any products form.

The forward rate depends on reactant concentration, which is at its maximum before any products form.

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2. Which change indicates a reversible reaction is moving toward equilibrium from the product side?

Product concentration decreases while reactant concentration increases.

The total mass of the system increases.

The reaction rate stays at zero throughout.

Both concentrations increase at the same time.

Net reverse reaction consumes products and forms reactants, so product concentration falls while reactant concentration rises.

Net reverse reaction consumes products and forms reactants, so product concentration falls while reactant concentration rises.

Net reverse reaction consumes products and forms reactants, so product concentration falls while reactant concentration rises.

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3. A double arrow in a chemical equation signals that:

the reaction occurs in two separate containers.

the products are twice as concentrated as reactants.

the reaction needs two different catalysts.

the reaction can proceed in both forward and reverse directions.

The double arrow denotes a reversible reaction in which both the forward and reverse processes occur.

The double arrow denotes a reversible reaction in which both the forward and reverse processes occur.

The double arrow denotes a reversible reaction in which both the forward and reverse processes occur.

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4. A reversible reaction left undisturbed in a sealed bottle will:

move toward the equilibrium position regardless of starting materials.

stop at whatever composition was present first.

oscillate forever between the two sides.

always end with only products present.

The faster of the two rates dominates until they equalize, which always moves the system toward equilibrium.

The faster of the two rates dominates until they equalize, which always moves the system toward equilibrium.

The faster of the two rates dominates until they equalize, which always moves the system toward equilibrium.

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5. A chemist starts a reversible reaction with a mixture already at the equilibrium composition. The expected observation is:

no net change in any concentration over time.

a slow drift toward pure products.

alternating forward and backward swings.

a slow drift toward pure reactants.

At the equilibrium composition the forward and reverse rates are already equal, so no net change occurs.

At the equilibrium composition the forward and reverse rates are already equal, so no net change occurs.

At the equilibrium composition the forward and reverse rates are already equal, so no net change occurs.

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6. A reversible reaction can be approached from either side, which means:

catalysts work only when both sides are present at once.

the reaction vessel must be flipped during the experiment.

starting with all reactants or all products leads to the same equilibrium state.

temperature must change halfway through the reaction.

The same equilibrium composition results whether the system starts from reactants, products, or any mixture of them.

The same equilibrium composition results whether the system starts from reactants, products, or any mixture of them.

The same equilibrium composition results whether the system starts from reactants, products, or any mixture of them.

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7. If the forward rate is faster than the reverse rate in a closed system, over time:

reactant concentration falls and product concentration rises.

reactant concentration rises while product falls.

both concentrations fall at the same speed.

all concentrations remain exactly as they started.

Net forward reaction consumes reactants and forms products, so reactants decrease while products increase.

Net forward reaction consumes reactants and forms products, so reactants decrease while products increase.

Net forward reaction consumes reactants and forms products, so reactants decrease while products increase.

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8. Two containers hold the same reversible reaction; one starts with pure reactants, the other with pure products. Both eventually show identical composition because:

each side proceeds until forward and reverse rates match.

each mixture eventually becomes a single pure substance.

products always convert fully back to reactants.

the reactants migrate from one container to the other.

Either starting point drives net reaction toward the composition where the two rates equalize, giving the same equilibrium state.

Either starting point drives net reaction toward the composition where the two rates equalize, giving the same equilibrium state.

Either starting point drives net reaction toward the composition where the two rates equalize, giving the same equilibrium state.

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9. When the reverse rate is initially larger than the forward rate, the system:

shows no observable change of any kind.

undergoes net conversion of products into reactants.

requires a catalyst before anything changes.

undergoes net conversion of reactants into products.

The faster reverse reaction dominates, so products are consumed and reactants accumulate until the rates match.

The faster reverse reaction dominates, so products are consumed and reactants accumulate until the rates match.

The faster reverse reaction dominates, so products are consumed and reactants accumulate until the rates match.

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10. Starting with only NO₂ in a sealed container, the reaction 2 NO₂ <=> N₂O₄ shows N₂O₄ forming. This tells us:

NO₂ molecules cannot collide with each other.

the net direction was reverse until the rates equalized.

equilibrium is impossible when starting from products.

the forward reaction can never occur in this system.

N₂O₄ is the product of the reverse reaction written as shown, so its formation means net reverse reaction occurred until equilibrium.

N₂O₄ is the product of the reverse reaction written as shown, so its formation means net reverse reaction occurred until equilibrium.

N₂O₄ is the product of the reverse reaction written as shown, so its formation means net reverse reaction occurred until equilibrium.

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11. When a reversible system at equilibrium is disturbed by removing some product, the immediate response is:

both rates dropping to zero instantly.

a net forward reaction to replace the lost product.

a net reverse reaction consuming more product.

no change because equilibrium cannot be disturbed.

Removing product lowers the reverse rate, so the forward reaction temporarily dominates and rebuilds product.

Removing product lowers the reverse rate, so the forward reaction temporarily dominates and rebuilds product.

Removing product lowers the reverse rate, so the forward reaction temporarily dominates and rebuilds product.

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12. A system starts with concentrations far from equilibrium on the reactant side. The reverse rate will:

decrease steadily until it vanishes.

remain at zero for the entire experiment.

increase as products accumulate until it matches the forward rate.

exceed the forward rate at every moment.

As products form, their concentration rises, increasing the reverse rate until it equals the forward rate.

As products form, their concentration rises, increasing the reverse rate until it equals the forward rate.

As products form, their concentration rises, increasing the reverse rate until it equals the forward rate.

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13. The direction a reversible reaction initially favors depends most directly on:

the size of the container holding the mixture.

the color of the chemicals in the mixture.

the name assigned to the reaction vessel.

the relative amounts of reactants and products present at the start.

Starting composition sets the initial forward and reverse rates, and the faster one determines the net direction.

Starting composition sets the initial forward and reverse rates, and the faster one determines the net direction.

Starting composition sets the initial forward and reverse rates, and the faster one determines the net direction.

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14. A sealed flask contains only H₂ and I₂. The reaction H₂ + I₂ <=> 2 HI is observed to form HI over time. The net direction is forward because:

iodine gas is colored and hydrogen is not.

the reverse reaction requires sunlight to occur.

HI is more stable than any possible compound.

the forward rate initially exceeds the reverse rate.

Starting with no HI, the reverse rate begins at zero, so the forward reaction dominates and HI accumulates.

Starting with no HI, the reverse rate begins at zero, so the forward reaction dominates and HI accumulates.

Starting with no HI, the reverse rate begins at zero, so the forward reaction dominates and HI accumulates.

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15. For a reversible reaction with only reactants present initially, the reaction will:

immediately move backward to remake more reactants.

proceed in the forward direction until equilibrium is reached.

stop after the first product molecule forms.

remain unchanged because no catalyst is present.

With only reactants present, the forward rate exceeds the reverse rate, so net conversion to products occurs until the rates equalize.

With only reactants present, the forward rate exceeds the reverse rate, so net conversion to products occurs until the rates equalize.

With only reactants present, the forward rate exceeds the reverse rate, so net conversion to products occurs until the rates equalize.

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16. When only products are placed in a closed vessel, the initial net change is:

a complete halt of both reactions.

formation of entirely new chemical species.

conversion of products back into reactants.

a steady increase in product concentration.

With only products present, the reverse rate dominates, so the system undergoes net conversion back toward reactants.

With only products present, the reverse rate dominates, so the system undergoes net conversion back toward reactants.

With only products present, the reverse rate dominates, so the system undergoes net conversion back toward reactants.

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17. Which observation best supports the claim that equilibrium can be reached from either direction?

One flask heats up while the other cools down.

Neither flask ever shows any color change.

Both flasks change color at exactly the same speed.

Flasks starting from pure reactants and pure products end with identical mixtures.

Reaching the same final composition from opposite starting points proves equilibrium is a state independent of direction.

Reaching the same final composition from opposite starting points proves equilibrium is a state independent of direction.

Reaching the same final composition from opposite starting points proves equilibrium is a state independent of direction.

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18. For a reversible reaction, the direction that is favored initially is determined by comparing:

the odors of the gases in the vessel.

the molar masses of reactants and products.

the forward and reverse rates under the current conditions.

the alphabetical order of the formulas.

Whichever rate is larger determines the net direction; the system favors the direction with the greater initial rate.

Whichever rate is larger determines the net direction; the system favors the direction with the greater initial rate.

Whichever rate is larger determines the net direction; the system favors the direction with the greater initial rate.

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Direction Means Net Change

Whenever both sides of a reversible reaction are present, both reactions run at the same time: reactants turn into products while products turn back into reactants.

The direction of the reaction names the net change: which of the two conversions is winning. During a net forward reaction, product molecules still break back down, but they are formed faster.

Only the difference between the two rates shows in the concentrations, and a system not at equilibrium always has a net direction, toward equilibrium.

The Decision Rule: Compare the Two Rates

If the forward rate is greater, the net reaction is forward: reactants are consumed faster than they are re-formed, so reactant concentrations fall and product concentrations rise. If the reverse rate is greater, everything runs in mirror image: products are consumed net, reactant concentrations rise and product concentrations fall. If the two rates are equal, there is no net change: that is equilibrium.

Net forward reaction drains reactants and builds products, so the forward rate falls and the reverse rate rises.

Starting Composition Sets the Opening Rates

Each rate is driven by its own side: the forward rate needs reactants, the reverse rate needs products.

Start with reactants only, and the reverse rate is zero, because there is no product to react back: the forward rate, at its maximum, forces a net forward start. Start with products only, and the forward rate is zero instead, so the system turns products back into reactants. Start with a mixture, and neither rate is zero: compare the two rates rather than asking which side holds more. A mixture already at the equilibrium composition shows no net change.

Which Concentrations Rise, and by How Much

In a net forward reaction every reactant concentration falls and every product concentration rises; net reverse flips both.

Changes follow the coefficients, so species do not change by equal amounts. In N₂(g) + 3H₂(g) ⇌ 2NH₃(g), every 0.10 mol L⁻¹ of N₂ consumed takes 0.30 mol L⁻¹ of H₂ with it and produces 0.20 mol L⁻¹ of NH₃.

Read the direction off any one species whose change is known and the equation fixes every other change; changes that refuse to fit the ratios cannot belong to one closed-vessel reaction.

Reading Direction From Particle Pictures

Two snapshots of one sealed box show the direction directly: the species whose count rose was formed net; the species whose count fell was consumed net.

The counts are not free numbers: every snapshot of the same box holds the same total of each element, and changes between snapshots must fit the coefficients. For 2NO₂(g) ⇌ N₂O₄(g), a fall of four NO₂ particles pairs with a rise of two N₂O₄; identical counts mean no net change, so with both sides present the system was at equilibrium.

Forward and Reverse Belong to the Equation as Written

Forward means left to right as written; reverse means right to left.

Rewrite the reaction the other way round and the labels swap, while nothing inside the flask changes. NO₂ dimerizing, 2NO₂(g) ⇌ N₂O₄(g), is net forward as written; the same flask written as N₂O₄(g) ⇌ 2NO₂(g) is undergoing a net reverse reaction. Take the direction from the printed equation.

A Disturbed Equilibrium, Read From Rates

Remove some product from a mixture at equilibrium: the reverse rate drops at once, because there is less product to react back, while the forward rate is untouched at that instant. The forward rate is now the larger, so the system runs net forward until the rates match again at a new equilibrium. Removing reactant flips the story: the forward rate drops and the system runs net reverse. These responses are named as a general principle in Topic 7.9.

Worked Examples

Example 1: direction and concentration changes from two rates. For N₂(g) + 3H₂(g) ⇌ 2NH₃(g), suppose the forward reaction consumes N₂ at 0.30 mol L⁻¹ min⁻¹ while the reverse reaction produces N₂ at 0.10 mol L⁻¹ min⁻¹. The forward rate is larger, so the net reaction is forward, with a net rate of 0.30 − 0.10 = 0.20 mol L⁻¹ min⁻¹. Following the coefficients, [N₂] falls at 0.20, [H₂] at 3 × 0.20 = 0.60, and [NH₃] rises at 2 × 0.20 = 0.40 mol L⁻¹ min⁻¹. With the reverse rate at 0.45 instead, the net direction would be reverse at 0.15 mol L⁻¹ min⁻¹: [NH₃] falling at 0.30, [N₂] rising at 0.15, [H₂] rising at 0.45.

Example 2: direction from particle snapshots. Snapshots of a sealed box holding 2NO₂(g) ⇌ N₂O₄(g) show 12 NO₂ particles and 2 N₂O₄ particles, then 8 NO₂ and 4 N₂O₄. Nitrogen is conserved: 12 + 2 × 2 = 16 atoms, then 8 + 2 × 4 = 16. The changes, NO₂ down by 4 and N₂O₄ up by 2, fit the 2 : 1 coefficients, and since NO₂ is consumed net, the direction is forward as written. Equilibrium for this box is 6 NO₂ and 5 N₂O₄.

Example 3: starting from products only. A flask is charged with NH₃ alone at [NH₃] = 1.20 mol L⁻¹, for the reaction written N₂(g) + 3H₂(g) ⇌ 2NH₃(g). With no N₂ and no H₂ present, the forward rate starts at zero and the reverse rate at its maximum, so the net reaction is reverse. It settles at [NH₃] = 0.80, [N₂] = 0.20 and [H₂] = 0.60 mol L⁻¹: the 0.40 mol L⁻¹ of NH₃ consumed formed 0.20 of N₂ and 0.60 of H₂, in the 2 : 1 : 3 ratio of the equation read backwards. A flask starting from [N₂] = 0.60 and [H₂] = 1.80 mol L⁻¹ reaches the same equilibrium from the forward side.

Example 4: a disturbed equilibrium. A mixture at equilibrium for A ⇌ B holds [A] = 0.30 mol L⁻¹ and [B] = 0.70 mol L⁻¹. Extra B is injected, raising [B] to 1.00 mol L⁻¹ at once. The reverse rate jumps because [B] rose, while the forward rate is unchanged at that instant, so the net reaction runs reverse until the rates meet again. The new equilibrium holds [A] = 0.39 mol L⁻¹ and [B] = 0.91 mol L⁻¹, keeping [B] / [A] = 2.33 with [A] + [B] = 1.30 mol L⁻¹.

Traps That Cost Marks in Topic 7.2

“More product than reactant means the net direction is reverse.” Not necessarily. A flask with [H₂] = [I₂] = 0.50 mol L⁻¹ and [HI] = 1.00 mol L⁻¹ holds more product than either reactant, yet its equilibrium sits at [HI] = 1.56 mol L⁻¹, so it still runs net forward. Direction compares rates, not head counts.

“During a net forward reaction, no reactants are being formed.” The reverse reaction runs whenever any product exists; net forward means only that product formation outpaces breakdown, with reactants re-formed throughout.

“All the concentrations change by the same amount.” Changes follow the coefficients: in the ammonia reaction, a fall of 0.20 mol L⁻¹ in [N₂] comes with 0.60 down in [H₂] and 0.40 up in [NH₃].

“Removing product makes the system react backwards to restore it.” Removing product lowers the reverse rate, so the forward reaction wins and replaces the product: the opposite of the claim.

“The system overshoots equilibrium and swings back and forth.” As the system moves, the winning rate falls and the losing rate rises; when they meet, net change stops: no overshoot, no oscillation.

“This reaction runs in the forward direction, full stop.” Direction belongs to a starting mixture and a written equation, not to the reaction alone: one system can run forward from one start and reverse from another.

Key Relationships

• Net direction follows the larger rate, forward or reverse
• Forward rate greater: reactant concentrations fall, product concentrations rise; reverse rate greater: the mirror image
• Equal rates mean no net change: the system is at equilibrium
• An absent side has a rate of zero, so a one-sided start forces the opening direction
• Concentration changes in a net reaction always follow the coefficients
• Snapshots of a closed system conserve element totals, and count changes obey the coefficients
• Forward and reverse are defined by the equation as written; rewriting swaps the labels
• Comparing a mixture with the equilibrium composition is made quantitative with Q and K in Topic 7.3