AP Chemistry Quiz: Direction Of Reversible Reactions
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Direction Of Reversible ReactionsQuestion 1 of 14

A reversible reaction occurs in a sealed container at constant temperature:

2NO(g)+Cl2(g)⇌2NOCl(g)\mathrm{2NO(g) + Cl_2(g) \rightleftharpoons 2NOCl(g)}

A concentration-versus-time graph shows that at the indicated time (before equilibrium), [NOCl][\mathrm{NOCl}] is decreasing while [NO][\mathrm{NO}] and [Cl2][\mathrm{Cl_2}] are increasing. In which direction will the net reaction proceed to reach equilibrium?

Toward products
Toward reactants
No net change
Toward products because the reactant side has more terms
No net change because the graph shows all species present
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AP Chemistry Quiz

AP Chemistry Quiz: Direction Of Reversible Reactions

Practice Direction Of Reversible Reactions in AP Chemistry with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Direction Of Reversible Reactions, giving you a quick way to practice the rules, question types, and explanations that matter most for AP Chemistry.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

A reversible reaction occurs in a sealed container at constant temperature:

2NO(g)+Cl2(g)⇌2NOCl(g)\mathrm{2NO(g) + Cl_2(g) \rightleftharpoons 2NOCl(g)}

A concentration-versus-time graph shows that at the indicated time (before equilibrium), [NOCl][\mathrm{NOCl}] is decreasing while [NO][\mathrm{NO}] and [Cl2][\mathrm{Cl_2}] are increasing. In which direction will the net reaction proceed to reach equilibrium?

  1. Toward products
  2. Toward reactants (correct answer)
  3. No net change
  4. Toward products because the reactant side has more terms
  5. No net change because the graph shows all species present

Explanation: This question tests the skill of determining the direction of reversible reactions. The snapshot from the concentration-versus-time graph shows [NOCl] decreasing while [NO] and [Cl₂] are increasing, indicating that products are currently favored relative to equilibrium. This trend suggests the reverse reaction is faster, consuming NOCl to produce more NO and Cl₂. Therefore, the net reaction proceeds toward reactants to reduce the overrepresentation of products and approach equilibrium. A tempting distractor is D, which is incorrect due to the misconception that 'more terms on one side determine direction,' but direction depends on current concentration trends, not the number of species in the equation. To predict the direction in such scenarios, determine which side is currently overrepresented, then predict the direction that restores equilibrium.

Question 2

A sealed container at constant temperature contains the reversible reaction

CO(g)+H2O(g)⇌CO2(g)+H2(g)\mathrm{CO(g) + H_2O(g) \rightleftharpoons CO_2(g) + H_2(g)}

A concentration-versus-time graph shows that at the indicated time (before equilibrium), [CO2][\mathrm{CO_2}] and [H2][\mathrm{H_2}] are increasing, while [CO][\mathrm{CO}] and [H2O][\mathrm{H_2O}] are decreasing. In which direction will the net reaction proceed to reach equilibrium?

  1. Toward reactants
  2. No net change
  3. Toward products (correct answer)
  4. No net change because the lines are not equal in height
  5. Toward reactants because there are two products listed

Explanation: This question tests the skill of determining the direction of reversible reactions. The snapshot from the concentration-versus-time graph shows [CO₂] and [H₂] increasing while [CO] and [H₂O] are decreasing, indicating that reactants are currently favored relative to equilibrium. This trend suggests the forward reaction is faster, consuming CO and H₂O to produce more CO₂ and H₂. Therefore, the net reaction proceeds toward products to reduce the overrepresentation of reactants and approach equilibrium. A tempting distractor is E, which is incorrect due to the misconception that 'multiple products imply shift to reactants,' but direction depends on current concentration trends, not the number of species in the equation. To predict the direction in such scenarios, determine which side is currently overrepresented, then predict the direction that restores equilibrium.

Question 3

A reversible reaction occurs in a sealed container at constant temperature:

PCl5(g)⇌PCl3(g)+Cl2(g)\mathrm{PCl_5(g) \rightleftharpoons PCl_3(g) + Cl_2(g)}

A concentration-versus-time plot shows that at the displayed time (before equilibrium), [PCl5][\mathrm{PCl_5}] is increasing while both [PCl3][\mathrm{PCl_3}] and [Cl2][\mathrm{Cl_2}] are decreasing. In which direction will the net reaction proceed to reach equilibrium?

  1. Toward products
  2. Toward reactants (correct answer)
  3. No net change
  4. Toward products because there are more gas particles on the product side
  5. No net change because the concentrations are changing at the same time

Explanation: This question tests the skill of determining the direction of reversible reactions. The snapshot from the concentration-versus-time plot shows [PCl₅] increasing while [PCl₃] and [Cl₂] are decreasing, indicating that products are currently favored relative to equilibrium. This trend suggests the reverse reaction is faster, consuming PCl₃ and Cl₂ to produce more PCl₅. Therefore, the net reaction proceeds toward reactants to reduce the overrepresentation of products and approach equilibrium. A tempting distractor is D, which is incorrect due to the misconception that 'more gas particles determine direction,' but direction is based on current concentration changes, not the number of particles in the equation. To predict the direction in such scenarios, determine which side is currently overrepresented, then predict the direction that restores equilibrium.

Question 4

A sealed container at constant temperature contains the reversible reaction

Fe3+(aq)+SCN−(aq)⇌FeSCN2+(aq)\mathrm{Fe^{3+}(aq) + SCN^-(aq) \rightleftharpoons FeSCN^{2+}(aq)}

At a moment before equilibrium is reached, the concentrations are:

SpeciesCurrent concentration (M)
Fe3+\mathrm{Fe^{3+}}0.40
SCN−\mathrm{SCN^-}0.40
FeSCN2+\mathrm{FeSCN^{2+}}0.001

In which direction will the net reaction proceed to reach equilibrium?

  1. No net change
  2. Toward reactants
  3. Toward products (correct answer)
  4. No net change because [Fe3+][\mathrm{Fe^{3+}}] equals [SCN−][\mathrm{SCN^-}]
  5. Toward reactants because the product has a larger charge magnitude

Explanation: This question tests the skill of determining the direction of reversible reactions. The snapshot shows high concentrations of Fe³⁺ (0.40 M) and SCN⁻ (0.40 M) compared to a very low concentration of FeSCN²⁺ (0.001 M), indicating that reactants are currently favored relative to equilibrium. With more reactants present, the forward reaction rate is higher than the reverse, driving the system to produce more FeSCN²⁺. Therefore, the net reaction proceeds toward products to restore balance by decreasing the reactants and increasing the product. A tempting distractor is D, which is incorrect due to the misconception of 'equal concentrations at equilibrium,' but equilibrium occurs when rates are equal, not necessarily when reactant concentrations match. To predict the direction in such scenarios, determine which side is currently overrepresented, then predict the direction that restores equilibrium.

Question 5

In a closed vessel at constant temperature, the reversible reaction occurs:

2SO2(g)+O2(g)⇌2SO3(g)\mathrm{2SO_2(g) + O_2(g) \rightleftharpoons 2SO_3(g)}

At a moment before equilibrium is reached, the following concentrations are measured:

SpeciesCurrent concentration (M)
SO2\mathrm{SO_2}0.05
O2\mathrm{O_2}0.10
SO3\mathrm{SO_3}0.90

In which direction will the net reaction proceed to reach equilibrium?

  1. No net change
  2. Toward products
  3. Toward reactants (correct answer)
  4. Toward products because SO3\mathrm{SO_3} has the largest concentration
  5. No net change because the reaction is reversible

Explanation: This question tests the skill of determining the direction of reversible reactions. The snapshot shows a high concentration of SO₃ (0.90 M) compared to low concentrations of SO₂ (0.05 M) and O₂ (0.10 M), indicating that products are currently favored relative to equilibrium. With more product present, the reverse reaction rate is higher than the forward, driving the system to produce more SO₂ and O₂. Therefore, the net reaction proceeds toward reactants to restore balance by decreasing the product and increasing the reactants. A tempting distractor is D, which is incorrect due to the misconception that 'highest concentration determines direction,' but direction is based on overall imbalance relative to equilibrium, not just the largest individual concentration. To predict the direction in such scenarios, determine which side is currently overrepresented, then predict the direction that restores equilibrium.

Question 6

A closed container at constant temperature contains the reversible reaction

2NH3(g)⇌N2(g)+3H2(g)\mathrm{2NH_3(g) \rightleftharpoons N_2(g) + 3H_2(g)}

At a moment before equilibrium is reached, the measured concentrations are:

SpeciesCurrent concentration (M)
NH3\mathrm{NH_3}0.02
N2\mathrm{N_2}0.60
H2\mathrm{H_2}0.90

In which direction will the net reaction proceed to reach equilibrium?

  1. Toward products
  2. No net change
  3. Toward reactants (correct answer)
  4. Toward products because there are more moles of gas on the product side
  5. No net change because the reaction is reversible

Explanation: This question tests the skill of determining the direction of reversible reactions. The snapshot shows a low concentration of NH₃ (0.02 M) compared to high concentrations of N₂ (0.60 M) and H₂ (0.90 M), indicating that products are currently favored relative to equilibrium. With more products present, the reverse reaction rate is higher than the forward, driving the system to produce more NH₃. Therefore, the net reaction proceeds toward reactants to restore balance by decreasing the products and increasing the reactant. A tempting distractor is E, which is incorrect due to the misconception that 'reversibility implies no change,' but reversible reactions can shift direction based on current concentrations to reach equilibrium. To predict the direction in such scenarios, determine which side is currently overrepresented, then predict the direction that restores equilibrium.

Question 7

A closed container at constant temperature contains the reversible reaction

CH3COOH(aq)⇌H+(aq)+CH3COO−(aq)\mathrm{CH_3COOH(aq) \rightleftharpoons H^+(aq) + CH_3COO^-(aq)}

At a moment before equilibrium is reached, the following concentrations are measured:

SpeciesCurrent concentration (M)
CH3COOH\mathrm{CH_3COOH}0.010
H+\mathrm{H^+}0.90
CH3COO−\mathrm{CH_3COO^-}0.90

In which direction will the net reaction proceed to reach equilibrium?

  1. No net change because equilibrium requires all concentrations to be the same
  2. Toward reactants (correct answer)
  3. No net change
  4. Toward products
  5. Toward products because there are two products in the balanced equation

Explanation: This question tests the skill of determining the direction of reversible reactions. The snapshot shows a low concentration of CH₃COOH (0.010 M) compared to high concentrations of H⁺ (0.90 M) and CH₃COO⁻ (0.90 M), indicating that products are currently favored relative to equilibrium. With more products present, the reverse reaction rate is higher than the forward, driving the system to produce more CH₃COOH. Therefore, the net reaction proceeds toward reactants to restore balance by decreasing the products and increasing the reactant. A tempting distractor is E, which is incorrect due to the misconception of 'equal concentrations at equilibrium,' but equilibrium is achieved when forward and reverse rates are equal, not when all concentrations are identical. To predict the direction in such scenarios, determine which side is currently overrepresented, then predict the direction that restores equilibrium.

Question 8

In a closed vessel at constant temperature, the reversible reaction occurs:

PCl5(g)⇌PCl3(g)+Cl2(g)\text{PCl}_5(g) \rightleftharpoons \text{PCl}_3(g) + \text{Cl}_2(g).

At a particular moment (not at equilibrium), the concentrations are [PCl5]=0.02 M[\text{PCl}_5]=0.02\,\text{M}, [PCl3]=0.50 M[\text{PCl}_3]=0.50\,\text{M}, and [Cl2]=0.50 M[\text{Cl}_2]=0.50\,\text{M}. In which direction will the net reaction proceed to reach equilibrium?

  1. Toward products
  2. Toward reactants (correct answer)
  3. No net change (the system is already at equilibrium)
  4. Toward products because there are more moles on the right
  5. Toward reactants because equilibrium requires equal concentrations

Explanation: This question tests understanding of direction of reversible reactions. The system has [PCl₅] = 0.02 M, [PCl₃] = 0.50 M, and [Cl₂] = 0.50 M, showing products are much more concentrated than reactants. The reaction quotient Q = ([PCl₃][Cl₂])/[PCl₅] = (0.50×0.50)/0.02 = 12.5 is large, indicating an excess of products relative to equilibrium. To establish equilibrium, the net reaction must proceed in reverse (toward reactants) to decrease Q by combining PCl₃ and Cl₂ to form PCl₅. Choice E incorrectly assumes equilibrium requires equal concentrations, when equilibrium actually occurs when Q equals K regardless of individual concentrations. To predict reaction direction, calculate Q—when Q exceeds equilibrium requirements, the reaction shifts toward reactants to decrease Q.

Question 9

A sealed flask at constant temperature contains the reversible reaction

N2O4(g)⇌2NO2(g)\mathrm{N_2O_4(g) \rightleftharpoons 2NO_2(g)}

At a particular moment before equilibrium is reached, the measured concentrations are shown:

SpeciesCurrent concentration (M)
N2O4\mathrm{N_2O_4}0.80
NO2\mathrm{NO_2}0.10

Based on this snapshot, in which direction will the net reaction proceed to reach equilibrium?

  1. Toward products (correct answer)
  2. Toward reactants
  3. No net change
  4. Toward products until N2O4\mathrm{N_2O_4} is completely consumed
  5. No net change because equilibrium requires equal concentrations of reactant and product

Explanation: This question tests the skill of determining the direction of reversible reactions. The snapshot shows a high concentration of N₂O₄ (0.80 M) compared to a low concentration of NO₂ (0.10 M), indicating that reactants are currently favored relative to equilibrium. With more reactant present, the forward reaction rate is higher than the reverse, driving the system to produce more NO₂. Therefore, the net reaction proceeds toward products to restore balance by decreasing the reactant and increasing the product. A tempting distractor is E, which is incorrect due to the misconception of 'equal concentrations at equilibrium,' but equilibrium occurs when forward and reverse rates are equal, not when concentrations are equal. To predict the direction in such scenarios, determine which side is currently overrepresented, then predict the direction that restores equilibrium.

Question 10

A reaction mixture in a closed container undergoes the reversible reaction

H2(g)+I2(g)⇌2HI(g)\mathrm{H_2(g) + I_2(g) \rightleftharpoons 2HI(g)}

A concentration-versus-time plot (not yet at equilibrium) shows that at the displayed time, [HI][\mathrm{HI}] is decreasing while [H2][\mathrm{H_2}] and [I2][\mathrm{I_2}] are increasing. In which direction will the net reaction proceed to reach equilibrium?

  1. Toward products because there are 2 moles of product in the balanced equation
  2. Toward products
  3. No net change because changing concentrations imply the reaction has stopped
  4. Toward reactants (correct answer)
  5. No net change

Explanation: This question tests the skill of determining the direction of reversible reactions. The snapshot from the concentration-versus-time plot shows [HI] decreasing while [H₂] and [I₂] are increasing, indicating that products are currently favored relative to equilibrium. This trend suggests the reverse reaction is faster, consuming HI to produce more H₂ and I₂. Therefore, the net reaction proceeds toward reactants to reduce the overrepresentation of products and approach equilibrium. A tempting distractor is D, which is incorrect due to the misconception that 'stoichiometric coefficients determine direction,' but direction depends on current concentrations relative to equilibrium, not mole ratios in the equation. To predict the direction in such scenarios, determine which side is currently overrepresented, then predict the direction that restores equilibrium.

Question 11

In a closed flask at constant temperature, the reversible reaction is established:

H2(g)+Br2(g)⇌2HBr(g)\mathrm{H_2(g) + Br_2(g) \rightleftharpoons 2HBr(g)}

At a moment before equilibrium is reached, the concentrations are:

SpeciesCurrent concentration (M)
H2\mathrm{H_2}0.60
Br2\mathrm{Br_2}0.60
HBr\mathrm{HBr}0.02

In which direction will the net reaction proceed to reach equilibrium?

  1. Toward reactants because the balanced equation shows 2 moles on the product side
  2. Toward products (correct answer)
  3. No net change because [H2][\mathrm{H_2}] equals [Br2][\mathrm{Br_2}]
  4. No net change
  5. Toward reactants

Explanation: This question tests the skill of determining the direction of reversible reactions. The snapshot shows high concentrations of H₂ (0.60 M) and Br₂ (0.60 M) compared to a very low concentration of HBr (0.02 M), indicating that reactants are currently favored relative to equilibrium. With more reactants present, the forward reaction rate is higher than the reverse, driving the system to produce more HBr. Therefore, the net reaction proceeds toward products to restore balance by decreasing the reactants and increasing the product. A tempting distractor is D, which is incorrect due to the misconception of 'equal concentrations at equilibrium,' but equilibrium occurs when rates are equal, regardless of whether specific species concentrations match. To predict the direction in such scenarios, determine which side is currently overrepresented, then predict the direction that restores equilibrium.

Question 12

In a closed flask at constant temperature, the reversible reaction below occurs:

H2(g)+I2(g)⇌2HI(g)\text{H}_2(g) + \text{I}_2(g) \rightleftharpoons 2\text{HI}(g).

At a particular time (not yet at equilibrium), the concentrations are [H2]=0.20 M[\text{H}_2]=0.20\,\text{M}, [I2]=0.20 M[\text{I}_2]=0.20\,\text{M}, and [HI]=1.60 M[\text{HI}]=1.60\,\text{M}. In which direction will the net reaction proceed to reach equilibrium?

  1. Toward reactants (correct answer)
  2. Toward products
  3. No net change (the system is already at equilibrium)
  4. Toward products because there are two moles of product
  5. Toward reactants because equilibrium requires equal concentrations

Explanation: This question tests understanding of direction of reversible reactions. The system has [H₂] = 0.20 M, [I₂] = 0.20 M, and [HI] = 1.60 M, indicating products are currently much more concentrated than reactants. The reaction quotient Q = [HI]²/([H₂][I₂]) = (1.60)²/(0.20×0.20) = 64 is very large, suggesting the system has too much product relative to equilibrium. To reach equilibrium, the net reaction must proceed in reverse (toward reactants) to decrease Q by converting some HI back into H₂ and I₂. Choice E incorrectly assumes equilibrium requires equal concentrations, but equilibrium depends on Q equaling K, not on concentration equality. To determine reaction direction, calculate Q and recognize that when Q is too large, the reaction shifts toward reactants to decrease Q.

Question 13

A sealed container at constant temperature contains the reversible reaction

Fe3+(aq)+SCN−(aq)⇌FeSCN2+(aq)\text{Fe}^{3+}(aq) + \text{SCN}^-(aq) \rightleftharpoons \text{FeSCN}^{2+}(aq).

At a particular instant (before equilibrium), the concentrations are [Fe3+]=0.0010 M[\text{Fe}^{3+}]=0.0010\,\text{M}, [SCN−]=0.0010 M[\text{SCN}^-]=0.0010\,\text{M}, and [FeSCN2+]=0.80 M[\text{FeSCN}^{2+}]=0.80\,\text{M}. In which direction will the net reaction proceed to reach equilibrium?​

  1. No net change (the system is already at equilibrium)
  2. Toward products
  3. Toward reactants (correct answer)
  4. Toward products until all reactants are consumed
  5. Toward reactants because equilibrium requires equal concentrations

Explanation: This question tests understanding of direction of reversible reactions. The system has [Fe³⁺] = 0.0010 M, [SCN⁻] = 0.0010 M, and [FeSCN²⁺] = 0.80 M, indicating the product complex is much more concentrated than the reactant ions. The reaction quotient Q = [FeSCN²⁺]/([Fe³⁺][SCN⁻]) = 0.80/(0.0010×0.0010) = 8.0×10⁵ is extremely large, suggesting a massive excess of products relative to equilibrium. To reach equilibrium, the net reaction must proceed in reverse (toward reactants) to decrease Q by dissociating FeSCN²⁺ back into Fe³⁺ and SCN⁻. Choice E incorrectly assumes equilibrium requires equal concentrations, when equilibrium actually occurs at a specific Q value. To determine reaction direction, calculate Q and recognize that when Q greatly exceeds typical K values, the reaction shifts toward reactants.

Question 14

A sealed container at constant temperature contains the reversible reaction

2SO2(g)+O2(g)⇌2SO3(g)2\text{SO}_2(g) + \text{O}_2(g) \rightleftharpoons 2\text{SO}_3(g).

At one instant before equilibrium is reached, the concentrations are [SO2]=0.90 M[\text{SO}_2]=0.90\,\text{M}, [O2]=0.90 M[\text{O}_2]=0.90\,\text{M}, and [SO3]=0.05 M[\text{SO}_3]=0.05\,\text{M}. In which direction will the net reaction proceed to reach equilibrium?​

  1. Toward reactants because O2\text{O}_2 has a coefficient of 1
  2. Toward products (correct answer)
  3. No net change (the system is already at equilibrium)
  4. Toward reactants
  5. Toward products until all reactants are consumed

Explanation: This question tests understanding of direction of reversible reactions. The system has [SO₂] = 0.90 M, [O₂] = 0.90 M, and [SO₃] = 0.05 M, indicating reactants are currently much more concentrated than products. The reaction quotient Q = [SO₃]²/([SO₂]²[O₂]) = (0.05)²/((0.90)²×0.90) = 0.0034 is very small, suggesting insufficient product formation relative to equilibrium. To reach equilibrium, the net reaction must proceed forward (toward products) to increase Q by converting SO₂ and O₂ into SO₃. Choice A incorrectly focuses on the coefficient of O₂ rather than the concentration imbalance. To determine reaction direction, calculate Q and recognize that when Q is too small, the reaction shifts toward products to increase Q until it equals K.