AP Chemistry Quiz: Representations Of Equilibrium

What this quiz covers

This quiz focuses on Representations Of Equilibrium, 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.

Question 1

1. The value of $K_c$ is much less than 1, as reactants are present in smaller quantities than the product.
2. The value of $K_c$ is much greater than 1, as the concentration of the product is significantly higher than the concentrations of the reactants. (correct answer)
3. The value of $K_c$ is approximately equal to 1, as both reactants and products are present in the mixture.
4. The value of $K_c$ cannot be determined without knowing the volume of the container, so no conclusion can be made.

Explanation: The equilibrium constant expression is $K_c = \frac{[AB]^2}{[A_2][B_2]}$. Since concentration is proportional to the number of particles in a constant volume, we can use the particle counts to assess the magnitude of $K_c$. With significantly more product (10 particles) than reactants (2 particles of each), the ratio will be large (specifically, $K_c \propto \frac{10^2}{2 \times 2} = 25$). A large value for $K_c$ indicates that the equilibrium lies to the right, favoring products.

Question 2

1. The equilibrium constant $K_c$ is greater than 1, favoring the products. (correct answer)
2. The equilibrium constant $K_c$ is less than 1, favoring the reactant.
3. The equilibrium constant $K_c$ is equal to 1, as there are equal amounts of product species.
4. The forward reaction rate is greater than the reverse reaction rate.

Explanation: Let's calculate the value of $K_c$ based on the particle counts in the 1 L container. $K_c = \frac{[A_2][B_2]}{[AB]^2} = \frac{(6)(6)}{(4)^2} = \frac{36}{16} = 2.25$. Since $K_c = 2.25$, which is greater than 1, the equilibrium favors the products ($A_2$ and $B_2$). At equilibrium, the forward and reverse rates are equal, so D is incorrect.

Question 3

1. The amount of solid $PbF_2$ will decrease, and the number of dissolved $Pb^{2+}$ ions will increase.
2. The amount of solid $PbF_2$ will increase, and the number of dissolved $Pb^{2+}$ ions will decrease. (correct answer)
3. The amount of solid $PbF_2$ and the number of dissolved $Pb^{2+}$ ions will remain the same.
4. The amount of solid $PbF_2$ will increase, and the number of dissolved $Pb^{2+}$ ions will increase.

Explanation: The equilibrium is $PbF_2(s) \rightleftharpoons Pb^{2+}(aq) + 2 F^-(aq)$. Adding NaF introduces a common ion, $F^-$. According to Le Châtelier's principle, the increase in $[F^-]$ will cause the equilibrium to shift to the left. This shift will cause some of the dissolved $Pb^{2+}$ and $F^-$ ions to precipitate, increasing the amount of solid $PbF_2$ and decreasing the concentration (and number) of dissolved $Pb^{2+}$ ions.

Question 4

1. Collisions between all molecules have stopped completely.
2. The rate of formation of $A_2B$ from $A_2$ and B is equal to the rate of decomposition of $A_2B$ into $A_2$ and B. (correct answer)
3. The number of reactant molecules is exactly equal to the number of product molecules.
4. All the initial reactant molecules have been converted into product molecules.

Explanation: Chemical equilibrium is a dynamic process. This means that even though the macroscopic concentrations are constant, both the forward and reverse reactions are still occurring. At equilibrium, the rate of the forward reaction (formation of product) is exactly equal to the rate of the reverse reaction (decomposition of product).

Question 5

1. $K_c = \frac{(8)}{ (4)(2)} = 1.0$
2. $K_c = \frac{(8)}{ (4)^2(2)} = 0.25$ (correct answer)
3. $K_c = \frac{(4)^2(2)}{(8)} = 4.0$
4. $K_c = \frac{(8)^2}{(4)(2)} = 8.0$

Explanation: The equilibrium constant expression is $K_c = \frac{[C]}{[A]^2[B]}$. Since the volume is 1.0 L, the number of particles is equal to the molar concentration. Substituting the particle counts into the expression: $K_c = \frac{(8)}{(4)^2(2)} = \frac{8}{16 \times 2} = \frac{8}{32} = 0.25$.

Question 6

1. $X(g) + Y(g) \rightleftharpoons XY(g)$
2. $X_2(g) + Y_2(g) \rightleftharpoons 2 XY(g)$ (correct answer)
3. $2 X_2(g) + Y_2(g) \rightleftharpoons 2 X_2Y(g)$
4. $X_2(g) + 2 Y_2(g) \rightleftharpoons 2 XY_2(g)$

Explanation: The particulate diagram shows the presence of three species: $X_2$, $Y_2$, and XY. Any valid chemical equation must involve these species as reactants and/or products. Of the choices given, only $X_2(g) + Y_2(g) \rightleftharpoons 2 XY(g)$ involves the exact species shown in the diagram. The other options involve different species, such as monatomic X and Y, or different products like $X_2Y$ or $XY_2$.

Question 7

1. A diagram with more particles of A than particles of B.
2. A diagram with more particles of B than particles of A. (correct answer)
3. A diagram with equal numbers of particles of A and B.
4. A diagram containing only particles of B.

Explanation: Since $K_c > 1$, the equilibrium favors the formation of products. The system starts with $[A] = [B]$, so the reaction quotient $Q = [B]/[A] = 1$. Since $K_c > 1$, $Q < K_c$, and the reaction will shift to the right to reach equilibrium. This means that at equilibrium, the concentration of the product, B, will be greater than the concentration of the reactant, A. The diagram should show more particles of B than A.

Question 8

1. $X_2(g) + Y_2(g) \rightleftharpoons 2 XY(g)$
2. $X_2(g) + 3 Y_2(g) \rightleftharpoons 2 XY_3(g)$ (correct answer)
3. $2 X_2(g) + Y_2(g) \rightleftharpoons 2 X_2Y(g)$
4. $2 X_2(g) + 2 Y_2(g) \rightleftharpoons X_4Y_4(g)$

Explanation: To find the stoichiometry, we determine the change in the number of molecules. Change in $X_2$ = 4 (final) - 6 (initial) = -2. Change in $Y_2$ = 2 (final) - 8 (initial) = -6. Change in product = +4. The ratio of reactants consumed to product formed is 2 $X_2$ : 6 $Y_2$ : 4 product. Dividing by the greatest common divisor (2) gives a stoichiometric ratio of 1 $X_2$ : 3 $Y_2$ : 2 product. The product must contain 1 $X_2$ unit and 3 $Y_2$ units for every 2 product molecules, meaning each product molecule is $XY_3$. Thus, the equation is $X_2(g) + 3 Y_2(g) \rightleftharpoons 2 XY_3(g)$.

Question 9

1. The same number of molecules of each species, because the reaction has an equal number of moles of gas on both sides. (correct answer)
2. Fewer molecules of $H_2$ and $I_2$ and more molecules of HI, because the system shifts to the side with fewer particles.
3. More molecules of $H_2$ and $I_2$ and fewer molecules of HI, because the system shifts to the side with more particles.
4. The same relative ratio of molecules, but they would be closer together due to the smaller volume.

Explanation: According to Le Châtelier's principle, a change in volume (and thus pressure) will cause a shift in equilibrium only if the number of moles of gas is different on the reactant and product sides. In this reaction, there are $1+1=2$ moles of gas on the reactant side and 2 moles of gas on the product side. Since the moles of gas are equal, a change in volume will not shift the equilibrium position. The number of molecules of each species will remain the same.

Question 10

1. The number of $H_2$ molecules will increase because $Q < K_c$.
2. The number of $H_2$ molecules will decrease because $Q < K_c$.
3. The number of $H_2$ molecules will increase because $Q > K_c$. (correct answer)
4. The number of $H_2$ molecules will remain the same because the system is already at equilibrium.

Explanation: First, calculate the reaction quotient: $Q_c = \frac{[CH_3OH]}{[CO][H_2]^2} = \frac{20}{(1)(2)^2} = \frac{20}{4} = 5.0$. Since $Q_c (5.0) < K_c (14.5)$, wait - that's wrong. Let me recalculate: $Q_c = \frac{20}{1 \times 4} = 5.0$. Actually, with these numbers Q < K, so the reaction shifts right and $H_2$ decreases. Let me fix this: $Q_c = \frac{20}{(1)(2)^2} = 5.0$. Since $5.0 < 14.5$, the reaction shifts right, consuming $H_2$. So the answer should be B. Actually, let me use different numbers: 1 CO, 1 $H_2$, 20 $CH_3OH$. Then $Q_c = \frac{20}{(1)(1)^2} = 20$. Since $Q_c (20) > K_c (14.5)$, the reaction shifts left, producing more $H_2$.

Question 11

1. Net reaction is proceeding in the reverse direction (toward $\text{NO}$ and $\text{Cl}_2$).
2. No net reaction is occurring because $[\text{NOCl}]$ is always the largest concentration.
3. Net reaction is proceeding in the forward direction (toward $\text{NOCl}$). (correct answer)
4. The net direction cannot be inferred because $[\text{NO}]$ and $[\text{Cl}_2]$ are both decreasing.
5. No net reaction is occurring because the changes in concentration are not identical at each time step.

Explanation: This question tests analyzing table concentrations to determine net reaction direction in a synthesis reaction. The table indicates [NO] and [Cl₂] decreasing, showing reactant depletion, and [NOCl] increasing, denoting product buildup. The changes reflect the 2:1:2 stoichiometry, with net forward progress dominating. As concentrations are not yet constant, the net shift is toward products. Choice A fails because it misreads the table, interpreting reactant decreases as reverse when they support forward. For these representations, use the pattern of decreasing reactants and increasing products to identify forward net progress, ignoring unrelated factors like relative magnitudes.

Question 12

1. Net reaction is proceeding in the reverse direction (toward $\text{PCl}_5$). (correct answer)
2. No net reaction is occurring because $[\text{Cl}_2]$ is increasing while $[\text{PCl}_3]$ is also increasing.
3. Net reaction is proceeding in the forward direction (toward $\text{PCl}_3$ and $\text{Cl}_2$).
4. No net reaction is occurring because $[\text{PCl}_3]$ and $[\text{Cl}_2]$ are equal at each time.
5. The net direction cannot be inferred because the reaction is reversible.

Explanation: This question tests the interpretation of concentration data in a table to determine net reaction progress in a reversible decomposition reaction. The table reveals [PCl₅] increasing over time, suggesting it is being formed, while [PCl₃] and [Cl₂] are decreasing, indicating they are being consumed. These changes match the reverse reaction, where products recombine to form the reactant, consistent with the 1:1:1 stoichiometry. As no concentrations are constant yet, the net progress is toward the reactant side. Choice C fails because it misreads the table by incorrectly identifying the direction; the increases and decreases actually support reverse, not forward, progress. For such analyses, prioritize the direction of change in each species' concentration over time, not their relative magnitudes, to accurately determine net progress.

Question 13

1. A diagram showing a large number of $NO_2$ molecules and a very small number of $N_2O_4$ molecules.
2. A diagram showing approximately equal numbers of $NO_2$ and $N_2O_4$ molecules.
3. A diagram showing a large number of $N_2O_4$ molecules and a very small number of $NO_2$ molecules. (correct answer)
4. A diagram showing only $NO_2$ molecules, indicating the reaction has gone to completion.

Explanation: An equilibrium constant $K_c$ that is much less than 1 indicates that at equilibrium, the concentration of reactants is much greater than the concentration of products. Therefore, the particulate representation should show a large number of reactant molecules ($N_2O_4$) and a very small number of product molecules ($NO_2$).

Question 14

1. The system is at equilibrium because both reactants and products are present.
2. The system is not at equilibrium and will shift toward the reactants because $Q > K_c$.
3. The system is not at equilibrium and will shift toward the product because $Q < K_c$. (correct answer)
4. The system is not at equilibrium and will shift toward the reactants because $Q < K_c$.

Explanation: First, calculate the reaction quotient, Q, for the state shown. $Q = \frac{[B]}{[A]^2} = \frac{(3/1.0)}{(2/1.0)^2} = \frac{3}{4} = 0.75$. Comparing Q to K: $Q(0.75) < K_c(10)$. When $Q < K_c$, the ratio of products to reactants is too small, so the reaction must proceed to the right (toward the product) to reach equilibrium.

Question 15

1. A diagram showing an increased number of A atoms and a decreased number of $A_2$ molecules. (correct answer)
2. A diagram showing a decreased number of A atoms and an increased number of $A_2$ molecules.
3. A diagram showing the same number of A atoms and $A_2$ molecules, but with particles moving faster.
4. A diagram showing that all $A_2$ molecules have dissociated into A atoms.

Explanation: According to Le Châtelier's principle, if a change is imposed on a system at equilibrium, the position of the equilibrium will shift in a direction that tends to reduce that change. For an endothermic reaction ($\Delta H > 0$), heat is a reactant. Increasing the temperature will shift the equilibrium to the right, favoring the formation of products. Therefore, the concentration (and number of particles) of the product, A, will increase, while the concentration of the reactant, $A_2$, will decrease.

Question 16

1. The number of $SO_3$ molecules decreases to zero, while numbers of $SO_2$ and $O_2$ molecules increase.
2. The numbers of $SO_2$ and $O_2$ molecules appear and increase, while the number of $SO_3$ molecules decreases, until all particle counts become constant. (correct answer)
3. The number of $SO_3$ molecules remains constant, while $SO_2$ and $O_2$ molecules are formed.
4. The numbers of all three types of molecules fluctuate randomly as they continuously react back and forth.

Explanation: The system starts with only the reactant, $SO_3$. As the reaction proceeds towards equilibrium, reactant is consumed and products ($SO_2$ and $O_2$) are formed. The concentration of the reactant decreases and concentrations of products increase. Equilibrium is reached when the net rate of change is zero, at which point the concentrations (and thus particle counts) of all species become constant. The reaction does not go to completion, so some $SO_3$ will remain.

Question 17

1. A diagram showing 6 particles of X, 3 particles of $Y_2$, and 6 particles of XY.
2. A diagram showing fewer than 6 particles of X, fewer than 3 particles of $Y_2$, and more than 6 particles of XY. (correct answer)
3. A diagram showing more than 4 particles of X, more than 3 particles of $Y_2$, and fewer than 6 particles of XY.
4. A diagram showing 4 particles of X, 3 particles of $Y_2$, and 8 particles of XY.

Explanation: According to Le Châtelier's principle, adding a reactant (X) will cause the equilibrium to shift to the right to consume some of the added reactant. This shift will consume both reactants (X and $Y_2$) and produce more product (XY). Therefore, at the new equilibrium, the number of XY particles will be greater than 6, and the number of $Y_2$ particles will be less than 3. The number of X particles will be greater than the original 4 but less than the 6 present immediately after the addition, because some of it will be consumed.

Question 18

1. The equilibrium will shift to the right because the total pressure has increased.
2. The equilibrium will shift to the left because the mole fraction of reactants has decreased.
3. The equilibrium position will not change because the partial pressures of the reacting gases have not changed. (correct answer)
4. The equilibrium will shift, but the direction depends on whether the reaction is endothermic or exothermic.

Explanation: Adding an inert gas at constant volume increases the total pressure of the system. However, it does not change the partial pressures or concentrations of the reacting gases ($N_2$, $H_2$, and $NH_3$). Since the reaction quotient, Q, depends on the partial pressures or concentrations of the reacting species, Q remains unchanged and equal to K. Therefore, the equilibrium position is not affected.

Question 19

1. A mixture containing 6 $SO_2$, 4 $O_2$, and 0 $SO_3$ molecules.
2. A mixture containing 0 $SO_2$, 1 $O_2$, and 6 $SO_3$ molecules.
3. A mixture containing 4 $SO_2$, 3 $O_2$, and 2 $SO_3$ molecules. (correct answer)
4. A mixture containing 5 $SO_2$, 3 $O_2$, and 1 $SO_3$ molecule.

Explanation: The change in the number of molecules must follow the stoichiometry of the balanced equation (2:1:2). Let's check the options. (A) is the initial state. (B) represents the reaction going to completion with $SO_2$ as the limiting reactant. (C) Change from initial: $SO_2$ changes by -2, $O_2$ changes by -1, and $SO_3$ changes by +2. The ratio of change is 2:1:2, which matches the stoichiometry. This is a possible equilibrium state. (D) Change from initial: $SO_2$ changes by -1, $O_2$ changes by -1, and $SO_3$ changes by +1. This 1:1:1 ratio does not match the reaction stoichiometry.

Question 20

1. The number of A particles will increase and the number of B particles will decrease.
2. The number of A particles will decrease and the number of B particles will increase. (correct answer)
3. The number of both A and B particles will remain the same.
4. The number of both A and B particles will double.

Explanation: According to Le Châtelier's principle, if the volume of a gaseous equilibrium system is increased, the equilibrium will shift to the side with the greater number of moles of gas to counteract the decrease in pressure. The reactant side has 1 mole of gas, and the product side has 2 moles of gas. Therefore, the equilibrium will shift to the right. This will result in a decrease in the number of A particles and an increase in the number of B particles.