Representations of Equilibrium - AP Chemistry

1. A particulate diagram of an equilibrium mixture for the reaction $2X(g) + Y_2(g) \rightleftharpoons 2XY(g)$ shows 4 particles of X, 3 particles of $Y_2$, and 6 particles of XY. If two particles of X are added to the container at constant volume and temperature, which diagram represents the new equilibrium?
   Answer: A diagram showing fewer than 6 particles of X, fewer than 3 particles of $Y_2$, and more than 6 particles of XY. 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.
2. A closed system contains the reversible gas-phase reaction $2\text{NO}(g)+\text{Cl}_2(g) \rightleftharpoons 2\text{NOCl}(g)$. A student measures concentrations early in the run, before any concentrations become constant. Based only on the table, what is the direction of the net reaction progress during the interval shown?
   Answer: Net reaction is proceeding in the forward direction (toward $\text{NOCl}$). 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.
3. The dissociation of dinitrogen tetroxide is represented by the equation $N_2O_4(g) \rightleftharpoons 2 NO_2(g)$. The equilibrium constant for the reaction is very small ($K_c \ll 1$). Which particulate diagram best represents a mixture of the two gases at equilibrium in a closed vessel?
   Answer: A diagram showing a large number of $N_2O_4$ molecules and a very small number of $NO_2$ molecules. 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 ($NO_2$).
4. A particulate diagram shows an equilibrium mixture for the reaction $A(g) \rightleftharpoons 2B(g)$ in a container of volume V. It contains 3 particles of A and 6 particles of B. If the volume of the container is increased to 2V at constant temperature, which statement describes the new equilibrium?
   Answer: The number of A particles will decrease and the number of B particles will increase. 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.
5. Based on the particulate representations of the initial and equilibrium states, what is the balanced chemical equation for the reaction?
   Answer: $X_2(g) + 3 Y_2(g) \rightleftharpoons 2 XY_3(g)$ 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 $X_2(g) + 3 Y_2(g) \rightleftharpoons 2 XY_3(g)$.
6. Which statement accurately describes the system represented in the diagram?
   Answer: The system is not at equilibrium and will shift toward the product because $Q < K_c$. 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.
7. A sealed tube contains the reversible reaction $\mathrm{PCl_5(g) \rightleftharpoons PCl_3(g) + Cl_2(g)}$. Concentrations are recorded at early times before equilibrium is reached. What is the direction of net reaction progress during the interval shown?
   Answer: Net reaction proceeds in the forward direction (toward $\mathrm{PCl_3}$ and $\mathrm{Cl_2}$). This question tests the skill of determining reaction direction from concentration trends in decomposition equilibria. The data shows [PCl₅] decreasing while [PCl₃] and [Cl₂] are increasing, indicating that PCl₅ is decomposing into PCl₃ and Cl₂. Since the forward reaction converts PCl₅ into products, the net reaction proceeds in the forward direction. Choice C incorrectly assumes equal product concentrations mean no net reaction, but the key is whether concentrations are changing, not their relative values. The strategy is to identify the decomposition pattern: one species decreasing while multiple species increase indicates forward decomposition.
8. Which of the following is a possible particulate representation of the mixture at equilibrium?
   Answer: A mixture containing 4 $SO_2$, 3 $O_2$, and 2 $SO_3$ molecules. 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.
9. The reaction $A_2(g) \rightleftharpoons 2 A(g)$ is endothermic. A particulate diagram shows the system at equilibrium at 300 K. The temperature is then increased to 500 K. Which of the following particulate representations would show the new equilibrium state?
   Answer: A diagram showing an increased number of A atoms and a decreased number of $A_2$ molecules. 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.
10. The reaction $A_2(g) + B(g) \rightleftharpoons A_2B(g)$ is at equilibrium. A particulate diagram of the mixture shows several molecules of each species. What must be true about the system at the molecular level?
    Answer: 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. 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).
11. Consider the reversible reaction: $A_2(g) + B_2(g) \rightleftharpoons 2 AB(g)$. A particulate representation of a sealed container at equilibrium shows 10 particles of AB, 2 particles of $A_2$, and 2 particles of $B_2$. Which statement correctly describes the equilibrium constant, $K_c$, for this reaction?
    Answer: 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. 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.
12. A sealed container has the reversible reaction $\text{Br}_2(g) \rightleftharpoons 2\text{Br}(g)$. Concentrations are recorded during the first few seconds after heating begins; the data shown are from a period before any concentration becomes constant. Based only on the table, what is the direction of the net reaction progress during the interval shown?
    Answer: Net reaction is proceeding in the forward direction (toward $\text{Br}$). This question examines using table data to determine net progress in a dissociation reaction. The table displays [Br₂] decreasing and [Br] increasing, consistent with forward dissociation where one Br₂ yields two Br. The net increase in [Br] reflects stoichiometric production exceeding consumption. Before constants, net forward progress occurs. Choice A fails due to misreading the table, confusing product increase with reverse direction. When reviewing such tables, emphasize the net accumulation of products or reactants through trends, not comparisons of absolute concentrations, for accurate direction assessment.
13. A particulate diagram representing the equilibrium state for the gas-phase reaction $2 AB(g) \rightleftharpoons A_2(g) + B_2(g)$ is contained in a 1 L box. The diagram shows 4 molecules of AB, 6 molecules of $A_2$, and 6 molecules of $B_2$. Which statement is consistent with this representation?
    Answer: The equilibrium constant $K_c$ is greater than 1, favoring the products. 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.
14. The reaction $CO(g) + 2H_2(g) \rightleftharpoons CH_3OH(g)$ has an equilibrium constant $K_c = 14.5$. A particulate representation of a mixture contains 1 CO molecule, 2 $H_2$ molecules, and 20 $CH_3OH$ molecules in a 1.0 L container. How will the number of $H_2$ molecules change as the system proceeds to equilibrium?
    Answer: The number of $H_2$ molecules will increase because $Q > K_c$. 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 $H_2$.
15. The reaction $N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g)$ is at equilibrium. A particulate diagram shows He atoms, an inert gas, have been added to the mixture at constant volume. Which statement describes the effect on the equilibrium?
    Answer: The equilibrium position will not change because the partial pressures of the reacting gases have not changed. 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.
16. A container initially holds only $\text{SO}_3$ gas, which decomposes according to the equation $2\text{SO}_3(g) \rightleftharpoons 2\text{SO}_2(g) + \text{O}_2(g)$. Which description best represents the contents of the container over time as the system approaches equilibrium?
    Answer: The numbers of $\text{SO}_2$ and $\text{O}_2$ molecules appear and increase, while the number of $\text{SO}_3$ molecules decreases, until all particle counts become constant. The system starts with only the reactant, $\text{SO}_3$. As the reaction proceeds towards equilibrium, reactant is consumed and products ($\text{SO}_2$ and $\text{O}_2$) 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 $\text{SO}_3$ will remain.
17. A system represented by $H_2(g) + I_2(g) \rightleftharpoons 2 HI(g)$ is at equilibrium. A particulate diagram shows 3 molecules of $H_2$, 3 molecules of $I_2$, and 6 molecules of HI. If the volume of the container is decreased at constant temperature, what would a new particulate diagram at equilibrium show?
    Answer: The same number of molecules of each species, because the reaction has an equal number of moles of gas on both sides. 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.
18. A particulate representation of a saturated aqueous solution of $PbF_2(s)$ at equilibrium shows a small number of dissolved $Pb^{2+}$ and $F^-$ ions. If a solution containing $Na^+(aq)$ and $F^-(aq)$ is added, how will the particulate representation change once equilibrium is re-established?
    Answer: The amount of solid $PbF_2$ will increase, and the number of dissolved $Pb^{2+}$ ions will decrease. 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.
19. Consider the equilibrium $A(g) \rightleftharpoons B(g)$ with $K_c > 1$. An initial mixture is prepared with equal moles of A and B. Which particulate diagram best represents the mixture after equilibrium is established?
    Answer: A diagram with more particles of B than particles of A. 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.
20. The reaction $CaCO_3(s) \rightleftharpoons CaO(s) + CO_2(g)$ reaches equilibrium in a closed container. Which statement best describes a particulate representation of this system at equilibrium?
    Answer: The diagram would show solid lattices of $CaCO_3$ and $CaO$, and dispersed $CO_2$ gas molecules. The number of $CO_2$ molecules per unit volume is constant. This is a heterogeneous equilibrium. For equilibrium to be established, all three substances must be present. $CaCO_3$ and $CaO$ are solids and would be represented as ordered lattices. $CO_2$ is a gas and would be represented as randomly moving molecules. At a given temperature, the pressure (and thus concentration) of $CO_2$ at equilibrium is constant, so the number of gas particles per unit volume is fixed, as long as both solids are present.

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