Representations of Equilibrium
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AP Chemistry › Representations of Equilibrium
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?
The value of $$K_c$$ is approximately equal to 1, as both reactants and products are present in the mixture.
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 value of $$K_c$$ is much less than 1, as reactants are present in smaller quantities than the product.
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.
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?
The amount of solid $$PbF_2$$ will increase, and the number of dissolved $$Pb^{2+}$$ ions will decrease.
The amount of solid $$PbF_2$$ will decrease, and the number of dissolved $$Pb^{2+}$$ ions will increase.
The amount of solid $$PbF_2$$ will increase, and the number of dissolved $$Pb^{2+}$$ ions will increase.
The amount of solid $$PbF_2$$ and the number of dissolved $$Pb^{2+}$$ ions will remain the same.
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.
Based on the particulate representations of the initial and equilibrium states, what is the balanced chemical equation for the reaction?
$$X_2(g) + 3 Y_2(g) \rightleftharpoons 2 XY_3(g)$$
$$2 X_2(g) + Y_2(g) \rightleftharpoons 2 X_2Y(g)$$
$$2 X_2(g) + 2 Y_2(g) \rightleftharpoons X_4Y_4(g)$$
$$X_2(g) + Y_2(g) \rightleftharpoons 2 XY(g)$$
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 $$X_2(g) + 3 Y_2(g) \rightleftharpoons 2 XY_3(g)$$.
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?
The number of $$H_2$$ molecules will decrease because $$Q < K_c$$.
The number of $$H_2$$ molecules will remain the same because the system is already at equilibrium.
The number of $$H_2$$ molecules will increase because $$Q < K_c$$.
The number of $$H_2$$ molecules will increase because $$Q > K_c$$.
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 $$H_2$$.
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?
No net reaction is occurring because the changes in concentration are not identical at each time step.
No net reaction is occurring because $[\text{NOCl}]$ is always the largest concentration.
Net reaction is proceeding in the reverse direction (toward $\text{NO}$ and $\text{Cl}_2$).
Net reaction is proceeding in the forward direction (toward $\text{NOCl}$).
The net direction cannot be inferred because $[\text{NO}]$ and $[\text{Cl}_2]$ are both decreasing.
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.
A sealed vessel contains the reversible reaction $\text{PCl}_5(g) \rightleftharpoons \text{PCl}_3(g)+\text{Cl}_2(g)$. A student records concentrations shortly after the reaction begins (before any concentration becomes constant). Using only the table, what is the direction of the net reaction progress over the interval shown?
No net reaction is occurring because $[\text{PCl}_3]$ and $[\text{Cl}_2]$ are equal at each time.
Net reaction is proceeding in the forward direction (toward $\text{PCl}_3$ and $\text{Cl}_2$).
No net reaction is occurring because $[\text{Cl}_2]$ is increasing while $[\text{PCl}_3]$ is also increasing.
Net reaction is proceeding in the reverse direction (toward $\text{PCl}_5$).
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.
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?
The diagram would only show $$CO_2$$ gas molecules, because solids are not included in the equilibrium expression and are therefore not present in the container.
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.
The diagram would show solid lattices of $$CaCO_3$$ and $$CaO$$. No $$CO_2$$ molecules would be visible, as they have completely reacted to form the solids.
The diagram must show equal numbers of $$CaCO_3$$, $$CaO$$, and $$CO_2$$ particles to represent equilibrium.
Explanation
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.
A container initially holds only $$SO_3$$ gas, which decomposes according to the equation $$2 SO_3(g) \rightleftharpoons 2 SO_2(g) + O_2(g)$$. Which sequence of particulate diagrams best represents the contents of the container over time as the system approaches equilibrium?
The number of $$SO_3$$ molecules decreases to zero, while numbers of $$SO_2$$ and $$O_2$$ molecules increase.
The number of $$SO_3$$ molecules remains constant, while $$SO_2$$ and $$O_2$$ molecules are formed.
The numbers of all three types of molecules fluctuate randomly as they continuously react back and forth.
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.
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.
Which of the following is a possible particulate representation of the mixture at equilibrium?
A mixture containing 0 $$SO_2$$, 1 $$O_2$$, and 6 $$SO_3$$ molecules.
A mixture containing 5 $$SO_2$$, 3 $$O_2$$, and 1 $$SO_3$$ molecule.
A mixture containing 6 $$SO_2$$, 4 $$O_2$$, and 0 $$SO_3$$ molecules.
A mixture containing 4 $$SO_2$$, 3 $$O_2$$, and 2 $$SO_3$$ molecules.
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.
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?
Net reaction is proceeding in the forward direction (toward $\text{Br}$).
Net reaction is proceeding in the reverse direction (toward $\text{Br}_2$).
No net reaction is occurring because both species are present at all times.
The net direction cannot be inferred because the reaction produces twice as much $\text{Br}$ as $\text{Br}_2$ consumed.
No net reaction is occurring because $[\text{Br}]$ is always less than $[\text{Br}_2]$.
Explanation
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.