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Chapter 15: Problem 11

Suppose that the gas-phase reactions \(\mathrm{A} \longrightarrow \mathrm{B}\) and \(\mathrm{B} \longrightarrow \mathrm{A}\) are both elementary processes with rate con- stants of \(3.8 \times 10^{-2} \mathrm{~s}^{-1}\) and \(3.1 \times 10^{-1} \mathrm{~s}^{-1}\), respectively. (a) What is the value of the equilibrium constant for the equilibrium \(\mathrm{A}(g) \rightleftharpoons \mathrm{B}(g) ?\) (b) Which is greater at equilibrium, the partial pressure of A or the partial pressure of B? Explain.

Short Answer

Expert verified
The equilibrium constant (K) for the given gas-phase reactions A 鈬� B is 0.1226. Since K < 1, the reverse reaction (B 鈫� A) is more favored, and thus at equilibrium, the partial pressure of A is greater than the partial pressure of B.

Step by step solution

01

Overall Reaction

The overall equilibrium reaction is given by: A (g) 鈬� B (g)
02

Finding the equilibrium constant

For an elementary reaction, the equilibrium constant (K) can be calculated as the ratio of the rate constants of the forward and reverse reactions: K = k1 / k2 Substitute the given rate constants: K = (3.8 脳 10鈦宦� s鈦宦�) / (3.1 脳 10鈦宦� s鈦宦�) K = 0.1226 So, the equilibrium constant (K) for this equilibrium reaction is 0.1226.
03

Comparing partial pressures

To compare the partial pressures of A and B at equilibrium, we can simply look at the equilibrium constant value. If K > 1, it means the forward reaction (A 鈫� B) is more favored, so the concentration (and partial pressure) of B will be higher at equilibrium. If K < 1, it means the reverse reaction (B 鈫� A) is more favored, so the concentration (and partial pressure) of A will be higher at equilibrium. Since K = 0.1226, which is less than 1, the reverse reaction (B 鈫� A) is more favored. Therefore, at equilibrium, the partial pressure of A is greater than the partial pressure of B.

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Most popular questions from this chapter

When \(2.00 \mathrm{~mol}\) of \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\) is placed in a 2.00-L flask at \(303 \mathrm{~K}, 56 \%\) of the \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\) decomposes to \(\mathrm{SO}_{2}\) and \(\mathrm{Cl}_{2}\) : $$ \mathrm{SO}_{2} \mathrm{Cl}_{2}(g) \rightleftharpoons \mathrm{SO}_{2}(g)+\mathrm{Cl}_{2}(g) $$ Calculate \(K_{c}\) for this reaction at this temperature.

For a certain gas-phase reaction, the fraction of products in an equilibrium mixture is increased by increasing the temperature and increasing the volume of the reaction vessel. (a) What can you conclude about the reaction from the influence of temperature on the equilibrium? (b) What can you conclude from the influence of increasing the volume?

The following equilibria were attained at \(823 \mathrm{~K}\) : $$ \begin{array}{ll} \mathrm{CoO}(s)+\mathrm{H}_{2}(g) & \rightleftharpoons \mathrm{Co}(s)+\mathrm{H}_{2} \mathrm{O}(g) & K_{c}=67 \\ \mathrm{CoO}(s)+\mathrm{CO}(g) & \rightleftharpoons \mathrm{Co}(s)+\mathrm{CO}_{2}(g) & K_{c}=490 \end{array} $$ Based on these equilibria, calculate the equilibrium constant for \(\mathrm{H}_{2}(\mathrm{~g})+\mathrm{CO}_{2}(g) \rightleftharpoons \mathrm{CO}(g)+\mathrm{H}_{2} \mathrm{O}(g)\) at \(823 \mathrm{~K}\)

The reaction \(\mathrm{PCl}_{3}(g)+\mathrm{Cl}_{2}(g) \rightleftharpoons \mathrm{PCl}_{5}(g)\) has \(K_{p}=0.0870\) at \(300^{\circ} \mathrm{C}\). A flask is charged with \(0.50\) atm \(\mathrm{PCl}_{3}, 0.50 \mathrm{~atm} \mathrm{Cl}_{2}\), and \(0.20 \mathrm{~atm} \mathrm{PCl}_{5}\) at this tempera- ture. (a) Use the reaction quotient to determine the direction the reaction must proceed to reach equilibrium. (b) Calculate the equilibrium partial pressures of the gases. (c) What effect will increasing the volume of the system have on the mole fraction of \(\mathrm{C}_{2}\) in the equilibrium mixture? (d) The reaction is exothermic. What effect will increasing the temperature of the system have on the mole fraction of \(\mathrm{Cl}_{2}\) in the equilibrium mixture?

For the reaction \(\mathrm{I}_{2}+\mathrm{Br}_{2}(g) \rightleftharpoons 2 \operatorname{IBr}(g), K_{c}=280\) at \(150^{\circ} \mathrm{C}\). Suppose that \(0.500 \mathrm{~mol}\) IBr in a \(1.00\) -L flask is allowed to reach equilibrium at \(150^{\circ} \mathrm{C}\). What are the equilibrium concentrations of \(\mathrm{IBr}, \mathrm{I}_{2}\), and \(\mathrm{Br}_{2} ?\)
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