Question

Suppose that the gas-phase reactions $\mathrm{A}⟶\mathrm{B}$ and $\mathrm{B}⟶\mathrm{A}$ 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

The equilibrium constant (K) for the given gas-phase reactions A ⇌ B is 0.1226. Since K < 1, the reverse reaction (B → 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

Overall Reaction

The overall equilibrium reaction is given by: A (g) ⇌ B (g)

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 × 10⁻² s⁻¹) / (3.1 × 10⁻¹ s⁻¹) K = 0.1226 So, the equilibrium constant (K) for this equilibrium reaction is 0.1226.

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 → 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 → 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 → A) is more favored. Therefore, at equilibrium, the partial pressure of A is greater than the partial pressure of B.