Question

Suppose that the gas-phase reactions \(A \longrightarrow B\) and \(\mathrm{B} \longrightarrow \mathrm{A}\) are both elementary processes with rate constants of \(4.7 \times 10^{-3} \mathrm{~s}^{-1}\) and \(5.8 \times 10^{-1} \mathrm{~s}^{-1}\), respectively. (a) What is the value of the equilibrium constant for the equilibrium \(\mathrm{A}(\mathrm{g}) \rightleftharpoons \mathrm{B}(\mathrm{g})\) ? (b) Which is greater at equilibrium, the partial pressure of \(A\) or the partial pressure of \(B\) ?

Short answer

(a) The equilibrium constant \(K_{eq}\) for the reaction A(g) ⇌ B(g) is \(8.1 \times 10^{-3}\). (b) At equilibrium, the partial pressure of A is greater than the partial pressure of B.

Step-by-step solution

Recall the relationship between rate constants and equilibrium constant

For a reversible reaction, the equilibrium constant is given as the ratio of the forward rate constant (k_forward) to the reverse rate constant (k_reverse): \[ K_{eq} = \frac{k_{forward}}{k_{reverse}} \]

Calculate the equilibrium constant

Using the provided rate constants, determine the equilibrium constant (K_eq) for the reaction A(g) ⇌ B(g): \[ K_{eq} = \frac{4.7 \times 10^{-3} \mathrm{~s}^{-1}}{5.8 \times 10^{-1} \mathrm{~s}^{-1}} \] Now, calculate K_eq value: \[ K_{eq} = \frac{4.7 \times 10^{-3}}{5.8 \times 10^{-1}} = 8.1 \times 10^{-3} \]

Compare the partial pressures at equilibrium

Since K_eq is less than 1, this means that the reverse reaction (B → A) is more favored. Therefore, at equilibrium, the partial pressure of A will be greater than the partial pressure of B. (a) The equilibrium constant K_eq for the reaction A(g) ⇌ B(g) is 8.1 x 10^(-3). (b) At equilibrium, the partial pressure of A is greater than the partial pressure of B.

Key concepts

Rate Constant

In chemical kinetics, a rate constant is a crucial factor. It measures the speed of a chemical reaction. This constant, often denoted by the symbol 'k', changes depending on the reaction being examined and the conditions such as temperature.
For a simple reaction where species A is converted to species B, the rate of the reaction can be expressed as:\[ \text{Rate} = k \cdot [A] \]where

• \(k\) is the rate constant,
• \([A]\) is the concentration of A.

The rate constant has units that depend on the order of reaction. For first-order reactions, it is typically expressed in s^{-1}.
This means that k tells us how fast the reaction proceeds to convert A to B under a set of specified conditions without directly changing the concentrations of A or B.Interestingly, for elementary reactions, the rate constant can directly inform us about the kinetic behavior of reversible reactions by also considering the reverse reaction's rate constant as we will see next.

Reversible Reactions

Reversible reactions are like a two-way street in chemistry where reactants turn into products and back again. This happens at the same time, establishing a dynamic balance at equilibrium.
The reaction is described as:\[\mathrm{A} \rightleftharpoons \mathrm{B}\]Both forward \((\mathrm{A} \rightarrow \mathrm{B})\) and reverse reactions \((\mathrm{B} \rightarrow \mathrm{A})\) occur.
Equilibrium constant \(K_{eq}\) reflects the ratio of rate constants of these reactions:\[ K_{eq} = \frac{k_{forward}}{k_{reverse}} \]When \(K_{eq}\) is less than 1, it signifies that the reverse reaction is predominant, hence more reactants remain compared to products at equilibrium.
This tells us that in our study of \(A\) and \(B\), more \(A\) is present at equilibrium since \(K_{eq} = 8.1 \times 10^{-3}\), suggesting that the equilibrium position favors the reactants.

Partial Pressure

Partial pressure is the pressure contributed by a single gas in a mixture of gases. It plays a prominent role in understanding chemical equilibria in gas-phase reactions.
Each gas in a mixture exerts pressure as if it is alone in the container. This means gas A and B each contribute their partial pressure to the total pressure of the system. The total pressure can be represented as the sum of these individual pressures:\[ P_{total} = P_A + P_B \]Here,

• \(P_A\) is the partial pressure of gas A,
• \(P_B\) is the partial pressure of gas B.

In the equilibrium scenario we examined, since \(K_{eq}\) is less than 1, the partial pressure of \(A\) will be higher than that of \(B\) at equilibrium. This implies that the backward transformation of \(B\) into \(A\) is energetically more favorable under the given conditions, highlighting the dominance of A in its gaseous form.