Question

At equilibrium for the reaction $ A_2 (g) + B_2 (g) \rightleftharpoons 2AB (g) $, the concentrations of $ A_2 $, $ B_2 $, and $ AB $ respectively are $ 1.5 \times 10^{-3} M $, $ 2.1 \times 10^{-3} M $, and $ 1.4 \times 10^{-3} M $. What will be $ K_p $ for the decomposition of $ AB $ at the same temperature?

• A. \(0.62\)
• B. \(1.6\)
• C. \(0.44\)
• D. \(2.27\)

Hint:

For equilibrium calculations: - Use the balanced chemical equation. - Substitute equilibrium concentrations into the expression for \( K_c \).

Answer 1

The Correct Option is B

Step 1: Write the equilibrium constant expression The equilibrium constant \( K_c \) for the given reaction is: \[ K_c = \frac{[AB]^2}{[A_2][B_2]} \]
Step 2: Substitute the given concentrations \[ K_c = \frac{(1.4 \times 10^{-3})^2}{(1.5 \times 10^{-3}) \times (2.1 \times 10^{-3})} \] \[ K_c = \frac{1.96 \times 10^{-6}}{3.15 \times 10^{-6}} \] \[ K_c \approx 0.622 \]
Step 3: Relating \( K_p \) to \( K_c \) Using the relation: \[ K_p = K_c \times (RT)^{\Delta n} \] Since \( \Delta n = (2 - (1 + 1)) = 0 \), \[ K_p = K_c \times (RT)^0 = K_c \] Thus, \[ K_p \approx 0.62 \]
Step 4: Finding \( K_p \) for decomposition reaction For the decomposition of \( AB \): \[ 2AB (g) \rightleftharpoons A_2 (g) + B_2 (g) \] By reversing the reaction, \[ K'_p = \frac{1}{K_p} = \frac{1}{0.62} \approx 1.6 \]
Step 5: Final Answer The correct answer is \( 1.6 \).