Question

For the reaction A(g) $\rightleftharpoons$ 2B(g), the backward reaction rate constant is higher than the forward reaction rate constant by a factor of 2500, at 1000 K.
[Given: R = 0.0831 atm $mol^{–1} K^{–1}$]
$K_p$ for the reaction at 1000 K is:

• A. \(2.077 \times 10^5\)
• B. 0.033
• C. 0.021
• D. 83.1

Hint:

When backward rate constant is given as a multiple of the forward rate constant, use \( K_c = \frac{k_f}{k_b} \), and relate it to \( K_p \) using \( (RT)^{\Delta n} \).

Answer 1

The Correct Option is B

We are given: - Backward rate constant, \( k_b = 2500 \times k_f \) -
Therefore, \( \frac{k_f}{k_b} = \frac{1}{2500} \) - Also, for a gaseous equilibrium reaction: \[ K_p = K_c(RT)^{\Delta n} \] But since we are dealing with rate constants, we can use the relation: \[ K_c = \frac{k_f}{k_b} = \frac{1}{2500} \] Given reaction: \( A(g) \rightleftharpoons 2B(g) \), so \( \Delta n = 2 - 1 = 1 \) Use the formula: \[ K_p = K_c(RT)^{\Delta n} \] Substitute values: \[ K_p = \frac{1}{2500} \times (0.0831 \times 1000)^1 = \frac{1}{2500} \times 83.1 \] \[ K_p = \frac{83.1}{2500} = 0.03324 \approx 0.033 \] Hence, the value of \( K_p \) is approximately 0.033.