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

The backward rate constant is provided as \( k_b = 2500 \times k_f \). This implies that \( \frac{k_f}{k_b} = \frac{1}{2500} \). For a gaseous equilibrium reaction, the relationship between \( K_p \) and \( K_c \) is given by \( K_p = K_c(RT)^{\Delta n} \). Since we are working with rate constants, we can equate \( K_c \) with the ratio of the forward and backward rate constants: \( K_c = \frac{k_f}{k_b} = \frac{1}{2500} \). The given reaction is \( A(g) \rightleftharpoons 2B(g) \), which means \( \Delta n = 2 - 1 = 1 \). Using the formula \( K_p = K_c(RT)^{\Delta n} \) and substituting the values: \[ K_p = \frac{1}{2500} \times (0.0831 \times 1000)^1 = \frac{1}{2500} \times 83.1 \] Therefore, \[ K_p = \frac{83.1}{2500} = 0.03324 \approx 0.033 \] The approximate value of \( K_p \) is 0.033.