Question

Consider the general reaction given below at 400 K: xA(g) ⇌ yB(g). The values of K___p and K___c are studied under the same condition of temperature but variation in x and y.
(i) K___p = 85.87 and K___c = 2.586
(ii) K___p = 0.862 and K___c = 28.62.
The values of x and y in (i) and (ii) respectively are :

• A. 4,1 4,1
• B. 3,1 3,1
• C. 1,3 2,1
• D. 1,2 2,1

Hint:

If $K_p>K_c$, then $\Delta n_g$ is positive (moles of gas increase). If $K_p<K_c$, then $\Delta n_g$ is negative (moles of gas decrease).

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
The relationship between $K_p$ and $K_c$ is given by the formula $K_p = K_c(RT)^{\Delta n_g}$, where $\Delta n_g = y - x$.
Step 2: Key Formula or Approach:
1. $K_p = K_c(RT)^{\Delta n_g}$.
2. Calculate $RT$ for $T = 400$ K: $R = 0.0821$ L·atm/(K·mol).
Step 3: Detailed Explanation:
$RT = 0.0821 \times 400 = 32.84$.
Case (i): $\frac{K_p}{K_c} = \frac{85.87}{2.586} \approx 33.2$.
Since $RT \approx 32.84$, then $(RT)^{\Delta n_g} \approx 33.2 \implies \Delta n_g = 1$.
$y - x = 1$. Among options, (1, 2) fits this ($2 - 1 = 1$).
Case (ii): $\frac{K_p}{K_c} = \frac{0.862}{28.62} \approx 0.030$.
Since $\frac{1}{RT} = \frac{1}{32.84} \approx 0.030$, then $(RT)^{\Delta n_g} \approx 0.030 \implies \Delta n_g = -1$.
$y - x = -1$. Among options, (2, 1) fits this ($1 - 2 = -1$).
Step 4: Final Answer:
The values of x and y are (1,2) for (i) and (2,1) for (ii).