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

To find the values of \(x\) and \(y\) for the given chemical reaction \(xA(g) \leftrightarrow yB(g)\) at 400 K, we need to use the relationship between the equilibrium constants \(K_p\) and \(K_c\). The relationship is given by:

\(K_p = K_c (RT)^{\Delta n}\)

where \(\Delta n\) is the change in the number of moles of gas, \(R\) is the universal gas constant, and \(T\) is the temperature in Kelvin.

For the reaction given, \(\Delta n = y - x\).

Let's analyze each pair (i) and (ii) given in the problem:

1. For case (i): \(K_p = 85.87\), \(K_c = 2.586\).

Substituting these into the equation \(K_p = K_c (RT)^{\Delta n}\), we get:

\(85.87 = 2.586 (RT)^{\Delta n}\)

We calculate the logarithm of both sides to find \(\Delta n\):

\(\Delta n = \frac{\log(85.87/2.586)}{\log(RT)}\)

Assuming \(R = 0.0821 \, \text{L atm K}^{-1}\text{mol}^{-1}\) and \(T = 400 \, \text{K}\), compute:

\(\Delta n = 1.0\)

Hence, \(y - x = 1\). Supplying the pairs, we deduce x = 1 and y = 2.

1. For case (ii): \(K_p = 0.862\), \(K_c = 28.62\).

From the equation \(K_p = K_c (RT)^{\Delta n}\), we have:

\(0.862 = 28.62 (RT)^{\Delta n}\)

Taking logarithms of both ratios:

\(\Delta n = \frac{\log(0.862/28.62)}{\log(RT)}\)

Calculate it similarly as before:

\(\Delta n = -1.0\)

Hence, \(y - x = -1\). We conclude x = 2 and y = 1.

Thus, the correct pair of values is 1,2 and 2,1 respectively for cases (i) and (ii).