Consider the general reaction given below at 400 K
$$xA(g)\rightleftharpoons yB(g).
The values of $$K_{p}\text{ and }K_{c}$$ are studied under the same condition of temperature but variation in x and y
(i)$$K_{p}=85.87\text{ and }K_{c}=2.586$$ appropriate units
(ii)$$K_{p}=0.862\text{ and }K_{c}=28.62$$ appropriate units
The values of x and yin (i) and (ii) respectively are:

We are given the reaction $$xA(g) \rightleftharpoons yB(g)$$ at $$T = 400$$ K, with two cases.

$$K_p = K_c(RT)^{\Delta n}$$, where $$\Delta n = y - x$$ (moles of gaseous products minus reactants) and $$R = 0.0821$$ L·atm/(mol·K).

At $$T = 400$$ K: $$RT = 0.0821 \times 400 = 32.84$$.

$$\frac{K_p}{K_c} = \frac{85.87}{2.586} \approx 33.2 \approx 32.84^1$$

So $$(RT)^{\Delta n} \approx (32.84)^1$$, meaning $$\Delta n = y - x = 1$$.

With $$x = 1, y = 2$$: $$\Delta n = 2 - 1 = 1$$ ✓

$$\frac{K_p}{K_c} = \frac{0.862}{28.62} \approx 0.0301 \approx \frac{1}{33.2} \approx (32.84)^{-1}$$

So $$(RT)^{\Delta n} \approx (32.84)^{-1}$$, meaning $$\Delta n = y - x = -1$$.

With $$x = 2, y = 1$$: $$\Delta n = 1 - 2 = -1$$ ✓

(i) $$x = 1, y = 2$$ and (ii) $$x = 2, y = 1$$.

The answer is Option A: (i) 1, 2 and (ii) 2, 1.