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# For the following gaseous equilibria X, Y and Z at 300 K. $X: 2SO_2 + O_2 \rightleftharpoons 2SO_3$ $Y: PCl_5 \rightleftharpoons PCl_3 + Cl_2$ $Z: 2HI \rightleftharpoons H_2 + I_2$ ratio of $K_p$ and $K_c$ in the increasing order is:

(a) X=Y=Z

(b) X< Y < Z

(c) X< Z< Y

(d) Z< Y< X

Toolbox:
• Equilibrium constant determined from the partial pressures is $K_p = K_c (RT)^{\Delta n_g}$
Answer: X < Z < Y

For X: $2SO_2 + O_2 \rightleftharpoons 2SO_3$
$\Delta n_g = -1$

$\therefore \frac{K_p}{K_c} = (RT)^{-1} = (0.0821 \times 300)^{-1} = \frac{1}{24.63}$

For Y : $PCl_5 \rightleftharpoons PCl_3 + Cl_2$
$\Delta n_g = 2 - 1 = 1$

$\therefore \frac{K_p}{K_c} = RT = 24.63$

For Z : $2HI \rightleftharpoons H_2 + I_2$
$\Delta n_g = 0$

$\therefore \frac{K_p}{K_c} = (RT)^0 = 1$

Clearly, $\frac{K_p}{K_c}$ increases in the order X < Z < Y