H₂(g) + I₂(g) \(\rightleftharpoons \) 2HI(g)
At a particular instant, the concentration of H₂, I₂and HI are found to be 0.2 mol L^-1, 0.2 mol L^-1 and 0.6 mol L^-1 respectively. From the above information we can predict the direction of reaction as follows.
\(Q={[HI]^2\over[H_2][I_2]}={0.6\times 0.6\over 0.2\times 0.2}=9\)
Since Q < K_c, the reaction will proceed in the forward direction.

H_2(g) + I_2(g) \(\rightleftharpoons \) 2HI_(g)
At equilibrium, [HI] = 0.4 mol L^-1 Kc= ?

\(\therefore K_c={[HI]^2\over [H_2[I_2]]}={0.4\times 0.4\over 0.8\times 0.8}=0.25\)

T₁ = 25 + 273 = 298 K
T₂ = 37 + 273 = 310 K
ΔH = 32.4 KJmol^-1 = 32400 Jmol^-1
R = 8.314 JK^-1 mol^-1
K_P1 = 0.0260
K_p2 = ?
\(\log {K_2\over K_1}={\Delta H^o\over 2.303R}[{T_2-T_1\over T_2T_1}]\)
\(\log {K_2\over K_1}={32400\over 2.303\times 8.314}({310-298\over 310\times 298})\)
\(={32400\times 10\over 2.303\times 8.314\times 310\times 298}\)
= 0.2198
\({K_2\over K_1}=\) antilog 0.2198 = 1.6588
K₂ = 1.6588 x 0.026 = 0.0431

\( \therefore\left[\mathrm{PCl}_{5}\right]_{\mathrm{eq}}=0.4 \mathrm{~mol} \mathrm{dm}^{-3} ; \quad\left[\mathrm{PCl}_{3}\right]_{\mathrm{eq}}=0.6 \mathrm{~mol} \mathrm{dm}^{-3} ; \quad\left[\mathrm{Cl}_{2}\right]_{\mathrm{eq}}=0.6 \mathrm{~mol}^{-3} \)
\(\mathrm{K}_{c}=\frac{\left[\mathrm{PCl}_{3}\right]\left[\mathrm{Cl}_{2}\right]}{\left[\mathrm{PCl}_{5}\right]}=\frac{0.6 \times 0.6}{0.4}\)
K_c = 0.9 mol dm^-3

V = 3L
\([HI]_{initial}={0.3\ mol\over 3L}=0.1\ M\)
[HI]_eq = 0.05 M
2HI(g) ⇌ H₂(g) + I₂(g)

\(K_c={[H_2][I_2]\over [HI]^2}\)
\(={0.025\times 0.025\over 0.05\times 0.05}\)
K_c = 0.25
K_P = K_C (RT)^Δng
Δn_g = 2 – 2 = 0
K_P = 0.25 (RT)^o
K_P = 0.25