\(K_{eq}={[Fe(SCN)]^{2+}\over [Fe^{3+}][SCN^-]}\)
\(={2\times 10^{-4}M\over 8\times 10^{-4}M\times 6\times 10^{-4}M}\)
= 0.0416 x 10⁴
K_eq = 41.6 x 10² M^-1

N₂O₄(g) ⇌ 2NO₂(g)
T₁ = 298 K        K_P1 = 0.15
T₂ = 100o C = 100 + 273 = 373 K ;
K_P2 = ?
\(\log({K_2\over K_1})={\Delta H^o\over 2.303R}[{T_2-T_1\over T_1T_2}]\)
K₂ > K₁ and T₂ > T₁
\(\log({K_{p_2}\over 0.15})={57.2KJ\ mol^{-1}\over 2.303\times 8.314JK^{-1}\ mol^{-1}}[{373-298\over 373\times 298}]\)
\(\log({K_{P_2}\over 0.15})={57.2\times 10^{+3}\times 75\over 2.303\times 8.314\times 373\times 298}\)
\(\log({K_{p_2}\over 0.15})=2.02\)
\({K_{p_2}\over 0.15}=104.7\)
\(K_{p_2}=104.7\times 0.15\)
\(K_{p_2}=15.705\)

CO(g) + H₂O(g) ⇌ CO₂ (g) + H₂(g) Given K_P = 2.7
[CO] = 0.13, [H₂O] = 0.56
[CO₂] = 0.78 ; [H₂] = 0.28
V = 2L
K_P = K_C (RT)
2.7 = K_C (RT)^o
K_C = 2.7
\(Q_c={[CO_2][H_2]\over [CO][H_2O]}\)\(={({0.78\over 2})({0.28\over 2})\over ({0.13\over 2})({0.56\over 2})}\)
Q = 3
Q > K_c, Hence the reaction proceed in the reverse direction.

2XY₂(g) ⇌ 2XY(g) + Y₂(g)

Total No. of moles = 1 – x + x + x/2 = 1 + x/2 \(\cong \) 1
[∵ Given that x < < 1 ; 1 – x \(\cong \) 1 and 1 + x/2 \(\cong \) 1]
\(K_p={(P_{XY})^2(P_{Y_2})\over (P_{XY_2})^2}={({x\over 1}\times p)^2({x/2\over 1}\times P)\over ({1\over1}\times P)^2}\)

\(\Rightarrow\) 2K_p = x³P.

K_C = 4 x 10^–2 for the reaction,
H₂S(g) ⇌ H₂(g) + 1/2 S₂(g)
\(K_c={[H_2][S_2]^{1/2}\over [H_2S]}\)
\(\Rightarrow 4\times 10^{-2}={[H_2][S_2]^{1/2}\over [H_2S]}\)
For the reaction,
2H₂S(g) ⇌ 2H₂(g) + S₂(g)
\(K_c={[H_2]^2[S_2]\over [H_2S]}=(4\times 10^{-2})^2=16\times 10^{-4}\)
For the reaction,
3H₂S(g) ⇌ 3H₂(g) + 3/2 S₂(g)
\(K_c={[H_2]^3[S_2]^{3/2}\over [H_2S]^3}=(4\times 10^{-2})^3=64\times 10^{-6}\)