\(P_{CO_2}\) = 1.017 x 10^-3 atm T = 500^oC
K_p = \(P_{CO_2}\)
\(\therefore K_{P_1}\) = 1.017 x 10^-3  T = 500 + 273 = 773 K
\(K_{P_2}\) = ?    T = 600 + 273 = 873 K
\(\Delta H^o=181\ KJ\ mol^{-1}\)
\(\log({K_{P_1}\over K_{P_1}})={\Delta H^o\over 2.303\ R}({T_2-T_1\over T_1T_2})\)
\(\log({K_{P_2}\over 1.017\times 10^{-3}})={181\times 10^3\over 2.303\times 8.314}({873-773\over 873\times 773})\)
\(\log({K_{P_2}\over 1.017\times 10^{-3}})={181\times 10^3\times 100\over 2.303\times 8.314\times 873\times 773}\)
\({K_{P_2}\over 1.017\times 10^{-3}}=\) anti log of (1.40)
\({K_{P_2}\over 1.017\times 10^{-3}}=25.12\)
\(\Rightarrow K_{P_2}=\) 25.12 x 1.017 x 10^-3
\(K_{P_2}=\) 25.54 x 10^-3.

CH₄(g) + 2H₂S(g) ⇌ CS₂(g) + 4H₂(g)
K_C = 4 x 10^–2 mol lit^–2
Volume = 500 ml = 1/2 L
\(\left[\mathrm{CH}_{4}\right]_{\text {in }}=\frac{2 \mathrm{~mol}}{1 / 2 \mathrm{~L}} \)
= 2 mol L^-1
\(\left[\mathrm{CS}_{2}\right]_{\text {in }}=\frac{1 \mathrm{~mol}}{1 / 2 \mathrm{~L}}\)
= 2 mol L^-1
\( {\left[\mathrm{H}_{2} \mathrm{~S}\right]_{\text {in }}=\frac{2 \mathrm{~mol}}{1 / 2 \mathrm{~L}}=4 \mathrm{~mol} \mathrm{~L}^{-1} \quad\left[\mathrm{H}_{2}\right]=\frac{2 \mathrm{~mol}}{1 / 2 \mathrm{~L}}=4 \mathrm{~mol} \mathrm{~L}^{-1}} \)
\(\mathrm{Q}=\frac{\left[\mathrm{CS}_{2}\right]\left[\mathrm{H}_{2}\right]^{4}}{\left[\mathrm{CH}_{4}\right]\left[\mathrm{H}_{2} \mathrm{~S}\right]^{2}}=\frac{2 \times(4)^{4}}{(2) \times(4)^{2}}=16 \)
Q > K_c
\(\therefore \) The reaction will proceed in the reverse direction to reach the equilibrium.

PCl₅ ⇌ PCl₃ + Cl₂
Given that [PCl₅]_initial = 1 mol; V = 1 dm³; K_C = 2

\(K_c={[PCl_3][Cl_2]\over [PCl_5]}\)
\(2={x\times x\over (1-x)}\)
2 - 2x = x²|
x² + 2x - 2 = 0
Solution for a quadratic equation
\(a x^{2}+b x+c=0 \text { are }, x=\frac{-b-\sqrt{b^{2}-4 a c}}{2 a}\)
a- = 1 b = 2 c = -2
\(x=\frac{-2-\sqrt{4-4 \times 1 \times-2}}{2 \times 1} \)
\(x=\frac{-2-\sqrt{12}}{2}=\frac{-2-\sqrt{4 \times 3}}{2}\)
\( x =\frac{-2-2 \sqrt{3}}{2} \)
\(=\frac{-2+2 \sqrt{3}}{2}, \frac{-2-2 \sqrt{3}}{2} \)
\(x =-1+\sqrt{3} ;-1-\sqrt{3} \)
\(=-1-\sqrt{3} \text { not possible } \)
Since x is + ve, 
x = -1 + 1.732
x = 0.732
Equilibrium concentration of
\( {\left[\mathrm{PCl}_{5}\right]_{\mathrm{eq}}=\frac{1-x}{1}=1-0.732=0.268 \mathrm{M}} \)
\({\left[\mathrm{PCl}_{3}\right]_{\mathrm{eq}}=\frac{x}{1}=\frac{0.732}{1}=0.732} \)
\({\left[\mathrm{Cl}_{2}\right]_{\mathrm{eq}}=\frac{x}{1}=\frac{0.732}{1}=0.732} \)

A₂(g) + B₂(g) ⇌ 2AB(g)

Total no. of moles = 1 – x + 1 – x + 2x = 2
\(K_p={(P_{AB})^2\over (P_{A_2})(P_{B_2})}={({2x\over 2}\times p)^2\over ({(1-x)\over 2}\times p)({1-x\over 2}\times p)}\)
\(K_p={4x^2\over (1-x)^2}\)
Given that K_p = 1; \({4x^2\over (1-x)^2}=1\)
\(\Rightarrow \) 4x² = (1 - x)²
\(\Rightarrow \) 4x² = 1 + x² - 2x
3x² + 2x - 1 = 0
\(X={-2\pm\sqrt{4-4\times 3\times -1}\over 2(3)}\)
\(X={-2\pm\sqrt{4+12}\over 6}\)
\(={-2\pm\sqrt{16}\over 6}\)
\(={-2+4\over 6};{-2-4\over 6}\)
\(={2\over 6};{-6\over 6}\)
X = 0.33 ; -1 (not possible)
\(\therefore\) [A₂]_eq = 1 - x = 1 - 0.33 = 0.67
[B₂]_eq = 1 - x = 1 - 0.33 = 0.67
[AB]_eq = 2X = 2 x 0.33 = 0.66.

Given \(m_{N_2}\) = 28 g \(m_{H_2}\) = 6g
V = 1 L
\((n_{N_2})_{initial}={28\over 28}=1\ mol\)
\((n_{H_2})_{initial}={6\over 2}=3\ mol\)
N₂(g) + 3H₂(g) ⇌ 2 NH₃(g)

\([NH_3]=({17\over 17})=1\ mol=1\ mol\)
Weight of N₂ = (no. of moles of N₂) × molar mass of N₂
= 0.5 x 28 = 14 g
Weight of H₂ = (no. of moles of H₂) × molar mass of H₂
= 1.5 x 2 = 3 g