2NO (g) + O₂(g) ⇌ 2NO₂(g)

\(K_p={(p_{NO_2})^2\over (P_{NO})^2(P_{o_2})}={0.96\times 0.96\over 0.04\times 0.04\times 0.52}\)
K_p = 1.017 x 10³.

If we know the value of k_c and Q, the reaction quotient, we can predict the direction of a reaction
If Q = Kc; the reaction is in equilibrium state.
If Q > Kc: the reaction will proceed in the reverse direction i.e., formation of reactants.
If Q < Kc: the reaction will proceed in the forward direction i.e., formation of products.

\(\Delta n_{g}=\sum n p_{(g)}-\sum n R_{(g)}\)
As \(\Delta n_{p}(g)\) is greater \(\Delta n_{g}=+v e\)
\( \therefore K_{p}=K_{c}(R T)^{+v e} \)
\(\therefore K_{p}>K_{c} \)
So K is smaller than K_p.

i) \(K_{p}=K_{c}(R T)^{\Delta n_{g}}\)
K_p = Equilibrium constant in term of partial Pressures.
K_c = Equilibrium constant in term of concentration.
R = Gas constant; T = Temperature
\(\Delta \mathrm{n}_{\mathrm{g}}\) = Difference between the sum of number of moles of products and the sum of number of moles of reactants in gas phases.
ii) Synthesis of HI:
\( \mathrm{H}_{2(\mathrm{~g})}+\mathrm{I}_{2(\mathrm{~g})} \rightleftharpoons 2 \mathrm{HI}_{(\mathrm{g})} \)
\(\Delta n_{g}=0 \therefore K_{p}=K_{c}(R T) \Delta n_{g}\)
\(K_{p}=K_{c}(R T)^{\circ} \)
\(K_{p}=K_{c} \text {. }\)

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
\( {\left[\mathrm{PCl}_{5}\right]_{\mathrm{eq}}=\frac{1-x}{1}=1-0.732=0.268 \mathrm{M}} \)
\({\left[\mathrm{PCl}_{3}\right]_{\mathrm{eq}}=\frac{x}{1}=\frac{0.732}{1}=0.732} \)
\({\left[\mathrm{Cl}_{2}\right]_{\mathrm{eq}}=\frac{x}{1}=\frac{0.732}{1}=0.732} \)

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

N₂O₂(g) \(\rightleftharpoons \) 2NO₂(g)
Kc = 0.21 at 373 K. The concentrations N₂O₄ and NO₂ are found to be 0.125 mol dm^-3 and 0.5 mol dm^-3 respectively at a given time. From the above information we can predict the direction of reaction as follows.
\(Q={[NO_2]^2\over [N_2O_4]}={0.5\times 0.5\over 0.125}=2\)
The Q value is greater than K_c. Hence, the reaction will proceed in the reverse direction until the Q value reaches 0.21.

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.

The equilibrium constant is a constant and it is for equilibrium condition. But 'Q', the reaction quotient is not a constant as it is for non - equilibrium condition. 'Q' is the ratio of the product of active masses of a reaction products raised to the respective stoichiometric coefficients in the balanced chemical equation to that of the reactants, under non - equilibrium conditions.
i) If Q = K_c; it is equilibrium
ii) If Q > K_c.; the reaction will proceed in reverse direction
iii) If Q < K_c ; the reaction will proceed in forward direction