JEE Main 2021 — Chemical Equilibrium Question with Solution

[p = Total pressure at equilibrium = 1.9 atm]

Now, at equilibrium pV = (1 + 2x)RT

$$\Rightarrow 1+2x=\frac{pV}{RT}=\frac{1.9\times 25}{0.082\times 300}=1.93$$

[V = 25 L, R = 0.082 L atm mol^$$-$$1 K^$$-$$1 T = 300 K]

$$\Rightarrow x=\frac{1.93-1}{2}=0.465$$

$$\Rightarrow K_p=\frac{p_A\times p_B^2}{p_{A{B}_2}}\Rightarrow \frac{\left(\frac{x}{1+2x}p\right)\times {\left(\frac{2x}{1+2x}p\right)}^2}{\left(\frac{1-x}{1+2x}p\right)}$$

$$=\frac{4x^3\times p^3}{{(1+2x)}^3}\times \frac{(1+2x)}{(1-x)\times p}=\frac{4x^3\times p^2}{{(1+2x)}^2\times (1-x)}$$

$$=\frac{4\times {(0.465)}^3\times {(1.9)}^2}{{(1+2\times 0.465)}^2\times (1-0.465)}=0.7285$$ atm

$$=72.85\times 10^{-2}$$ atm $$\simeq 73\times 10^{-2}=x\times 10^{-2}$$

$$\therefore$$ $$x=73$$