JEE Main 2022 — Chemical Equilibrium Question with Solution

Here 4 moles of inert gas argon also present.

$$\therefore$$ Total moles of mixture present at equilibrium,

n_T = 5 + x + 4

= 9 + x

At equilibrium, total pressure (p_T) = 6 atm

Volume (v) = 100 L

Temperature = 610 K

$$\therefore$$ Using ideal gas equation,

$$P_{T}V=n_{T}RT$$

$$\Rightarrow 6\times 100=(9+x)\times 0.082\times 610$$

$$\Rightarrow x=3$$

Now,

$$K_P=\frac{P_{PC{l}_3}\times P_{C{l}_2}}{P_{PC{l}_5}}$$

$$=\frac{\left[\frac{3}{9+3}\times 6\right]\times \left[\frac{3}{9+3}\times 6\right]}{\left[\frac{5-3}{9+3}\times 6\right]}$$

$$=\frac{27}{12}$$

$$=\frac{9}{4}$$

= 2.25 atm

Note :

Inert gas always contribute to total mole and pressure calculation.