JEE Chemistry 2026 Question with Solution

Given:

• Reaction: $\mathrm{NH_3(g)} \rightleftharpoons \frac{1}{2}\mathrm{N_2(g)} + \frac{3}{2}\mathrm{H_2(g)}$
• Total pressure: $P = \sqrt{3} \, \text{atm}$
• Pressure equilibrium constant: $K_p = 9 \, \text{atm}$

Find: The value of $x$ in $(x \times 10^{-2})^{-1/2}$.

From the solution working, let the degree of dissociation be $\alpha$.

Initial moles are:

• $1$ for $\mathrm{NH_3}$
• $0$ for $\mathrm{N_2}$
• $0$ for $\mathrm{H_2}$

At equilibrium:

• $\mathrm{NH_3} = 1-\alpha$
• $\mathrm{N_2} = \alpha/2$
• $\mathrm{H_2} = 3\alpha/2$

Total moles at equilibrium:

$$(1-\alpha) + \frac{\alpha}{2} + \frac{3\alpha}{2} = 1 + \alpha$$

Therefore, the mole fractions are:

$$X_{\mathrm{NH_3}} = \frac{1-\alpha}{1+\alpha}, \qquad X_{\mathrm{N_2}} = \frac{\alpha/2}{1+\alpha}, \qquad X_{\mathrm{H_2}} = \frac{3\alpha/2}{1+\alpha}$$

Using partial pressures in the expression for $K_p$:

$$K_p = \frac{(P_{\mathrm{N_2}})^{1/2}(P_{\mathrm{H_2}})^{3/2}}{P_{\mathrm{NH_3}}}$$

Substituting the partial pressures:

$$K_p = \frac{\left[\frac{\alpha/2}{1+\alpha}P\right]^{1/2}\left[\frac{3\alpha/2}{1+\alpha}P\right]^{3/2}}{\frac{1-\alpha}{1+\alpha}P} = \frac{\sqrt{27}}{4}\frac{\alpha^2 P}{1-\alpha^2}$$

Now substitute $K_p = 9$ and $P = \sqrt{3}$:

$$9 = \frac{3\sqrt{3}}{4}\frac{\alpha^2\sqrt{3}}{1-\alpha^2}$$ $$9 = \frac{9}{4}\frac{\alpha^2}{1-\alpha^2}$$ $$4 = \frac{\alpha^2}{1-\alpha^2}$$ $$4 - 4\alpha^2 = \alpha^2$$ $$5\alpha^2 = 4$$ $$\alpha = \sqrt{\frac{4}{5}}$$

Rewrite this as:

$$\alpha = \left(\frac{5}{4}\right)^{-1/2} = (1.25)^{-1/2} = (125 \times 10^{-2})^{-1/2}$$

Comparing with the given form $(x \times 10^{-2})^{-1/2}$, we get $x = 125$.

Therefore, the value of $x$ is $125$.

Given: The dissociation of $\mathrm{NH_3}$ is represented in terms of $\alpha$.

Find: Express $\alpha$ in the form $(x \times 10^{-2})^{-1/2}$.

The key step is to first calculate total moles after dissociation. Because pressure terms depend on mole fraction, missing the factor $1+\alpha$ gives the wrong result.

From the equilibrium composition:

$$\text{Total moles} = 1+\alpha$$

Hence,

$$P_{\mathrm{NH_3}} = \frac{1-\alpha}{1+\alpha}P, \qquad P_{\mathrm{N_2}} = \frac{\alpha/2}{1+\alpha}P, \qquad P_{\mathrm{H_2}} = \frac{3\alpha/2}{1+\alpha}P$$

Substitute into the equilibrium expression exactly as shown in the solution:

$$K_p = \frac{(P_{\mathrm{N_2}})^{1/2}(P_{\mathrm{H_2}})^{3/2}}{P_{\mathrm{NH_3}}}$$

This reduces to:

$$K_p = \frac{\sqrt{27}}{4}\frac{\alpha^2 P}{1-\alpha^2}$$

Now use the given values:

$$9 = \frac{\sqrt{27}}{4}\frac{\alpha^2\sqrt{3}}{1-\alpha^2}$$

Since $\sqrt{27}\sqrt{3} = 9$,

$$9 = \frac{9}{4}\frac{\alpha^2}{1-\alpha^2}$$

So,

$$4 = \frac{\alpha^2}{1-\alpha^2}$$ $$5\alpha^2 = 4$$ $$\alpha = \sqrt{\frac{4}{5}}$$

Now convert to the required form:

$$\sqrt{\frac{4}{5}} = \left(\frac{5}{4}\right)^{-1/2} = (1.25)^{-1/2} = (125 \times 10^{-2})^{-1/2}$$

Therefore, $x = 125$.