Question

For the following gas phase equilibrium reaction at constant temperature, NH₃(g) = 1/2 N₂(g) + 3/2 H₂(g) if the total pressure is √3 atm and the pressure equilibrium constant (K___p) is 9 atm, then the degree of dissociation is given as (x × 10⁻²)⁻¹/². The value of x is_______ (nearest integer)}

Hint:

In $K_p$ expressions, always find the total moles first to determine the mole fraction, then multiply by total pressure ($P$) for partial pressures.

Answer 1

Correct Answer: 125

Step 1: Understanding the Concept:
We use the expression for $K_p$ in terms of the degree of dissociation ($\alpha$) and total pressure ($P$).
Step 2: Detailed Explanation:
Reaction: $NH_3(g) \rightleftharpoons \frac{1}{2}N_2(g) + \frac{3}{2}H_2(g)$ Initial moles: 1, 0, 0 Equilibrium: $(1-\alpha)$, $\alpha/2$, $3\alpha/2$ Total moles = $(1-\alpha) + \alpha/2 + 3\alpha/2 = 1 + \alpha$. Mole fractions: $X_{NH_3} = \frac{1-\alpha}{1+\alpha}$, $X_{N_2} = \frac{\alpha/2}{1+\alpha}$, $X_{H_2} = \frac{3\alpha/2}{1+\alpha}$. $K_p = \frac{(P_{N_2})^{1/2} (P_{H_2})^{3/2}}{P_{NH_3}} = \frac{[\frac{\alpha/2}{1+\alpha} P]^{1/2} [\frac{3\alpha/2}{1+\alpha} P]^{3/2}}{\frac{1-\alpha}{1+\alpha} P} = \frac{\sqrt{27}}{4} \frac{\alpha^2 P}{1-\alpha^2}$.
Step 3: Solving for \(\alpha\):
Given $K_p = 9$ and $P = \sqrt{3}$: $9 = \frac{3\sqrt{3}}{4} \frac{\alpha^2 \sqrt{3}}{1-\alpha^2} \implies 9 = \frac{9}{4} \frac{\alpha^2}{1-\alpha^2} \implies 4 = \frac{\alpha^2}{1-\alpha^2}$ $4 - 4\alpha^2 = \alpha^2 \implies 5\alpha^2 = 4 \implies \alpha = \sqrt{4/5}$. $\alpha = (5/4)^{-1/2} = (1.25)^{-1/2} = (125 \times 10^{-2})^{-1/2}$. Comparing with $(x \times 10^{-2})^{-1/2}$, we get $x = 125$.
Step 4: Final Answer:
The value of x is 125.