If a is the fraction of ammonia present by volume in an equilibrium mixture made from 1 volume of an — Chemical Equilibrium Chemistry Question

💡 Solution & Explanation

Step 1 - Set up the ICE table For \(\ce{N2(g) + 3H2(g) <=> 2NH3(g)}\), starting with 1 mol N2 and 3 mol H2: | | N2 | H2 | NH3 | |--|--|--|--| | Initial | 1 | 3 | 0 | | Change | -y | -3y | +2y | | Equilibrium | 1-y | 3(1-y) | 2y | Total moles at equilibrium: \(n_{\text{total}} = 4 - 2y\) Step 2 - Express y in terms of a Volume fraction = mole fraction for ideal gases, so: \[a = \frac{2y}{4 - 2y} \implies y = \frac{2a}{1+a}\] Step 3 - Find equilibrium mole fractions \[x_{\ce{N2}} = \frac{1-a}{4}, \quad x_{\ce{H2}} = \frac{3(1-a)}{4}, \quad x_{\ce{NH3}} = a\] Partial pressures: \[p_{\ce{NH3}} = aP, \quad p_{\ce{N2}} = \frac{(1-a)P}{4}, \quad p_{\ce{H2}} = \frac{3(1-a)P}{4}\] Step 4 - Calculate Kp \[K_p = \frac{p_{\ce{NH3}}^2}{p_{\ce{N2}} \cdot p_{\ce{H2}}^3} = \frac{(aP)^2}{\frac{(1-a)P}{4} \cdot \left(\frac{3(1-a)P}{4}\right)^3}\] \[K_p = \frac{a^2 P^2}{\frac{27(1-a)^4 P^4}{256}} = \frac{256 a^2}{27(1-a)^4 P^2}\] Step 5 - Match to options Taking square root: \(\sqrt{K_p} = \frac{16}{3\sqrt{3}} \cdot \frac{a}{(1-a)^2 P}\) Rearranging: \(\frac{a}{(1-a)^2} \propto P\), i.e., \(\frac{a}{(1-a)^2} = \text{const} \times P\) * **Option (A)**: Incorrect. Has (1-a) not (1-a)^2. * **Option (B)**: Correct. \(\frac{a}{(1-a)^2} \propto P\) — matches our derivation. * **Option (C)**: Incorrect. Has P^2 dependence, which is wrong. * **Option (D)**: Incorrect. Missing P dependence. \[\boxed{\text{B}}\]