In the reaction: (g) + 3(g) ⇌ 2(g) + Heat. One mole of reacts with three moles of . If at equilibriu — Chemical Equilibrium Chemistry Question
Question
In the reaction: $N_2$(g) + 3$H_2$(g) ⇌ 2$NH_3$(g) + Heat. One mole of $N_2$ reacts with three moles of $H_2$. If at equilibrium 'x' moles of $N_2$ combined, then the value of 'x' in terms of $K_p$ and the total pressure of gases at equilibrium, P, is:
💡 Solution & Explanation
Reaction: $\text{N}_2(g) + 3\text{H}_2(g) \rightleftharpoons 2\text{NH}_3(g)$ Starting: 1 mol N$_2$, 3 mol H$_2$ (stoichiometric ratio). Let $x$ mol N$_2$ react. \textbf{ICE table:} \begin{center} \begin{tabular}{lcccc} & N$_2$ & H$_2$ & NH$_3$ & Total \\ Initial & 1 & 3 & 0 & 4 \\ Change & $-x$ & $-3x$ & $+2x$ & \\ Equil. & $1-x$ & $3(1-x)$ & $2x$ & $4-2x$ \\ \end{tabular} \end{center} For $x \ll 1$: total moles $\approx 4$. \textbf{Mole fractions and partial pressures:} \[ x_{\text{N}_2} \approx \frac{1}{4},\quad x_{\text{H}_2} \approx \frac{3}{4},\quad x_{\text{NH}_3} \approx \frac{x}{2} \] \[ p_{\text{N}_2} = \frac{P}{4},\quad p_{\text{H}_2} = \frac{3P}{4},\quad p_{\text{NH}_3} = \frac{xP}{2} \] \textbf{Kp expression:} \[ K_p = \frac{p_{\text{NH}_3}^2}{p_{\text{N}_2} \cdot p_{\text{H}_2}^3} = \frac{(xP/2)^2}{(P/4)(3P/4)^3} = \frac{x^2P^2/4}{P^4 \cdot 27/256} = \frac{64x^2}{27P^2} \] \textbf{Solving for $x$:} \[ x^2 = \frac{27K_p P^2}{64} \implies x = \frac{3P\sqrt{3K_p}}{8} \] \textbf{Answer: A} — $x = \dfrac{3P\sqrt{3K_p}}{8}$