The process: 2A(g) ⇌ A2(g) has = 8 × 10^8 atm^-1. If 'A' atoms are taken at 1 atm pressure, what sho — Chemical Equilibrium Chemistry Question

💡 Solution & Explanation

Reaction: $2\text{A}(g) \rightleftharpoons \text{A}_2(g)$; \quad $K_p = 8\times10^8\ \text{atm}^{-1}$ $K_p$ is very large ($\gg 1$), so the equilibrium strongly favours $\text{A}_2$. Nearly all A atoms dimerise. Starting with A atoms at total pressure 1 atm (all as A atoms). If $x$ atm of A remains at equilibrium: \begin{align*} P_{\text{A}} &= x \\ P_{\text{A}_2} &= \frac{1-x}{2} \approx \frac{1}{2} = 0.5\ \text{atm} \quad (\text{since } x \ll 1) \end{align*} Note: each pair of A atoms at pressure 1 atm gives $\frac{1}{2}$ atm of $\text{A}_2$ (since pressure is proportional to moles). \[ K_p = \frac{P_{\text{A}_2}}{P_{\text{A}}^2} = \frac{0.5}{x^2} = 8\times10^8 \] \[ x^2 = \frac{0.5}{8\times10^8} = 6.25\times10^{-10} \] \[ x = P_{\text{A}} = 2.5\times10^{-5}\ \text{atm} \] \textbf{Answer: B} — $P_{\text{A}} = 2.5\times10^{-5}\ \text{atm}$