For the equilibrium: SrCl2.6(s) ⇌ SrCl2.2(s) + 4(g); = 8.1 × 10^-7 atm^4 at 27°C. If 1.642 L of air — Chemical Equilibrium Chemistry Question
Question
For the equilibrium: SrCl2.6$H_2O$(s) ⇌ SrCl2.2$H_2O$(s) + 4$H_2O$(g); $K_p$ = 8.1 × 10^-7 atm^4 at 27°C. If 1.642 L of air saturated with water vapour at 27°C is exposed to a large quantity of SrCl2.2$H_2O$(s), what mass of water vapour will be absorbed? Saturated vapour pressure of water at 27°C = 30.4 torr.
💡 Solution & Explanation
Reaction: $\text{SrCl}_2 \cdot 6\text{H}_2\text{O}(s) \rightleftharpoons \text{SrCl}_2 \cdot 2\text{H}_2\text{O}(s) + 4\text{H}_2\text{O}(g)$; \quad $K_p = 8.1\times10^{-7}\ \text{atm}^4$ \textbf{Step 1 — Equilibrium water vapour pressure:} \[ K_p = (P_{\text{H}_2\text{O}})^4 \implies P_{\text{H}_2\text{O}}^{(\text{eq})} = (8.1\times10^{-7})^{1/4} \] $(8.1\times10^{-7})^{1/4} = (81\times10^{-8})^{1/4} = 3\times10^{-2} = 0.03\ \text{atm}$ \textbf{Step 2 — Initial water vapour pressure:} \[ P_{\text{initial}} = 30.4\ \text{mmHg} = \frac{30.4}{760}\ \text{atm} = 0.04\ \text{atm} \] Since $P_{\text{initial}} = 0.04\ \text{atm} > P_{\text{eq}} = 0.03\ \text{atm}$, there is excess moisture that will be absorbed by SrCl$_2 \cdot 2$H$_2$O to drive the reverse reaction (forming SrCl$_2 \cdot 6$H$_2$O). \textbf{Step 3 — Pressure drop and moles absorbed:} \[ \Delta P = 0.04 - 0.03 = 0.01\ \text{atm} \] From $pV = nRT$ (V = 1.642 L, T = 300 K): \[ \Delta n_{\text{H}_2\text{O}} = \frac{\Delta P \cdot V}{RT} = \frac{0.01 \times 1.642}{0.0821 \times 300} = \frac{0.01642}{24.63} = 6.67\times10^{-4}\ \text{mol} \] \textbf{Step 4 — Mass of water absorbed:} \[ m = 6.67\times10^{-4} \times 18\ \text{g/mol} = 0.012\ \text{g} = 12\ \text{mg} \] \textbf{Answer: A} — 12 mg of water vapour is absorbed.