.5(s) ⇌ .3(s) + 2(g), for the equilibrium is 1.0 × 10^-4 atm^2 at 25°C. What is the maximum pressure — Chemical Equilibrium Chemistry Question
Question
$CuSO_4$.5$H_2O$(s) ⇌ $CuSO_4$.3$H_2O$(s) + 2$H_2O$(g), $K_p$ for the equilibrium is 1.0 × 10^-4 atm^2 at 25°C. What is the maximum pressure of water vapour (moisture) in the atmosphere, below which the pentahydrate is efflorescent?
💡 Solution & Explanation
Reaction: $\text{CuSO}_4 \cdot 5\text{H}_2\text{O}(s) \rightleftharpoons \text{CuSO}_4 \cdot 3\text{H}_2\text{O}(s) + 2\text{H}_2\text{O}(g)$; \quad $K_p = 1.0\times10^{-4}\ \text{atm}^2$ \textbf{Equilibrium water vapour pressure:} \[ K_p = (P_{\text{H}_2\text{O}})^2 \implies P_{\text{H}_2\text{O}} = \sqrt{1.0\times10^{-4}} = 0.01\ \text{atm} \] Converting to mm Hg: $P_{\text{H}_2\text{O}} = 0.01 \times 760 = 7.60\ \text{mm Hg}$ \textbf{Efflorescence condition:} A hydrate effloresces when the atmospheric partial pressure of water vapour is \emph{less than} the equilibrium vapour pressure established by the hydrate. If the atmosphere has less moisture than the hydrate releases at equilibrium, the hydrate loses water spontaneously. Therefore, CuSO$_4 \cdot 5$H$_2$O will effloresce when the atmospheric moisture pressure is below \textbf{7.60 mm Hg}. \textbf{Answer: A} — Maximum water vapour pressure for efflorescence is 7.60 mm