for the process: .5(s) ⇌ .3(s) + 2(g) is 1.21 × 10^-4 atm^2 at certain temperature. If aqueous tensi — Chemical Equilibrium Chemistry Question
Question
$K_p$ for the process: $CuSO_4$.5$H_2O$(s) ⇌ $CuSO_4$.3$H_2O$(s) + 2$H_2O$(g) is 1.21 × 10^-4 atm^2 at certain temperature. If aqueous tension at that temperature is 40 torr, then at what relative humidity of air will $CuSO_4$.5$H_2O$ effloresce?
💡 Solution & Explanation
Same reaction as Q87 with $K_p = 1.21\times10^{-4}\ \text{atm}^2$. Aqueous tension (saturated vapour pressure of water) = 40 torr. \textbf{Step 1 — Equilibrium water vapour pressure of the hydrate:} \[ P_{\text{H}_2\text{O}}^{(\text{eq})} = \sqrt{1.21\times10^{-4}} = 0.011\ \text{atm} = 0.011 \times 760\ \text{torr} = 8.36\ \text{torr} \] \textbf{Step 2 — Relative humidity (RH) condition for efflorescence:} \[ \text{RH} = \frac{P_{\text{H}_2\text{O}}(\text{air})}{\text{Aqueous tension}} \times 100\% = \frac{P_{\text{H}_2\text{O}}}{40} \times 100\% \] For efflorescence, the air must have $P_{\text{H}_2\text{O}} < 8.36\ \text{torr}$: \[ \text{RH} < \frac{8.36}{40} \times 100\% = 20.9\% \] \textbf{Answer: D} — CuSO$_4 \cdot 5$H$_2$O effloresces when RH is below 20.9\%