At what relative humidity will .10(s) be efflorescent when exposed to air at 27°C? — Chemical Equilibrium Chemistry Question

💡 Solution & Explanation

Step 1 - Understand the Chemical Basis of Efflorescence Efflorescence is the spontaneous loss of water of crystallization when a hydrated salt is exposed to air whose water vapor pressure is lower than the equilibrium water vapor pressure of the hydrated system: \[p_{\ce{H2O}, \text{atm}} < p_{\ce{H2O}, \text{eq}} \implies \text{efflorescence}\] Step 2 - Determine the Equilibrium Partial Pressure of Water Vapor The dehydration equilibrium of Glauber's salt: \[\ce{Na2SO4.10H2O(s) <=> Na2SO4(s) + 10H2O(g)}\] Both solids have activity = 1, so: \[K_p = (p_{\ce{H2O}, \text{eq}})^{10} = 1.024 \times 10^{-27} \text{ atm}^{10}\] Taking the tenth root: \[p_{\ce{H2O}, \text{eq}} = (1.024 \times 10^{-27})^{1/10}\] Rewriting: $1.024 \times 10^{-27} = 1024 \times 10^{-30} = 2^{10} \times 10^{-30}$ \[p_{\ce{H2O}, \text{eq}} = (2^{10} \times 10^{-30})^{1/10} = 2 \times 10^{-3}\text{ atm} = 0.002\text{ atm}\] Step 3 - Convert to Relative Humidity Relative humidity is defined as: \[\text{RH} = \frac{p_{\ce{H2O}, \text{atm}}}{p^\circ_{\ce{H2O}}} \times 100\%\] For efflorescence: $p_{\ce{H2O}, \text{atm}} < 0.002\text{ atm}$, and $p^\circ_{\ce{H2O}} = 0.04\text{ atm}$: \[\text{RH} < \frac{0.002}{0.04} \times 100\% = \frac{1}{20} \times 100\% = 5\%\] The salt will be efflorescent when $\text{RH} < 5\%$. Step 4 - Evaluate Each Option * **(A) below 50%**: Incorrect. At 40% RH (below 50%), efflorescence does NOT occur since $40\% > 5\%$. * **(B) above 50%**: Incorrect. High humidity prevents dehydration; $p_{\ce{H2O}}$ exceeds the equilibrium threshold. * **(C) above 5%**: Incorrect. Above 5% RH, the atmospheric water vapor pressure exceeds equilibrium, so the salt remains stable. * **(D) below 5%**: Correct. Efflorescence occurs only when RH $< 5\%$. \[\boxed{\text{D}}\]