Which is the most effective drying agent at 27°C? — Chemical Equilibrium Chemistry Question

💡 Solution & Explanation

The most effective drying agent is the one that establishes the \emph{lowest equilibrium water-vapour pressure}. The lower the $P_{\text{H}_2\text{O}}$, the more completely it absorbs moisture from the surroundings. For a solid hydrate equilibrium of the type: \[ \text{Hydrate}(s) \rightleftharpoons \text{Less-hydrated solid}(s) + n\,\text{H}_2\text{O}(g) \] \[ K_p = (P_{\text{H}_2\text{O}})^n \implies P_{\text{H}_2\text{O}} = K_p^{1/n} \] \textbf{Calculation:} \begin{enumerate} \item \textbf{SrCl$_2$.6H$_2$O $\rightleftharpoons$ SrCl$_2$.2H$_2$O + 4H$_2$O}: $K_p = 2.56\times10^{-10}$ atm$^4$ \[ P_{\text{H}_2\text{O}} = (2.56\times10^{-10})^{1/4} = (256\times10^{-12})^{1/4} = 4\times10^{-3}\ \text{atm} \] \item \textbf{Na$_2$HPO$_4$.12H$_2$O $\rightleftharpoons$ Na$_2$HPO$_4$.7H$_2$O + 5H$_2$O}: $K_p = 2.43\times10^{-13}$ atm$^5$ \[ P_{\text{H}_2\text{O}} = (2.43\times10^{-13})^{1/5} = (243\times10^{-15})^{1/5} = 3\times10^{-3}\ \text{atm} \] \item \textbf{Na$_2$SO$_4$.10H$_2$O $\rightleftharpoons$ Na$_2$SO$_4$(s) + 10H$_2$O}: $K_p = 1.024\times10^{-27}$ atm$^{10}$ \[ P_{\text{H}_2\text{O}} = (1.024\times10^{-27})^{1/10} = (1024\times10^{-30})^{1/10} = 2\times10^{-3}\ \text{atm} \] \end{enumerate} \textbf{Ranking of $P_{\text{H}_2\text{O}}$:} Na$_2$SO$_4$ ($2\times10^{-3}$) $<$ Na$_2$HPO$_4$.7H$_2$O ($3\times10^{-3}$) $<$ SrCl$_2$.2H$_2$O ($4\times10^{-3}$) Anhydrous Na$_2$SO$_4$ (option C) maintains the lowest water-vapour pressure — it is the most effective drying agent.