If $\begin{aligned}{\mathrm{CuSO}_{4} .5 \mathrm{H}_{2} \mathrm{O}_{(\mathrm{s})} ightleftharpoons}{ } &…
$\mathrm{K}_{\mathrm{p}}=1.086 \times 10^{-4} \mathrm{~atm}^{2}$ at $25^{\circ} \mathrm{C}$. The efflorescent nature of $\mathrm{CuSO}_{4} .5 \mathrm{H}_{2} \mathrm{O}$ can be noticed when vapour pressure of $\mathrm{H}_{2} \mathrm{O}$ in atmosphere is
- $>7.29 \mathrm{~mm}$
- $ < 7.92 \mathrm{~mm}$
- $\geq 7.92 \mathrm{~mm}$
- None
Solution
For the reaction
$\mathrm{CuSO}_{4} \cdot 5 \mathrm{H}_{2} \mathrm{O}_{(\mathrm{s})}+\mathrm{CuSO}_{4} \cdot 3 \mathrm{H}_{2} \mathrm{O}_{(\mathrm{s})}+2 \mathrm{H}_{2} \mathrm{O}_{(\mathrm{i})}$
$\mathrm{K}_{\mathrm{p}}=\left(\mathrm{p}_{\mathrm{H}_{2} \mathrm{O}}^{\prime}ight)^{2}=1.086 \times 10^{-4}$
$\mathrm{p}_{\mathrm{H}_{2} \mathrm{O}}^{\prime}=1.042 \times 10^{-2} \mathrm{~atm}=7.92 \mathrm{~mm}$
$\mathrm{Q}$ If $\mathrm{p}_{\mathrm{H}_{2} \mathrm{O}}^{\prime}$ at $25^{\circ} \mathrm{C} < 7.92 \mathrm{~mm}$ only then, reaction will proceed in forward direction, .
Asked in: JEE-TOPICTESTS-CHEMISTRY