Calculate heat required to convert 9 g of liquid water to water vapours from following equations.…
Calculate heat required to convert 9 g of liquid water to water vapours from following equations.
$\begin{aligned}
& \mathrm{H}_{2_{(\mathrm{g})}}+\frac{1}{2} \mathrm{O}_{2_{(\mathrm{g})}} \longrightarrow \mathrm{H}_2 \mathrm{O}_{(\mathrm{g})} \Delta \mathrm{H}=-57 \mathrm{kCal} \\
& \mathrm{H}_{2_{(\mathrm{g})}}+\frac{1}{2} \mathrm{O}_{2_{(\mathrm{g})}} \longrightarrow \mathrm{H}_2 \mathrm{O}_{(\mathrm{g})} \Delta \mathrm{H}=-68.3 \mathrm{kCal}
\end{aligned}$
5.65 k Cal
6.28 k Cal
$7.05 \mathrm{k} \mathrm{Cal}$
9.72 k Cal
Solution
Given:
- Mass of water: 9 g ,
- $\Delta \mathrm{H}$ for $\mathrm{H}_2 \mathrm{O}(\mathrm{l}) \rightarrow \mathrm{H}_2 \mathrm{O}(\mathrm{g})=68.3 \mathrm{kCal} / \mathrm{mol}-57 \mathrm{kCal} / \mathrm{mol}=11.3 \mathrm{kCal} / \mathrm{mol}$. Moles of Water:
$\text { Moles }=\frac{\text { Mass }}{\text { Molar Mass }}=\frac{9}{18}=0.5 \mathrm{~mol} .$ Heat Required:
$\text { Heat }=\text { Moles } \times \Delta H=0.5 \times 11.3=5.65 \mathrm{kCal}$ Answer: 5.65 kCal, Option 1.