Standard enthalpy of formation of water is $-286 \mathrm{~kJ} \mathrm{~mol}^{-1}$. When $1800 \mathrm{mg}$…
Standard enthalpy of formation of water is $-286 \mathrm{~kJ} \mathrm{~mol}^{-1}$. When $1800 \mathrm{mg}$ of water is formed from its constituent elements in their standard states the amount of energy liberated is
$2 \cdot 86 \mathrm{~kJ}$
$5 \cdot 72 \mathrm{~kJ}$
$57 \cdot 2 \mathrm{~kJ}$
$28 \cdot 6 \mathrm{~kJ}$
Solution
$\mathrm{H}_{2(\mathrm{~g})}+\frac{1}{2} \mathrm{O}_{2(\mathrm{~g})} \longrightarrow \mathrm{H}_{2} \mathrm{O}_{(\ell)} \quad \Delta \mathrm{H}_{\mathrm{f}}^{0}=-286 \mathrm{~kJ} \mathrm{~mol}^{-1}$
For $18 \mathrm{~g}$ of $\mathrm{H}_{2} \mathrm{O}$, amount of energy liberated $=286 \mathrm{~kJ} \mathrm{~mol}^{-1}$
$\therefore$ For $1.8 \mathrm{~g}$ of $\mathrm{H}_{2} \mathrm{O}$, amount of energy liberated $=\frac{1.8 \times 286}{18}=28.6 \mathrm{~kJ}$