JEE MAIN - Chemistry (2023 - 31st January Evening Shift - No. 17)
Enthalpies of formation of $\mathrm{CCl}_{4}(\mathrm{~g}), \mathrm{H}_{2} \mathrm{O}(\mathrm{g}), \mathrm{CO}_{2}(\mathrm{~g})$ and $\mathrm{HCl}(\mathrm{g})$ are $-105,-242,-394$ and $-92 ~\mathrm{kJ}$ $\mathrm{mol}^{-1}$ respectively. The magnitude of enthalpy of the reaction given below is _________ $\mathrm{kJ} ~\mathrm{mol}^{-1}$. (nearest integer)
$\mathrm{CCl}_{4}(\mathrm{~g})+2 \mathrm{H}_{2} \mathrm{O}(\mathrm{g}) \rightarrow \mathrm{CO}_{2}(\mathrm{~g})+4 \mathrm{HCl}(\mathrm{g})$
$\mathrm{CCl}_{4}(\mathrm{~g})+2 \mathrm{H}_{2} \mathrm{O}(\mathrm{g}) \rightarrow \mathrm{CO}_{2}(\mathrm{~g})+4 \mathrm{HCl}(\mathrm{g})$
Answer
173
Explanation
$$
\mathrm{CCl}_4(\mathrm{~g})+2 \mathrm{H}_2 \mathrm{O}(\mathrm{g}) \rightarrow \mathrm{CO}_2(\mathrm{~g})+4 \mathrm{HCl}(\mathrm{g})
$$
Enthalpy of above reaction
$$ \begin{aligned} & =\Delta \mathrm{H}_f\left(\mathrm{CO}_2(\mathrm{~g})\right)+4 \Delta \mathrm{H}_{\mathrm{f}}(\mathrm{HCl}(\mathrm{g}))-\Delta \mathrm{H}_{\mathrm{f}}\left(\mathrm{CCl}_4\right)-2 \Delta \mathrm{H}_{\mathrm{f}}\left(\mathrm{H}_2 \mathrm{O}\right) \\\\ & =-394-4 \times 92+105+2 \times 242 \\\\ & =-394-368+105+484 \\\\ & =-173 \mathrm{~kJ} \mathrm{~mol}^{-1} \end{aligned} $$
Hence the magnitude of this will be $173 \mathrm{~kJ} \mathrm{~mol}^{-1}$.
Enthalpy of above reaction
$$ \begin{aligned} & =\Delta \mathrm{H}_f\left(\mathrm{CO}_2(\mathrm{~g})\right)+4 \Delta \mathrm{H}_{\mathrm{f}}(\mathrm{HCl}(\mathrm{g}))-\Delta \mathrm{H}_{\mathrm{f}}\left(\mathrm{CCl}_4\right)-2 \Delta \mathrm{H}_{\mathrm{f}}\left(\mathrm{H}_2 \mathrm{O}\right) \\\\ & =-394-4 \times 92+105+2 \times 242 \\\\ & =-394-368+105+484 \\\\ & =-173 \mathrm{~kJ} \mathrm{~mol}^{-1} \end{aligned} $$
Hence the magnitude of this will be $173 \mathrm{~kJ} \mathrm{~mol}^{-1}$.
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