Verified Solution • 2025–26 Session moderate

QJEEM23Ch18da: JEE Main 2023 Chemistry - Class 11 JEEM23Ch18da Solution: Enthalpies of formation of and are and respectively. The…

Enthalpies of formation of CCl4( g),H2O(g),CO2( g)\mathrm{CCl}_{4}(\mathrm{~g}), \mathrm{H}_{2} \mathrm{O}(\mathrm{g}), \mathrm{CO}_{2}(\mathrm{~g}) and HCl(g)\mathrm{HCl}(\mathrm{g}) are −105,−242,−394-105,-242,-394 and −92 kJ-92 ~\mathrm{kJ} mol−1\mathrm{mol}^{-1} respectively. The magnitude of enthalpy of the reaction given below is _________ kJ mol−1\mathrm{kJ} ~\mathrm{mol}^{-1} . (nearest integer)

CCl4( g)+2H2O(g)→CO2( g)+4HCl(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})

Step-by-Step Verified Solution

CCl4( g)+2H2O(g)→CO2( g)+4HCl(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})

Enthalpy of above reaction

amp;=ΔHf(CO2( g))+4ΔHf(HCl(g))−ΔHf(CCl4)−2ΔHf(H2O)amp;=−394−4×92+105+2×242amp;=−394−368+105+484amp;=−173 kJ mol−1\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 kJ mol−1173 \mathrm{~kJ} \mathrm{~mol}^{-1} .

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