Verified Solution • 2025–26 Session moderate

QJEEM25Ch7ac4: JEE Main 2025 Chemistry - Class 11 JEEM25Ch7ac4 Solution: Consider the following data : Heat of formation of Heat of…

Consider the following data :

Heat of formation of CO2(g)=−393.5 kJ mol−1\mathrm{CO}_2(\mathrm{g})=-393.5 \mathrm{~kJ} \mathrm{~mol}{ }^{-1}

Heat of formation of H2O(l)=−286.0 kJ mol−1\mathrm{H}_2 \mathrm{O}(\mathrm{l})=-286.0 \mathrm{~kJ} \mathrm{~mol}{ }^{-1}

Heat of combustion of benzene =−3267.0 kJ mol−1=-3267.0 \mathrm{~kJ} \mathrm{~mol}^{-1}

The heat of formation of benzene is __________ kJmol−1\mathrm{kJ} \mathrm{mol}^{-1} . (Nearest integer)

Step-by-Step Verified Solution

amp;ΔHf[CO2( g)]=−393.5 kJ/moleamp;ΔHf[H2O(ℓ)]=−286.0 kJ/moleamp;ΔHc[C6H6]=−3267.0 kJ/moleamp;ΔHfC6H6=(?)amp;C6H6+152O2( g)⟶6CO2( g)+3H2O(ℓ)amp;ΔHR=ΔHC=ΣΔHf(P)−ΣΔHf(R)amp;−3267=6×(−393.5)+3(−286)−ΔHf(C6H6)amp;ΔHf(C6H6)=48 kJ/mole\begin{aligned}& \Delta \mathrm{H}_{\mathrm{f}}\left[\mathrm{CO}_2(\mathrm{~g})\right]=-393.5 \mathrm{~kJ} / \mathrm{mole} \\& \Delta \mathrm{H}_{\mathrm{f}}\left[\mathrm{H}_2 \mathrm{O}(\ell)\right]=-286.0 \mathrm{~kJ} / \mathrm{mole} \\& \Delta \mathrm{H}_{\mathrm{c}}\left[\mathrm{C}_6 \mathrm{H}_6\right]=-3267.0 \mathrm{~kJ} / \mathrm{mole} \\& \Delta \mathrm{H}_{\mathrm{f}} \mathrm{C}_6 \mathrm{H}_6=(?) \\& \mathrm{C}_6 \mathrm{H}_6+\frac{15}{2} \mathrm{O}_2(\mathrm{~g}) \longrightarrow 6 \mathrm{CO}_2(\mathrm{~g})+3 \mathrm{H}_2 \mathrm{O}(\ell) \\& \Delta \mathrm{H}_{\mathrm{R}}=\Delta \mathrm{H}_{\mathrm{C}}=\Sigma \Delta \mathrm{H}_{\mathrm{f}}(\mathrm{P})-\Sigma \Delta \mathrm{H}_{\mathrm{f}}(\mathrm{R}) \\& -3267=6 \times(-393.5)+3(-286)-\Delta \mathrm{H}_{\mathrm{f}}\left(\mathrm{C}_6 \mathrm{H}_6\right) \\& \Delta \mathrm{H}_{\mathrm{f}}\left(\mathrm{C}_6 \mathrm{H}_6\right)=48 \mathrm{~kJ} / \mathrm{mole}\end{aligned}

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