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JEE Main 202528 JanuaryEvening Shift
Exam
JEE Main
Year
2025
Date
28 January
Shift
Evening Shift
Subject
Chemistry
Chapter
Thermodynamics (C)

JEE Main 2025 · 28 January · Evening Shift · Chemistry · Thermodynamics (C)

Consider the following data : Heat of formation of CO_2(~g)=-393.5...

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)

Correct answer

0

Step-by-step solution

ΔHf[CO2( g)]=−393.5 kJ/moleΔHf[H2O(ℓ)]=−286.0 kJ/moleΔHc[C6H6]=−3267.0 kJ/moleΔHfC6H6=(?)C6H6+152O2( g)⟶6CO2( g)+3H2O(ℓ)ΔHR=ΔHC=ΣΔHf(P)−ΣΔHf(R)−3267=6×(−393.5)+3(−286)−ΔHf(C6H6)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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