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JEE Main 20245 AprilEvening Shift
Exam
JEE Main
Year
2024
Date
5 April
Shift
Evening Shift
Subject
Chemistry
Chapter
Thermodynamics (C)

JEE Main 2024 · 5 April · Evening Shift · Chemistry · Thermodynamics (C)

Combustion of 1 mole of benzene is expressed at (C_6...

Combustion of 1 mole of benzene is expressed at (\mathrm{C}_6 \mathrm{H}_6(\mathrm{l})+\frac{15}{2} \mathrm{O}_2(\mathrm{ g}) \rightarrow 6 \mathrm{CO}_2(\mathrm{ g})+3 \mathrm{H}_2 \mathrm{O}(\mathrm{l}) \text{. }) The standard enthalpy of combustion of (2 \mathrm{ mol}) of benzene is (-^{\prime} x^{\prime} \mathrm{kJ}). x=x= ______ Given: 1. standard Enthalpy of formation of (1 \mathrm{ mol}) of (\mathrm{C}_6 \mathrm{H}_6(\mathrm{l})), for the reaction (6 \mathrm{C}) (graphite) (+3 \mathrm{H}_2(\mathrm{ g}) \rightarrow \mathrm{C}_6 \mathrm{H}_6(\mathrm{l})) is (48.5 \mathrm{ kJ} \mathrm{ mol}^{-1}). 2. Standard Enthalpy of formation of (1 \mathrm{ mol}) of (\mathrm{CO}_2(\mathrm{ g})), for the reaction (\mathrm{C}) (graphite) (+\mathrm{O}_2(\mathrm{ g}) \rightarrow \mathrm{CO}_2(\mathrm{ g})) is (-393.5 \mathrm{ kJ} \mathrm{ mol}^{-1}). 3. Standard and Enthalpy of formation of (1 \mathrm{ mol}) of (\mathrm{H}_2 \mathrm{O}(\mathrm{l})), for the reaction (\mathrm{H}_2(\mathrm{ g})+\frac{1}{2} \mathrm{O}_2(\mathrm{ g}) \rightarrow \mathrm{H}_2 \mathrm{O}(\mathrm{l})) is (-286 \mathrm{ kJ} \mathrm{ mol}^{-1}).

Correct answer

0

Step-by-step solution

6C( graphite )+3H2(g)→C6H6(ℓ);ΔH=48.5kJ/molC( graphite )+O2(g)→CO2(g);ΔH=−393.5kJ/molH2(g)+12(g)⟶H2O(ℓ);ΔH=−286kJ/mol equation −(1)×1+(2)×6+(3)×3−48.5−6×393.5−3×286=−3267.5kJ for  1mol=−6535kJ for 2mol Ans. 6535kJ\begin{aligned} & 6 \mathrm{C}(\text{ graphite })+3 \mathrm{H}_2(\mathrm{ g}) \rightarrow \mathrm{C}_6 \mathrm{H}_6(\ell) ; \Delta \mathrm{H}=48.5 \mathrm{ kJ} / \mathrm{mol} \\ & \mathrm{C}(\text{ graphite })+\mathrm{O}_2(\mathrm{ g}) \rightarrow \mathrm{CO}_2(\mathrm{ g}) ; \Delta \mathrm{H}=-393.5 \mathrm{ kJ} / \mathrm{mol} \\ & \mathrm{H}_2^{(\mathrm{g})}+\frac{1}{2}(\mathrm{ g}) \longrightarrow \mathrm{H}_2 \mathrm{O}(\ell) ; \Delta \mathrm{H}=-286 \mathrm{ kJ} / \mathrm{mol} \\ & \text{ equation }-(1) \times 1+(2) \times 6+(3) \times 3 \\ & -48.5-6 \times 393.5-3 \times 286 \\ & =-3267.5 \mathrm{ kJ} \text{\text{ for } } 1 \mathrm{ mol} \\ & =-6535 \mathrm{ kJ} \text{ for } 2 \mathrm{ mol} \\ & \text{ Ans. } 6535 \mathrm{ kJ}\end{aligned}
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