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Thermodynamics question

2025 · 28 Jan · Shift 2 · Q24
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Thermodynamics question

2025 · 28 Jan · Shift 2 · Q24

JEE MainChemistryThermodynamicsNumerical+4 / −1
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}CO2​(g)=−393.5 kJ 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}H2​O(l)=−286.0 kJ mol−1 Heat of combustion of benzene =−3267.0 kJ mol−1=-3267.0 \mathrm{~kJ} \mathrm{~mol}^{-1}=−3267.0 kJ mol−1 The heat of formation of benzene is ‾\underline{\hspace{2cm}}​kJmol−1\mathrm{kJ} \mathrm{mol}^{-1}kJmol−1. (Nearest integer)
Numerical answer
View written solutionFree

Correct answer: 48

  1. Write the combustion reaction of benzene

For benzene, C6H6(l)\mathrm{C_6H_6(l)}C6​H6​(l), the combustion reaction is:

C6H6(l)+152O2(g)→6CO2(g)+3H2O(l)\mathrm{C_6H_6(l) + \frac{15}{2}O_2(g) \rightarrow 6CO_2(g) + 3H_2O(l)}C6​H6​(l)+215​O2​(g)→6CO2​(g)+3H2​O(l)

Given:

  • ΔHf∘[CO2(g)]=−393.5 kJ mol−1\Delta H_f^\circ[\mathrm{CO_2(g)}] = -393.5\,\text{kJ mol}^{-1}ΔHf∘​[CO2​(g)]=−393.5kJ mol−1
  • ΔHf∘[H2O(l)]=−286.0 kJ mol−1\Delta H_f^\circ[\mathrm{H_2O(l)}] = -286.0\,\text{kJ mol}^{-1}ΔHf∘​[H2​O(l)]=−286.0kJ mol−1
  • Heat of combustion of benzene =−3267.0 kJ mol−1= -3267.0\,\text{kJ mol}^{-1}=−3267.0kJ mol−1

We need to find ΔHf∘[C6H6(l)]\Delta H_f^\circ[\mathrm{C_6H_6(l)}]ΔHf∘​[C6​H6​(l)].

  1. Use Hess's law

For any reaction,

ΔHrxn=∑ΔHf∘(products)−∑ΔHf∘(reactants)\Delta H_{\text{rxn}} = \sum \Delta H_f^\circ(\text{products}) - \sum \Delta H_f^\circ(\text{reactants})ΔHrxn​=∑ΔHf∘​(products)−∑ΔHf∘​(reactants)

So for combustion of benzene:

−3267=[6ΔHf∘(CO2)+3ΔHf∘(H2O)]−[ΔHf∘(C6H6)+152ΔHf∘(O2)]-3267 = \left[6\Delta H_f^\circ(\mathrm{CO_2}) + 3\Delta H_f^\circ(\mathrm{H_2O})\right] - \left[\Delta H_f^\circ(\mathrm{C_6H_6}) + \frac{15}{2}\Delta H_f^\circ(\mathrm{O_2})\right]−3267=[6ΔHf∘​(CO2​)+3ΔHf∘​(H2​O)]−[ΔHf∘​(C6​H6​)+215​ΔHf∘​(O2​)]

Since ΔHf∘(O2)=0\Delta H_f^\circ(\mathrm{O_2})=0ΔHf∘​(O2​)=0,

−3267=[6(−393.5)+3(−286.0)]−ΔHf∘(C6H6)-3267 = \left[6(-393.5) + 3(-286.0)\right] - \Delta H_f^\circ(\mathrm{C_6H_6})−3267=[6(−393.5)+3(−286.0)]−ΔHf∘​(C6​H6​)
  1. Calculate the product term
6(−393.5)=−2361.06(-393.5) = -2361.06(−393.5)=−2361.0 3(−286.0)=−858.03(-286.0) = -858.03(−286.0)=−858.0

Thus,

−2361.0+(−858.0)=−3219.0-2361.0 + (-858.0) = -3219.0−2361.0+(−858.0)=−3219.0

So,

−3267=−3219−ΔHf∘(C6H6)-3267 = -3219 - \Delta H_f^\circ(\mathrm{C_6H_6})−3267=−3219−ΔHf∘​(C6​H6​)
  1. Solve for heat of formation of benzene
−3267+3219=−ΔHf∘(C6H6)-3267 + 3219 = -\Delta H_f^\circ(\mathrm{C_6H_6})−3267+3219=−ΔHf∘​(C6​H6​) −48=−ΔHf∘(C6H6)-48 = -\Delta H_f^\circ(\mathrm{C_6H_6})−48=−ΔHf∘​(C6​H6​) ΔHf∘(C6H6)=48 kJ mol−1\Delta H_f^\circ(\mathrm{C_6H_6}) = 48\,\text{kJ mol}^{-1}ΔHf∘​(C6​H6​)=48kJ mol−1
  1. Nearest integer
48\boxed{48}48​

The derived answer matches the stored correct answer.

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