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Thermodynamics question

2025 · 23 Jan · Shift 1 · Q22
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Thermodynamics question

2025 · 23 Jan · Shift 1 · Q22

JEE MainChemistryThermodynamicsNumerical+4 / −1
The standard enthalpy and standard entropy of decomposition of N2O4\mathrm{N}_2 \mathrm{O}_4N2​O4​ to NO2\mathrm{NO}_2NO2​ are 55.0 kJ mol−155.0 \mathrm{~kJ} \mathrm{~mol}^{-1}55.0 kJ mol−1 and 175.0 J/K/mol175.0 \mathrm{~J} / \mathrm{K} / \mathrm{mol}175.0 J/K/mol respectively. The standard free energy change for this reaction at 25∘C25^{\circ} \mathrm{C}25∘C in J mol−1\mathrm{mol}^{-1}mol−1 is ‾\underline{\hspace{2cm}}​ (Nearest integer)
Numerical answer
View written solutionFree

Correct answer: 2850

  1. Write the reaction

The decomposition is: N2O4(g)→2NO2(g)\mathrm{N_2O_4(g) \rightarrow 2NO_2(g)}N2​O4​(g)→2NO2​(g)

Given:

  • Standard enthalpy change: ΔH∘=55.0 kJ mol−1=55000 J mol−1\Delta H^\circ = 55.0\ \text{kJ mol}^{-1} = 55000\ \text{J mol}^{-1}ΔH∘=55.0 kJ mol−1=55000 J mol−1
  • Standard entropy change: ΔS∘=175.0 J K−1mol−1\Delta S^\circ = 175.0\ \text{J K}^{-1}\text{mol}^{-1}ΔS∘=175.0 J K−1mol−1
  • Temperature: T=25∘C=298 KT = 25^\circ\text{C} = 298\ \text{K}T=25∘C=298 K
  1. Use Gibbs free energy relation

ΔG∘=ΔH∘−TΔS∘\Delta G^\circ = \Delta H^\circ - T\Delta S^\circΔG∘=ΔH∘−TΔS∘

  1. Substitute the values

ΔG∘=55000−(298)(175)\Delta G^\circ = 55000 - (298)(175)ΔG∘=55000−(298)(175)

First calculate: 298×175=52150298 \times 175 = 52150298×175=52150

So, ΔG∘=55000−52150=2850 J mol−1\Delta G^\circ = 55000 - 52150 = 2850\ \text{J mol}^{-1}ΔG∘=55000−52150=2850 J mol−1

  1. Final answer

The standard free energy change is: 2850 J mol−1\boxed{2850\ \text{J mol}^{-1}}2850 J mol−1​

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