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Thermodynamics question

2020 · 5 Sep · Shift 2 · Q2
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Thermodynamics question

2020 · 5 Sep · Shift 2 · Q2

JEE MainChemistryThermodynamicsNumerical+4 / −1
For a dimerization reaction, 2A(g) →\to→ A2A_2A2​(g) at 298 K, Δ\DeltaΔ Uo = –20 kJ mol–1, Δ\DeltaΔ So = –30 JK–1 mol–1, then the Δ\DeltaΔ Go will be ‾\underline{\hspace{2cm}}​ J.
Numerical answer
View written solutionFree

Correct answer: -13540TO-13537

  1. Use the relation between standard Gibbs free energy, enthalpy, and entropy:

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

  1. Convert ΔU∘\Delta U^\circΔU∘ to ΔH∘\Delta H^\circΔH∘ using:

ΔH∘=ΔU∘+ΔngRT\Delta H^\circ = \Delta U^\circ + \Delta n_g RTΔH∘=ΔU∘+Δng​RT

For the reaction:

2A(g)→A2(g)2A(g) \to A_2(g)2A(g)→A2​(g)

Change in moles of gas:

Δng=1−2=−1\Delta n_g = 1 - 2 = -1Δng​=1−2=−1

So,

ΔH∘=−20 kJ mol−1+(−1)(8.314)(298) J mol−1\Delta H^\circ = -20\,\text{kJ mol}^{-1} + (-1)(8.314)(298)\,\text{J mol}^{-1}ΔH∘=−20kJ mol−1+(−1)(8.314)(298)J mol−1

Convert −20 kJ mol−1-20\,\text{kJ mol}^{-1}−20kJ mol−1 to J/mol:

−20 kJ mol−1=−20000 J mol−1-20\,\text{kJ mol}^{-1} = -20000\,\text{J mol}^{-1}−20kJ mol−1=−20000J mol−1

Thus,

ΔH∘=−20000−(8.314)(298)\Delta H^\circ = -20000 - (8.314)(298)ΔH∘=−20000−(8.314)(298)

ΔH∘=−20000−2477.6\Delta H^\circ = -20000 - 2477.6ΔH∘=−20000−2477.6

ΔH∘=−22477.6 J mol−1\Delta H^\circ = -22477.6\,\text{J mol}^{-1}ΔH∘=−22477.6J mol−1

  1. Now calculate TΔS∘T\Delta S^\circTΔS∘.

Given:

ΔS∘=−30 J K−1mol−1\Delta S^\circ = -30\,\text{J K}^{-1}\text{mol}^{-1}ΔS∘=−30J K−1mol−1

At T=298 KT=298\,\text{K}T=298K,

TΔS∘=298(−30)=−8940 J mol−1T\Delta S^\circ = 298(-30) = -8940\,\text{J mol}^{-1}TΔS∘=298(−30)=−8940J mol−1

  1. Therefore,

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

ΔG∘=−22477.6−(−8940)\Delta G^\circ = -22477.6 - (-8940)ΔG∘=−22477.6−(−8940)

ΔG∘=−22477.6+8940\Delta G^\circ = -22477.6 + 8940ΔG∘=−22477.6+8940

ΔG∘=−13537.6 J mol−1\Delta G^\circ = -13537.6\,\text{J mol}^{-1}ΔG∘=−13537.6J mol−1

  1. Final integer value:

ΔG∘≈−13538 J\Delta G^\circ \approx -13538\,\text{J}ΔG∘≈−13538J

This lies within the stored correct range −13540-13540−13540 to −13537-13537−13537.

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