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Thermodynamics question

2020 · 7 Jan · Shift 1 · Q2
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Thermodynamics question

2020 · 7 Jan · Shift 1 · Q2

JEE MainChemistryThermodynamicsNumerical+4 / −1
For the reaction : A(lll) →\to→ 2B(g) ΔU=2.1 kcal, ΔS=20 cal K−1\Delta U = 2.1\,kcal,\,\Delta S = 20\,cal\,{K^{ - 1}}ΔU=2.1kcal,ΔS=20calK−1 at 300 K Hence Δ\DeltaΔ G in kcal is :
Numerical answer
View written solutionFree

Correct answer: -2.7

  1. Given data

For the reaction A(l)→2B(g)A(l) \to 2B(g)A(l)→2B(g) we have at T=300 KT=300\,\text{K}T=300K:

  • ΔU=2.1 kcal\Delta U = 2.1\,\text{kcal}ΔU=2.1kcal
  • ΔS=20 cal K−1\Delta S = 20\,\text{cal K}^{-1}ΔS=20cal K−1

We need to find ΔG\Delta GΔG.

  1. Use the relation between ΔH\Delta HΔH and ΔU\Delta UΔU

For reactions involving gases, ΔH=ΔU+ΔngRT\Delta H = \Delta U + \Delta n_g RTΔH=ΔU+Δng​RT

Here,

  • product side has 222 moles of gas
  • reactant side has 000 moles of gas

So, Δng=2−0=2\Delta n_g = 2-0=2Δng​=2−0=2

Using R=2 cal mol−1K−1R = 2\,\text{cal mol}^{-1}\text{K}^{-1}R=2cal mol−1K−1 (approximately), ΔngRT=2×2×300=1200 cal=1.2 kcal\Delta n_g RT = 2 \times 2 \times 300 = 1200\,\text{cal} = 1.2\,\text{kcal}Δng​RT=2×2×300=1200cal=1.2kcal

Thus, ΔH=2.1+1.2=3.3 kcal\Delta H = 2.1 + 1.2 = 3.3\,\text{kcal}ΔH=2.1+1.2=3.3kcal

  1. Now use Gibbs free energy relation

ΔG=ΔH−TΔS\Delta G = \Delta H - T\Delta SΔG=ΔH−TΔS

Convert TΔST\Delta STΔS into kcal: TΔS=300×20=6000 cal=6.0 kcalT\Delta S = 300 \times 20 = 6000\,\text{cal} = 6.0\,\text{kcal}TΔS=300×20=6000cal=6.0kcal

Therefore, ΔG=3.3−6.0=−2.7 kcal\Delta G = 3.3 - 6.0 = -2.7\,\text{kcal}ΔG=3.3−6.0=−2.7kcal

  1. Final answer

ΔG=−2.7 kcal\boxed{\Delta G = -2.7\,\text{kcal}}ΔG=−2.7kcal​

  1. Comparison with stored answer

Stored correct answer: −2.7-2.7−2.7

This matches exactly.

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