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Thermodynamics question

2024 · 1 Feb · Shift 2 · Q25
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Thermodynamics question

2024 · 1 Feb · Shift 2 · Q25

JEE MainChemistryThermodynamicsNumerical+4 / −1
For a certain reaction at 300 K, K=10300 \mathrm{~K}, \mathrm{~K}=10300 K, K=10, then ΔG∘\Delta \mathrm{G}^{\circ}ΔG∘ for the same reaction is - ‾\underline{\hspace{2cm}}​×10−1 kJ mol−1\times 10^{-1} \mathrm{~kJ} \mathrm{~mol}^{-1}×10−1 kJ mol−1. (Given R=8.314JK−1 mol−1\mathrm{R}=8.314 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}R=8.314JK−1 mol−1 )
Numerical answer
View written solutionFree

Correct answer: 57

  1. Use the relation between standard Gibbs free energy and equilibrium constant:
ΔG∘=−RTln⁡K\Delta G^{\circ} = -RT \ln KΔG∘=−RTlnK
  1. Substitute the given values:
R=8.314 J K−1mol−1,T=300 K,K=10R = 8.314\ \text{J K}^{-1}\text{mol}^{-1},\quad T = 300\ \text{K},\quad K=10R=8.314 J K−1mol−1,T=300 K,K=10

So,

ΔG∘=−(8.314)(300)ln⁡(10)\Delta G^{\circ} = -(8.314)(300)\ln(10)ΔG∘=−(8.314)(300)ln(10)
  1. Use
ln⁡10=2.303\ln 10 = 2.303ln10=2.303

Therefore,

ΔG∘=−(8.314)(300)(2.303)\Delta G^{\circ} = -(8.314)(300)(2.303)ΔG∘=−(8.314)(300)(2.303)
  1. Calculate:
8.314×300=2494.28.314 \times 300 = 2494.28.314×300=2494.2 2494.2×2.303≈5744 J mol−12494.2 \times 2.303 \approx 5744\ \text{J mol}^{-1}2494.2×2.303≈5744 J mol−1

Thus,

ΔG∘≈−5744 J mol−1\Delta G^{\circ} \approx -5744\ \text{J mol}^{-1}ΔG∘≈−5744 J mol−1
  1. Convert to kJ mol−1^{-1}−1:
ΔG∘≈−5.744 kJ mol−1\Delta G^{\circ} \approx -5.744\ \text{kJ mol}^{-1}ΔG∘≈−5.744 kJ mol−1
  1. The question asks in the form:
ΔG∘=−‾×10−1 kJ mol−1\Delta G^{\circ} = -\underline{\hspace{1cm}} \times 10^{-1}\ \text{kJ mol}^{-1}ΔG∘=−​×10−1 kJ mol−1

Write −5.744-5.744−5.744 as:

−5.744=−57.44×10−1-5.744 = -57.44 \times 10^{-1}−5.744=−57.44×10−1

Hence the required integer is

575757
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