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Chemical Equilibrium question

2021 · 24 Feb · Shift 1 · Q21
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Chemical Equilibrium question

2021 · 24 Feb · Shift 1 · Q21

JEE MainChemistryChemical EquilibriumNumerical+4 / −1
For the reaction A(g) →\to→ B(g) the value of the equilibrium constant at 300 K and 1 atm is equal to 100.0. The value of Δ\DeltaΔ rG for the reaction at 300 K and 1 atm in J mol-1 is – xR, where x is ‾\underline{\hspace{2cm}}​. (Rounded off to the nearest integer) [R = 8.31 J mol–1K-1 and ln 10 = 2.3)
Numerical answer
View written solutionFree

Correct answer: 1380

  1. Use the relation between standard Gibbs free energy change and equilibrium constant:

ΔrG∘=−RTln⁡K\Delta_r G^\circ = -RT \ln KΔr​G∘=−RTlnK

Here,

  • T=300 KT = 300\,\text{K}T=300K
  • K=100K = 100K=100
  • R=8.31 J mol−1K−1R = 8.31\,\text{J mol}^{-1}\text{K}^{-1}R=8.31J mol−1K−1
  1. Evaluate ln⁡100\ln 100ln100:

ln⁡100=ln⁡(102)=2ln⁡10=2(2.3)=4.6\ln 100 = \ln(10^2) = 2\ln 10 = 2(2.3) = 4.6ln100=ln(102)=2ln10=2(2.3)=4.6

  1. Substitute into the formula:

ΔrG∘=−(8.31)(300)(4.6)\Delta_r G^\circ = -(8.31)(300)(4.6)Δr​G∘=−(8.31)(300)(4.6)

First,

8.31×300=24938.31 \times 300 = 24938.31×300=2493

Then,

2493×4.6=11467.82493 \times 4.6 = 11467.82493×4.6=11467.8

So,

ΔrG∘=−11467.8 J mol−1\Delta_r G^\circ = -11467.8\,\text{J mol}^{-1}Δr​G∘=−11467.8J mol−1

  1. Given that

ΔrG=−xR\Delta_r G = -xRΔr​G=−xR

we compare:

−xR=−11467.8-xR = -11467.8−xR=−11467.8

Thus,

x=11467.88.31=1380.0x = \frac{11467.8}{8.31} = 1380.0x=8.3111467.8​=1380.0

Hence, rounded to the nearest integer,

x=1380x = 1380x=1380

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