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Chemical Equilibrium question

2023 · 29 Jan · Shift 1 · Q19
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Chemical Equilibrium question

2023 · 29 Jan · Shift 1 · Q19

JEE MainChemistryChemical EquilibriumNumerical+4 / −1
Consider the following reaction approaching equilibrium at 27 ∘^\circ∘ C and 1 atm pressure A+B\mathrm{A+B}A+B ⇌kr=102kf=103\mathrel{\mathop{\kern0pt\rightleftharpoons} \limits_{{k_r} = {{10}^2}}^{{k_f} = {{10}^3}}}kr​=102⇌kf​=103​ C+D\mathrm{C+D}C+D The standard Gibb's energy change (ΔrGθ)\mathrm{(\Delta_r G^\theta)}(Δr​Gθ) at 27 ∘^\circ∘ C is (−-−) ‾\underline{\hspace{2cm}}​ kJ mol −1^{-1}−1(Nearest integer). (Given : R=8.3 J K−1 mol−1\mathrm{R=8.3~J~K^{-1}~mol^{-1}}R=8.3 J K−1 mol−1 and ln⁡10=2.3\mathrm{\ln 10=2.3}ln10=2.3)
Numerical answer
View written solutionFree

Correct answer: 6

  1. Relate rate constants to equilibrium constant

For the reversible reaction

A+B⇌C+D\mathrm{A+B \rightleftharpoons C+D}A+B⇌C+D

we are given:

At equilibrium, the equilibrium constant is

K=kfkr=103102=10K = \frac{k_f}{k_r} = \frac{10^3}{10^2} = 10K=kr​kf​​=102103​=10
  1. Use the relation between standard Gibbs energy and equilibrium constant

The standard Gibbs energy change is

ΔrGθ=−RTln⁡K\Delta_r G^\theta = -RT \ln KΔr​Gθ=−RTlnK

Given:

R=8.3 J K−1mol−1,T=27∘C=300 K,ln⁡10=2.3R = 8.3\ \text{J K}^{-1}\text{mol}^{-1}, \quad T = 27^\circ \text{C} = 300\ \text{K}, \quad \ln 10 = 2.3R=8.3 J K−1mol−1,T=27∘C=300 K,ln10=2.3

So,

ΔrGθ=−(8.3)(300)(2.3)\Delta_r G^\theta = -(8.3)(300)(2.3)Δr​Gθ=−(8.3)(300)(2.3)
  1. Calculate the value

First,

8.3×300=24908.3 \times 300 = 24908.3×300=2490

Then,

2490×2.3=5727 J mol−12490 \times 2.3 = 5727\ \text{J mol}^{-1}2490×2.3=5727 J mol−1

Thus,

ΔrGθ=−5727 J mol−1\Delta_r G^\theta = -5727\ \text{J mol}^{-1}Δr​Gθ=−5727 J mol−1

Convert into kJ mol−1^{-1}−1:

ΔrGθ=−5.727 kJ mol−1\Delta_r G^\theta = -5.727\ \text{kJ mol}^{-1}Δr​Gθ=−5.727 kJ mol−1
  1. Nearest integer
ΔrGθ≈−6 kJ mol−1\Delta_r G^\theta \approx -6\ \text{kJ mol}^{-1}Δr​Gθ≈−6 kJ mol−1

Since the question asks:

The standard Gibbs energy change is (−) ‾(-)\,\underline{\hspace{1cm}}(−)​ kJ mol−1^{-1}−1

the required integer is:

666
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