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Chemical Equilibrium question

2022 · 30 Jun · Shift 1 · Q3
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  5. /2022 · 30 Jun · Shift 1 · Q3

Chemical Equilibrium question

2022 · 30 Jun · Shift 1 · Q3

JEE MainChemistryChemical EquilibriumMCQ+4 / −1
The equilibrium constant for the reversible reaction 2A(g) ⇌\rightleftharpoons⇌ 2B(g) + C(g) is K1 32{3 \over 2}23​ A(g) ⇌\rightleftharpoons⇌ 32{3 \over 2}23​ B(g) + 34{3 \over 4}43​ C(g) is K2. K1 and K2 are related as :
  1. A
    K1=K2{K_1} = \sqrt {{K_2}}K1​=K2​​
  2. B
    K2=K1{K_2} = \sqrt {{K_1}}K2​=K1​​
  3. C
    K2=K13/4{K_2} = K_1^{3/4}K2​=K13/4​
  4. D
    K1=K23/4{K_1} = K_2^{3/4}K1​=K23/4​
View written solutionFree

Correct answer: C

  1. Write the given reaction and its equilibrium constant

The original reaction is: 2A(g)⇌2B(g)+C(g)2A(g) \rightleftharpoons 2B(g) + C(g)2A(g)⇌2B(g)+C(g) with equilibrium constant K1K_1K1​.

So, K1=[B]2[C][A]2K_1 = \frac{[B]^2[C]}{[A]^2}K1​=[A]2[B]2[C]​

  1. Write the second reaction

The second reaction is: 32A(g)⇌32B(g)+34C(g)\frac{3}{2}A(g) \rightleftharpoons \frac{3}{2}B(g) + \frac{3}{4}C(g)23​A(g)⇌23​B(g)+43​C(g) with equilibrium constant K2K_2K2​.

Notice that this reaction is obtained by multiplying the first reaction by: 3/22=34\frac{3/2}{2} = \frac{3}{4}23/2​=43​

Indeed, (2A⇌2B+C)×34\left(2A \rightleftharpoons 2B + C\right) \times \frac{3}{4}(2A⇌2B+C)×43​ gives 32A⇌32B+34C\frac{3}{2}A \rightleftharpoons \frac{3}{2}B + \frac{3}{4}C23​A⇌23​B+43​C

  1. Use the rule for changing stoichiometric coefficients

If all stoichiometric coefficients in a balanced reaction are multiplied by a factor nnn, then the new equilibrium constant becomes: K′=KnK' = K^nK′=Kn

Here, the factor is: n=34n = \frac{3}{4}n=43​

Therefore, K2=K13/4K_2 = K_1^{3/4}K2​=K13/4​

  1. Match with the options

Thus the correct relation is: K2=K13/4\boxed{K_2 = K_1^{3/4}}K2​=K13/4​​

So, Option C is correct.

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