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At 500 K, equilibrium constant, KXc, for the following reaction is 5. 12HX2(g)+12IX2(g)↽−−⇀HI(g) What would be the equilibrium constant KXc for the reaction 2HI(g)↽−−⇀HX2(g)+IX2(g) - Chemistry

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Question

At 500 K, equilibrium constant, \[\ce{K_c}\], for the following reaction is 5.

\[\ce{1/2 H2 (g) + 1/2 I2 (g) ⇌ HI (g)}\]

What would be the equilibrium constant \[\ce{K_c}\] for the reaction

\[\ce{2HI (g) ⇌ H2 (g) + I2 (g)}\]

Options

  • 0.04

  • 0.4

  • 25

  • 2.5

MCQ
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Solution

0.04

Explanation:

If the equation is multiplied by 2, the equilibrium constant for the new equation is the square of \[\ce{K}\] it means we must do the square of 5 that is 25. Then and on reversing the reaction the value of the equilibrium constant is inversed means the value of K comes out to be 1/25 = 0.04. 

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Law of Chemical Equilibrium and Equilibrium Constant
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Chapter 7: Equilibrium - Multiple Choice Questions (Type - I) [Page 89]

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NCERT Exemplar Chemistry [English] Class 11
Chapter 7 Equilibrium
Multiple Choice Questions (Type - I) | Q 18 | Page 89

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