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6.1. Equilibrium - 1 (chemical Equilibrium) — Chemistry STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceChemistry6.1. Equilibrium - 1 (chemical Equilibrium)1 Mark
MCQ
The equilibrium constant for the reversible reaction, ${N_2} + 3{H_2}$ $\rightleftharpoons$ $2N{H_3}$ is $K$ and for the reaction $\frac{1}{2}{N_2} + \frac{3}{2}{H_2}$ $\rightleftharpoons$ $N{H_3}$ the equilibrium constant is $K'$. $K$ and $K'$ will be related as
  • A
    $K = K'$
  • ✓
    $K' = \sqrt K $
  • C
    $K = \sqrt {K'} $
  • D
    $K \times K' = 1$
✓

Answer

Correct option: B.
$K' = \sqrt K $
(b) $K' = {K^n};$ Hence $n = \frac{1}{2}$

$\therefore \,\,K' = {K^{1/2}} = \sqrt K $

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