MCQ
If equilibrium constant for reaction $2AB$ $\rightleftharpoons$ ${A_2} + {B_2}$, is $49$, then the equilibrium constant for reaction $AB$ $\rightleftharpoons$ $\frac{1}{2}{A_2} + \frac{1}{2}{B_2}$, will be
  • $7$
  • B
    $20$
  • C
    $49$
  • D
    $21$

Answer

Correct option: A.
$7$
(a) $2AB$ $ \rightleftharpoons $ ${A_2} + {B_2}$

${K_c} = \frac{{[{A_2}]\,\,[{B_2}]}}{{{{[AB]}^2}}}$

For reaction $AB$ $ \rightleftharpoons $ $\frac{1}{2}{A_2} + \frac{1}{2}{B_2}$

${{K}_{c}}^{'}=\frac{{{[{{A}_{2}}]}^{{1}/{2}\;}}{{[{{B}_{2}}]}^{{1}/{2}\;}}}{[AB]}$;   ${{K}_{c}}^{'} =\sqrt{{{K}_{c}}} =\sqrt{49} =7$

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