MCQ
$\left( 1 \right)\,{N_2}\left( g \right) + 3{H_2}\left( g \right) \rightleftharpoons 2N{H_3}\left( g \right)\,,\,{K_1}$

$\left( 2 \right)\,{N_2}\left( g \right) + {O_2}\left( g \right) \rightleftharpoons 2NO\left( g \right)\,,\,{K_2}$

$\left( 3 \right)\,{H_2}\left( g \right) + \frac{1}{2}{O_2}\left( g \right) \rightleftharpoons {H_2}O\left( g \right)\,,\,{K_3}$

The equation for the equilibrium constart of the reaction

$2N{H_3}\left( g \right) + \frac{5}{2}{O_2}\left( g \right) \rightleftharpoons 2NO\left( g \right) + 3{H_2}O\left( g \right)$

$(K_4)$ in terms of $K_1 , K_2$ , and $K_3$ is

  • A
    $\frac{{{K_1}{K_2}}}{{{K_3}}}$
  • B
    $\frac{{{K_1}{K_3^2}}}{{{K_2}}}$
  • C
    ${K_1}{K_2}{K_3}$
  • $\frac{{{K_2}{K_3^3}}}{{{K_1}}}$

Answer

Correct option: D.
$\frac{{{K_2}{K_3^3}}}{{{K_1}}}$
d
To calculate the value of $K_4$ in the given equation we should apply

eqn. $(2)$ + eqn. $(3)\times3 $ - eqn. $(1)$ 

hence ${K_4} = \frac{{{K_2}{K_3^3}}}{{{K_1}}}$

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