For the reaction $2 {NH}_{3} \to N_{2} + 3 H_{2},$ If $\frac{- d [N H_{3}]}{d t} = K_{1} [N H_{3}],$ $\frac{d [N_{2}]}{d t} = K_{2} [N H_{3}],$ $\frac{d [H_{2}]}{d t} = K_{3} [N H_{3}]$ then the relation between $K_{1}, K_{2} & K_{3}$ is:

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$1.5 K_{1} = 3 K_{2} = K_{3}$

$2 K_{1} = K_{2} = 3 K_{3}$

$K_{1} = 3 K_{2} = 2 K_{3}$

$K_{1} = K_{2} = K_{3}$

answer is C.

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Detailed Solution

$2 {NH}_{3} \to N_{2} + 3 H_{2}$
$- \frac{1}{2} \frac{d [N H_{3}]}{d t} = \frac{d [N_{2}]}{d t} = \frac{1}{3} \frac{d [H_{2}]}{d t}$
$\frac{1}{2} \times K_{1} = K_{2} = \frac{K_{3}}{3}$

[-ve sing shows decrement in reactant concentration]
$\frac{3}{2} K_{1} = 3 K_{2} = K_{3} \Rightarrow 1.5 K_{1} = 3 K_{2} = K_{3}$

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