Decomposition of ammonia on platinum surface follows the change,2NH3(g)⟶N2(g)+3H2(g)If the rate obeys (-d[NH3])dt = k1 [NH3] 1+k2[NH3], what will be the order for decomposition of NH3 if [NH3] is very very less, K1 and K1K2 are constant?

Decomposition of ammonia,1+k2NH3≃1⇒−dNH3dt=k1NH3So it is 1st order reaction.

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Decomposition of ammonia on platinum surface follows the change,
$2 N H_{3} (g) ⟶ N_{2} (g) + 3 H_{2} (g)$
If the rate obeys $\frac{(- d [N H_{3}])}{d t} = \frac{k_{1} [N H_{3}]}{1 + k_{2} [N H_{3}]}$ , what will be the order for decomposition of ${NH}_{3}$ if $[N H_{3}]$ is very very less, $K_{1}$ and $K_{1}$ $K_{2}$ are constant?

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Decomposition of ammonia,
$1 + k_{2} [{NH}_{3}] ≃ 1$

$\Rightarrow - \frac{d [{NH}_{3}]}{d t}$

$= k_{1} [{NH}_{3}]$

So it is 1st order reaction.

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