The decomposition of NH3 on finely divided platinum follows the rate expression- Rate - k1 -NH3 -1 - k2 -NH3 -It is a first order reaction when concentration of NH3 is-

The correct option is B very lowAt very low concentration of NH_3 , k_2[NH_3] lt; lt;1 and 1+k_2[NH_3] ≈ 1 #160;So, Rate = k_1 [NH ...

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Order of Reaction

The decomposi...

Question

The decomposition of $N H_{3}$ on finely divided platinum follows the rate expression,
Rate $= \frac{k_{1} [N H_{3}]}{1 + k_{2} [N H_{3}]}$
It is a first order reaction when concentration of $N H_{3}$ is:

very low

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very high

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moderate

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never

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{2 N H}_{3}

⟶

N_{2}

{3 H}_{2}

10^{- 4}

{M S}^{- 1}

- \frac{d {[N H}_{3}]}{d t} = \frac{K_{1} {[N H}_{3}]}{1 + K_{2} {[N H}_{3}]}

K_{1}

k_{2}

4 \times 10^{- 3} m s^{- 1}

N H_{3}

2 N H_{3} ⟶ N_{2} + 3 H_{2}

\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}]

k_{1}, k_{2}

k_{3}

2 N H_{3} \to N_{2} + 3 H_{2}

- \frac{Δ [N H_{3}]}{Δ t} = K_{1} [N H_{3}]; \frac{Δ [N_{2}]}{Δ t} = K_{2} [N H_{3}]; \frac{Δ [H_{2}]}{Δ t} = K_{3} [N H_{3}]

k_{1}, k_{2}

k_{3}

N H_{3}

2.5 \times 10^{- 4}

m o l . d m^{- 3}

s^{- 1}

350

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