Decomposition of NH3 on the surface of platinum has a rate constant of 2-5- 10-4 mol-dm-3 s-1 at 350 K- The order of the reaction is-

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Question

Decomposition of $N H_{3}$ on the surface of platinum has a rate constant of $2.5 \times 10^{- 4}$ $m o l . d m^{- 3}$ $s^{- 1}$ at $350$ K. The order of the reaction is:

0

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Solution

The correct option is A $0$
Decomposition of $N H_{3}$ on the surface of platinum has a rate constant of $2.5 \times 10^{- 4} m o l / d m^{3} s e c^{- 1}$ at 350K. The order of the reaction is 0.
Unit of rate constant of nth order reaction is $(m o l / d m^{3})^{1 - n} s e c^{- 1}$
For zero order $n = 0$
$∴ k = m o l / d m^{3} s e c^{- 1}$

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Decomposition of NH3 on the surface of platinum has a rate constant of 2.5×10−4 mol.dm−3s−1 at 350 K. The order of the reaction is:

The correct option is A 0Decomposition of NH3 on the surface of platinum has a rate constant of 2.5×10−4mol/dm3sec−1 at 350K. The order of the reaction is 0.Unit of rate constant of nth order reaction is (mol/dm3)1−nsec−1For zero order n=0∴k=mol/dm3sec−1

Decomposition of $N H_{3}$ on the surface of platinum has a rate constant of $2.5 \times 10^{- 4}$ $m o l . d m^{- 3}$ $s^{- 1}$ at $350$ K. The order of the reaction is: