Ncert Exercises & Solutions

4.3 The decomposition of NH3 on platinum surface is zero order reaction. What are the rates of production of N2 and H2 if k = 2.5 × 10–4 mol–1 L s–1?

NEETprep Answer:

The decomposition of NH3 on platinum surface is represented by the following equation.

$2 N H_{3 (g)} \to N_{2 (g)} + 3 H_{2 (g)}$
$T h e r e f o r e,$
$R a t e = - \frac{1}{2} \frac{d [N H_{3}]}{d t} = \frac{d [N_{2}]}{d t} = \frac{1}{3} \frac{d [H_{2}]}{d t}$

However, it is given that the reaction is of zero order. Therefore,

$- \frac{1}{2} \frac{d [N H_{3}]}{d t} = \frac{d [N_{2}]}{d t} = \frac{1}{3} \frac{d [H_{2}]}{d t} = K$
$= 2.5 \times 10^{- 4} m o l L^{- 1} s^{- 1}$

Therefore, the rate of production of N $_{2}$ is

$\frac{d [N H_{3}]}{d t} = 2.5 \times 10^{- 4} m o l L^{- 1} s^{- 1}$

And, the rate of production of H $_{2}$ is

$\frac{d [H_{2}]}{d t} = 3.2.5 \times 10^{- 4} m o l L^{- 1} s^{- 1}$
$= 7.5 \times 10^{- 4} m o l L^{- 1} s^{- 1}$

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