The decomposition of NH₃ on platinum surface is zero order. What are the rates of production of N₂ and H₂ if k = 2.5 x 10^–4 mol^–1Ls^–1?

Solution

The decomposition of NH₃ on plantinum surface is given as,

2 {NH}_{3} \to N_{2} + 3 H_{2} Rate of reaction, \frac{dx}{dt} = - \frac{1}{2} \frac{d ({NH}_{3})}{dt} = \frac{d [N_{2}]}{dt} = \frac{1}{3} \frac{d [H_{2}]}{dt} = k .

k is rate constant and reaction is of zero order.
Therefore, rate of reaction,

\frac{dx}{dt} = \frac{d [H_{2}]}{dt} = 2.5 \times 10^{- 4} {Ms}^{- 1}

Rate of production of

H_{2},

is given by

\frac{d [H_{2}]}{dt} = \frac{3 d [H_{2}]}{dt} = 3 \times 2.5 \times 10^{- 4} {Ms}^{- 1} = 7.5 \times 10^{- 4} {Ms}^{- 1} .

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