Question

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?

Answer 1

Answer:

Explanation:

Rate of reaction can be

$\frac{dx}{dt} - \frac{1}{2} \frac{d(NH^{3} )}{dt} = \frac{d(N_{2}) }{dt} =\frac{1}{3} \frac{d[H_2]}{dt} =k$

k is rate constant and reaction is of 0 order reaction. there4 rate of reaction is

$\frac{d}{dx} =\frac{d[H_2]}{dt} =  2.5× 10^{-4} Ms^{-1}$

rate of producction of H^2 is

$\frac{d[H_2]}{dt} =\frac{3d[H_2]}{dt}$

3× 2.5×$10^{-4} Ms^{-1}$

= 7.5×$10^{-4} Ms^{-1}$

        $LaiLa$          

Answer 2

Explanation:

The reaction is zero order. Hence, Rate of the reaction is equal to rate constant.

The reaction is,

2NH3→N2+3H2

Therefore, d[N2]/dt=2.5×10^−4

d[H2]/dt=2.5×10^−4×3=7.5×10^−4.

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