4.3     The decomposition of  NH_{3} on platinum surface is zero order reaction. What are the rates of production of N_{2} and H_{2} if k=2.5\times 10^{-4}mol^{-1}L s^{-1}?

Answers (1)
M manish

The decomposition of  NH_{3} on the platinum surface reaction 

2NH_{3}(s)\overset{Pt}{\rightarrow}N_{2}(g)+3H_{2}(g)

therefore,   

Rate = -\frac{1}{2}\frac{d[NH_{3}]}{dt}=\frac{d[N_{2}]}{dt}=\frac{1}{3}\frac{d[H_{2}]}{dt}

For zero order reaction rate = k

therefore, -\frac{1}{2}\frac{d[NH_{3}]}{dt}=\frac{d[N_{2}]}{dt}=\frac{1}{3}\frac{d[H_{2}]}{dt}=k

So   \frac{d[N_{2}]}{dt}= 2.5\times 10^{-4}mol\ L^{-1}\ s^{-1}

and the rate of production of dihydrogen (H_{2}) = 3\times(2.5\times 10^{-4}mol\ L^{-1}\ s^{-1}

                                                                           = 7.5\times 10^{-4}mol\ L^{-1}\ s^{-1}

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