The decomposition of \(NH_3\) on platinum surface is a zero order reaction . The rates of production of \(N_2\) and \(H_2\) respectively are: (in \(molL^{-1}s^{-1}\)). Given that \(2.5 \times 10^{-4}\, \ molL^{-1}s^{-1}\) is the rate of reaction.

\(2.5 \times 10^{-4}\), \(7.5 \times 10^{-4}\)

The correct answer is option 3. \(2.5 \times 10^{-4}\), \(7.5 \times 10^{-4}\).

The reaction can be written as:

\(2NH_3 \longrightarrow N_2 + 3H_2\)

The rate of reaction can be written as:

\(Rate\, \ = \, \ -\frac{1}{2}\frac{d[NH_3]}{dt}\, \ = \, \ \frac{d[N_2]}{dt}\, \ =\, \ \frac{1}{3}\frac{d[H_2]}{dt}\)

For a zero order reaction, \(rate\, \ = k\)

\(∴ Rate\, \ = \, \ -\frac{1}{2}\frac{d[NH_3]}{dt}\, \ = \, \ \frac{d[N_2]}{dt}\, \ =\, \ \frac{1}{3}\frac{d[H_2]}{dt} = 2.5 \times 10^{-4}\, \ molL^{-1}s^{-1}\)

Rate of production of \(N_2 = \frac{d[N_2]}{dt} =  2.5 \times 10^{-4}\, \ molL^{-1}s^{-1}\)

Rate of production of \(H_2 = \frac{d[H_2]}{dt} = 3 \times 2.5 \times 10^{-4}\, \ molL^{-1}s^{-1} = 7.5 \times 10^{-4} \ molL^{-1}s^{-1}\)