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NCERT Class XII Chemistry
Chapter - Chemical Kinetics
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Question : 3 of 39
Marks: +1, -0
The decomposition of NH3 on platinum surface is zero order reaction. What are the rates of production of N2\mathrm{N}_2 and H2\mathrm{H}_2 if k=2.5×10−4 mol L−1 s−1 ?k = 2.5 \times 10^{-4} \, \mathrm{mol \, L^{-1} \, s^{-1}} \, ?
Solution:  
For the reaction, 2NH3→N2+3H22\mathrm{NH}_3 \rightarrow \mathrm{N}_2 + 3\mathrm{H}_2
Rate=−12d[NH3]dt=d[N2]dt=13d[H2]dt          =- \frac{1}{2} \frac{d[\mathrm{NH}_3]}{dt} = \frac{d[\mathrm{N}_2]}{dt} = \frac{1}{3} \frac{d[\mathrm{H}_2]}{dt}\;\;\;\;\;...(1)
For zero order reaction
Rate = k                                                        \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; ...(2)
From equation (1) and (2)
Rate=−12d[NH3]dt=d[N2]dt=13d[H2]dt=k            =- \frac{1}{2} \frac{d[\mathrm{NH}_3]}{dt} = \frac{d[\mathrm{N}_2]}{dt} = \frac{1}{3} \frac{d[\mathrm{H}_2]}{dt} = k \;\;\;\;\;\;...(3)
∴  \therefore\; Rate of production of N2=d[N2]dt=k\mathrm{N}_2 = \frac{d[\mathrm{N}_2]}{dt} = k
k=2.5×10−4 mol L−1 s−1k = 2.5 \times 10^{-4} \, \mathrm{mol} \, \mathrm{L}^{-1} \, \mathrm{s}^{-1}
∴  \therefore\; Rate of production of N2=k=2.5×10−4 mol L−1 s−1\mathrm{N}_2 = k = 2.5 \times 10^{-4} \, \mathrm{mol} \, \mathrm{L}^{-1} \, \mathrm{s}^{-1}
From equation (3)(3)
Rate of production of H2=d[H2]dt\mathrm{H}_2 = \frac{d[\mathrm{H}_2]}{dt}
=3k= 3k
=3×2.5×10−4 mol L−1 s−1= 3 \times 2.5 \times 10^{-4} \, \mathrm{mol} \, \mathrm{L}^{-1} \, \mathrm{s}^{-1}
=7.5×10−4 mol L−1 s−1= 7.5 \times 10^{-4} \, \mathrm{mol} \, \mathrm{L}^{-1} \, \mathrm{s}^{-1}
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