The decomposition of NH3 on the platinum surface is zero order reaction. If k = 2.5 × 10−4 mol L−1 s−1 the rate of production of H2 is
Answer & explanation
Correct answer: option 2
Given rate constant = 2.5 × 10−4 mol L−1 s−1
The decomposition of NH3 on the platinum surface takes place as
\(2NH_3 ————→ N_2 + 3H_2\)
So, Rate of reaction = \(−\frac{1}{2}\frac{d[NH_3]}{dt} = \frac{d[N_2]}{dt} = \frac{1}{3}\frac{d[H_2]}{dt}\)
The rate of production of \(H_2 = \frac{d[H_2]}{dt}\)
\(= 3 × 2.5 × 10^{− 4} \text{ mol L}^{−1}\text{ s}^{−1}\)
\( = 7.5 × 10^{− 4} \text{ mol L}^{−1}\text{ s}^{−1}\)