The decomposition of $\mathrm{NH}_{3}$ on Mo surface follows Langmuir-Hinshelwood mechanism. The decomposition was carried out at low pressures. The initial pressure of $\mathrm{NH}_{3}$ was $10^{-2}$ torr. The pressure of $\mathrm{NH}_{3}$ was reduced to $10^{-4}$ torr in 10 minutes. The rate constant of decomposition of $\mathrm{NH}_{3}$ is:
(a)$9.9 \times 10^{-4}$ torr $\mathrm{min}^{-1}$
(b)$0.4606 \mathrm{min}^{-1}$
(c)$9.9 \times 10^{-3}$ torr $\mathrm{min}^{-1}$
(d)$0.693 \mathrm{min}^{-1}$
Answer
Answer: B ✓ checked by 4AB · confidence high
The source book printed no answer; this one was worked out and checked — see the explanation.
Explanation
At low pressure the Langmuir–Hinshelwood decomposition is first order: k = (1/t) ln(p₀/p) = (1/10) ln(100) = 0.4606 min⁻¹.