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)$9.9 \times 10^{-4}\,\mathrm{torr\,min^{-1}}$
(b)$0.4606\,\mathrm{min^{-1}}$
(c)$9.9 \times 10^{-3}\,\mathrm{torr\,min^{-1}}$
(d)$0.693\,\mathrm{min^{-1}}$
Answer
Answer: B ✓ checked by 4AB · confidence high
Explanation
At low pressure the Langmuir–Hinshelwood rate is first order: k = ln(10⁻²/10⁻⁴)/10 min = 4.606/10 = 0.4606 min⁻¹ (b).