The decomposition of phosphine $\mathrm{PH}{}_3$ on tungsten at low pressure is a first-order reaction. It is because the

Solution & Explanation

\textbf{Answer:} (A) rate is proportional to the surface coverage \textbf{Solution:} The decomposition of $\mathrm{PH}{}_3$ on a tungsten surface is a heterogeneous (surface) catalytic reaction. At low pressure the amount of $\mathrm{PH}{}_3$ adsorbed on the tungsten surface (the surface coverage) is small and is directly proportional to the gas-phase pressure (concentration) of $\mathrm{PH}{}_3$ — this is the linear, low-pressure region of the adsorption isotherm. Since the reaction occurs on the adsorbed layer, the rate is proportional to the surface coverage, which in turn is proportional to $[\mathrm{PH}{}_3]$. Hence $\text{Rate}\propto [\mathrm{PH}{}_3{]}^1$, making the reaction first order. At high pressure the surface becomes saturated, coverage becomes constant, and the reaction turns zero order.