Question

The decomposition of phosphine $[P{H}_3]$ on tungsten at low pressure is a first-order reaction. It is because the

• A. rate is proportional to the surface coverage
• B. rate is inversely proportional to the surface coverage
• C. rate is independent of the surface coverage
• D. rate of decomposition is very slow

Answer 1

Answer: (A)

$P{H}_3\to WP+\frac{3}{2}{H}_2$
Rate = $k[P{H}_3]$
It is dependent on surface coverage because at low pressure, surface area covered is proportional to partial pressure of $P{H}_3$ and it is first order w.r.t $P{H}_3$.
$\theta =\frac{kP}{1+kP}$
$\theta$ = fraction of surface area covered at low pressure $\theta$ = k P