The decomposition of phosphine `[PH_(3)]` on tungsten at low pressure is a first-order reaction. It is because the

Options

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

A

Explanation

For the decomposition of gas in the presence of heterogenous catalyst
`A(g)underset(k_(-1))overset(k_(1))hArrA_(ads) overset(k_(2))tounderset(k_(1) gt gt k_(-1))("Product")`
Rate of formation of product `r=k_(2)theta_(A)`
`theta_(A)=`surface coverage of solid catalyst
`therefore`According to Langmuir adsorption isotherm
rate`=k_(2)theta_(A)`
or rate `=k_(2)(k_(A)P_(A))/(1+k_(A)P_(A))`
at low pressure `P_(A) to 0 , theta_(A)` is very small and proportional to `P_(A)` (pressure of gas)
`therefore` reaction becaue 1st order
rate law, rate`=k_(2)k_(A)P_(A)`
`therefore`rate is dependent of `theta_(A)` i.e. surface area covered.