Question

The decomposition of phosphine $\left( {P{H_3}} \right)$on tungsten at low pressure is a first order reaction. It is because the:
A) Rate is independent of the surface coverage
B) Rate of decomposition is very slow
C) Rate is proportional to the surface coverage
D) Rate is inversely proportional to the surface coverage

Answer 1

Hint:‌ In case of gases (like $P{H_3}$) pressure of gas on metal/catalyst surface acts as activity/concentration hence the rate of reaction depends upon the pressure of gas. On increasing the pressure, adsorption of gas on the metal surface increases hence the rate of reaction also increases.
But at high pressures, the surface of metal/catalyst becomes saturated and thus pressure no longer affects the rate of reaction.


Complete answer:
The decomposition of$P{H_3}$on tungsten at low pressure is a first order reaction because surface area coverage is proportional to partial pressure of $P{H_3}$.
$Rate = k\left[ {{P_{P{H_3}}}} \right]$
According to Freundlich adsorption isotherm:
$\dfrac{x}{m} = k{P^{\dfrac{1}{n}}}$
Here $\dfrac{x}{m}$ is mass of gas absorbed per unit mass of metal/catalyst.
At low pressures n = 1 and it keeps increasing on increasing the pressure.
So, the correct option is (C).

Additional Information:As the pressure increases the dependence of adsorption on pressure decreases and hence the given statement in not correct for higher pressures

Note: We can use differential rate law in order to solve the given problem. In rate law we generally use concentration terms, so in this case concentration term can be correlated with the pressure of PH3. Accordingly, our rate law will be as follows:
$Rate = k\left[ {{P_{P{H_3}}}} \right]$