Question

Decomposition of phosphine on tungsten at low pressure is a first order reaction. This is because the

Answer 1

ANSWER : The rate of decomposition of phosphine on tungsten is proportional to surface coverage.

EXPLANATION :

Lets write the equation for the decomposition.

PH₃ ---- Tungsten → P + 3/2 H₂

The rate of decomposition is :

R = k [ PH₃ ] where k is a constant.

The rate of of decomposition of phosphine is dependent on surface coverage since :

At low pressure ,the surface area covered is proportional to partial pressure of PH₃.

And it is also first order with respect to PH₃.

DETAILS:

∅ = kP / 1 +kp where:

∅ = Fraction of the surface area

k = constant

p = pressure


∅ = kP

thus the answer is proved.

Answer 2

Since the options are not given, I am giving the answer based on the options from another website.

The correct answer should be - Rate is proportional to the surface coverage 

Rate = k[PH₃]

The rate is dependent on the surface coverage as under conditions of low pressure, the surface area covered will be proportional to the partial pressure of phosphine (PH₃). So, it will be a first order reaction with respect to Phosphine (PH₃).

The equations are here -

θ = 1+kP1+kP

θ = fraction of surface area covered at low pressure

θ = k P