# Ammonia decomposition on platinum surface

I was given a problem I have to solve, unfortunately I cant wrap my head about it. I've searched the page for answers but they doesnt seem to fulfill my needs, so i'm asking for little guidance.

The problem:

When the decomposition of ammonia is carried out on platinum surface at 1000°C the hydrogen binds strongly on the surface. Prove the that the rate is equal to $$\frac{dp(NH_3)}{dt}=-k\frac{p(NH_3)}{p(H_2)}$$.

I dont understand why we have to divide with the partial pressure of hydrogen. In my understanding the rate is $$\frac{dp(NH_3)}{dt}=-kp(NH_3)$$. I know I have to use the fact that hydrogen binds strongly at the surface, but I'm lost. Can you please explain to me how should I derive this, or just hint me to the right direction.

I lack of good understanding on this topic, so sorry for this novice question.

What I've searched:

Why is decomposition of ammonia gas on quartz a 1st order reaction?

What is the molecularity of the RDS of a zero order complex reaction?

$$\ce{2NH_3 ->N_2 + 3H_2}$$ $$\theta_{NH_3}=\frac{K_{NH_3}p(NH_3)}{1+K_{NH_3}p(NH_3)+K_{N_2}p(N_2)+K_{H_2}p(H_2)}$$ $$\theta_{N_2}=\frac{K_{N_2}p(N_2)}{1+K_{NH_3}p(NH_3)+K_{N_2}p(N_2)+K_{H_2}p(H_2)}$$ $$\theta_{H_2}=\frac{K_{H_2}p(H_2)}{1+K_{NH_3}p(NH_3)+K_{N_2}p(N_2)+K_{H_2}p(H_2)}$$
Since the hydrogen binds strongly on the surface, we can safely determine that $$K_{H_2}>>1$$, wich means the denominator $$1+K_{NH_3}p(NH_3)+K_{N_2}p(N_2)+K_{H_2}p(H_2)\approx K_{H_2}p(H_2)$$. So the surface coverage of ammonia can be written as $$\theta_{NH_3}=\frac{K_{NH_3}p(NH_3)}{K_{H_2}p(H_2)}$$. By definition $$v=-\frac{dp(NH_3)}{dt}=k_0\theta_{NH_3}=k_0\frac{K_{NH_3}p(NH_3)}{K_{H_2}p(H_2)}$$, defining $$k=k_0\frac{K_{NH_3}}{K_{H_2}}$$ we arrive at the desired equation.