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?

Thank you for your time.
Thanks to Poutnik's comment, I arrived to the answer and I would like to share it.

$$\ce{2NH_3 -&gt;N_2 + 3H_2}$$
$$\theta_{NH_3}=\frac{K_{NH_3}p(NH_3)}{1+K_{NH_3}p(NH_3)+K_{N_2}p(N_2)+K_{H_2}p(H_2)}$$
$$\theta_{N_2}=\frac{K_{N_2}p(N_2)}{1+K_{NH_3}p(NH_3)+K_{N_2}p(N_2)+K_{H_2}p(H_2)}$$
$$\theta_{H_2}=\frac{K_{H_2}p(H_2)}{1+K_{NH_3}p(NH_3)+K_{N_2}p(N_2)+K_{H_2}p(H_2)}$$

Since the hydrogen binds strongly on the surface, we can safely determine that $K_{H_2}&gt;&gt;1$, 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.

Feel free to edit.