Question

(B) The decomposition of $NH_3$ on a platinum surface is a zero-order reaction. What are the rates of production of $N_2$ and $H_2$ if $k = 2.5 \times 10^{-4} \, \text{mol L}^{-1} \, \text{s}^{-1}$?

Hint:

In zero-order reactions, the rate of product formation remains constant and is equal to the rate constant, irrespective of the reactant concentration.

Answer 1

In a zero-order reaction, the rate of reaction remains constant and is equal to the rate constant $k$. The rate of product formation is directly proportional to this rate constant. In this case, the rate constant is $k = 2.5 \times 10^{-4} \, \text{mol L}^{-1} \, \text{s}^{-1}$. For the decomposition of $NH_3$ on the platinum surface, the products $N_2$ and $H_2$ are produced in a 1:3 molar ratio. As a result, the rates of production of both $N_2$ and $H_2$ will be equal to the rate constant $k$. Thus, the rates of production of $N_2$ and $H_2$ are both: $$\text{Rate of production of } N_2 = \text{Rate of production of } H_2 = 2.5 \times 10^{-4} \, \text{mol L}^{-1} \, \text{s}^{-1}.$$ \bigskip