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