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:

For zero-order reactions, the rate of product formation is constant and equal to the rate constant. This applies regardless of the concentration of reactants.

Answer 1

For a zero-order reaction, the rate of reaction is constant and equal to the rate constant \( k \). The rate of production of products is the same as the rate constant. In this case, the rate constant \( k = 2.5 \times 10^{-4} \, \text{mol L}^{-1} \, \text{s}^{-1} \). The decomposition of \( NH_3 \) on the platinum surface produces \( N_2 \) and \( H_2 \) in a 1:3 molar ratio. Therefore, the rates of production of both \( N_2 \) and \( H_2 \) will be equal to \( k \). Thus, the rates of production of \( N_2 \) and \( H_2 \) are both: \[ \text{Rate of production of } N_2 = \text{Rate of production of } H_2 = 2.5 \times 10^{-4} \, \text{mol L}^{-1} \, \text{s}^{-1}. \]