Question

The decomposition of NH$_3$ on 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$^{-4}$ mol L$^{-1}$ s$^{-1}$?

Hint:

For zero-order reactions, the rate of product formation is constant and independent of reactant concentration.

Answer 1

The rate of reaction for a zero-order reaction is given by: \[ \text{rate} = k \, \text{[A]}_0 \] Here, \(k = 2.5 \times 10^{-4} \, \text{mol L}^{-1} \, \text{s}^{-1}\). The rate of formation of products (N$_2$ and H$_2$) will be directly proportional to the rate of the decomposition of NH$_3$.
Since each mole of NH$_3$ produces 1 mole of N$_2$ and 3 moles of H$_2$, the rates are: \[ \text{rate of N}_2 = \frac{1}{2} \times \text{rate of NH}_3 \] \[ \text{rate of H}_2 = \frac{3}{2} \times \text{rate of NH}_3 \] Therefore: \[ \text{rate of N}_2 = 1.25 \times 10^{-4} \, \text{mol L}^{-1} \, \text{s}^{-1}, \quad \text{rate of H}_2 = 3.75 \times 10^{-4} \, \text{mol L}^{-1} \, \text{s}^{-1} \]