Question

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

To address this problem, we must determine the production rates of \( N_2 \) and \( H_2 \) from the zero-order decomposition of \( NH_3 \) on a platinum surface. The rate constant is given as \( k = 2.5 \times 10^{-4} \, \text{mol L}^{-1} \, \text{s}^{-1} \).

1. Balanced Reaction Equation:
The decomposition of \( NH_3 \) proceeds as follows:

\( 2NH_3 \rightarrow N_2 + 3H_2 \).

2. Zero-Order Kinetics Interpretation:
In a zero-order reaction, the rate of \( NH_3 \) decomposition is constant and equal to the rate constant, \( \text{Rate} = k = 2.5 \times 10^{-4} \, \text{mol L}^{-1} \, \text{s}^{-1} \), irrespective of the \( [NH_3] \) concentration. This represents the rate of \( NH_3 \) consumption.

3. Rate Relationships Based on Stoichiometry:
The overall reaction rate can be expressed in terms of the consumption of \( NH_3 \) and the production of \( N_2 \) and \( H_2 \): \( -\frac{1}{2} \frac{d[NH_3]}{dt} = \frac{d[N_2]}{dt} = \frac{1}{3} \frac{d[H_2]}{dt} \). For a zero-order reaction, the rate of \( NH_3 \) consumption is \( -\frac{d[NH_3]}{dt} = k \). Therefore, the rate of \( NH_3 \) consumption is:

\( -\frac{d[NH_3]}{dt} = 2.5 \times 10^{-4} \, \text{mol L}^{-1} \, \text{s}^{-1} \).

4. Calculation of \( N_2 \) Production Rate:
According to the stoichiometry, 2 moles of \( NH_3 \) yield 1 mole of \( N_2 \). Thus, the rate of \( N_2 \) production is half the rate of \( NH_3 \) consumption: \( \frac{d[N_2]}{dt} = \frac{1}{2} \left( -\frac{d[NH_3]}{dt} \right) \):

\( \frac{d[N_2]}{dt} = \frac{1}{2} \times 2.5 \times 10^{-4} = 1.25 \times 10^{-4} \, \text{mol L}^{-1} \, \text{s}^{-1} \).

5. Calculation of \( H_2 \) Production Rate:
The stoichiometry indicates that 2 moles of \( NH_3 \) produce 3 moles of \( H_2 \). Consequently, the rate of \( H_2 \) production is 3/2 times the rate of \( NH_3 \) consumption: \( \frac{d[H_2]}{dt} = \frac{3}{2} \left( -\frac{d[NH_3]}{dt} \right) \):

\( \frac{d[H_2]}{dt} = \frac{3}{2} \times 2.5 \times 10^{-4} = 3.75 \times 10^{-4} \, \text{mol L}^{-1} \, \text{s}^{-1} \).

Final Result:
The production rate of \( N_2 \) is \( 1.25 \times 10^{-4} \, \text{mol L}^{-1} \, \text{s}^{-1} \), and the production rate of \( H_2 \) is \( 3.75 \times 10^{-4} \, \text{mol L}^{-1} \, \text{s}^{-1} \).