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 solve the problem, we need to determine the rates of production of \( N_2 \) and \( H_2 \) in the zero-order decomposition of \( NH_3 \) on a platinum surface, given \( k = 2.5 \times 10^{-4} \, \text{mol L}^{-1} \, \text{s}^{-1} \).

1. Write the Balanced Reaction:
The decomposition of \( NH_3 \) is:

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

2. Understand Zero-Order Kinetics:
For a zero-order reaction, the rate of decomposition of \( NH_3 \) is \( \text{Rate} = k = 2.5 \times 10^{-4} \, \text{mol L}^{-1} \, \text{s}^{-1} \), independent of \( [NH_3] \). This is the rate of consumption of \( NH_3 \).

3. Relate Rates Using Stoichiometry:
The reaction rate in terms of \( NH_3 \) consumption is \( -\frac{1}{2} \frac{d[NH_3]}{dt} = \frac{d[N_2]}{dt} = \frac{1}{3} \frac{d[H_2]}{dt} \). For zero-order, \( -\frac{d[NH_3]}{dt} = k \). Thus, the rate of consumption of \( NH_3 \) is:

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

4. Calculate Rate of Production of \( N_2 \):
From the stoichiometry, \( 2 \, \text{mol} \) of \( NH_3 \) produce \( 1 \, \text{mol} \) of \( N_2 \). So, \( \frac{d[N_2]}{dt} = \frac{1}{2} \left( -\frac{d[NH_3]}{dt} \right) \):

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

5. Calculate Rate of Production of \( H_2 \):
From the stoichiometry, \( 2 \, \text{mol} \) of \( NH_3 \) produce \( 3 \, \text{mol} \) of \( H_2 \). So, \( \frac{d[H_2]}{dt} = \frac{3}{2} \left( -\frac{d[NH_3]}{dt} \right) \):

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

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