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 find the rates of production of \( N_2 \) and \( H_2 \) for the zero-order decomposition of \( NH_3 \) on a platinum surface, given the rate constant \( k = 2.5 \times 10^{-4} \, \text{mol L}^{-1} \, \text{s}^{-1} \).

1. Understanding the Reaction and Zero-Order Kinetics:
The decomposition of ammonia (\( NH_3 \)) on a platinum surface can be represented by the balanced chemical equation:

\( 2 NH_3 \rightarrow N_2 + 3 H_2 \)
For a zero-order reaction, the rate of reaction is independent of the concentration of the reactant and is given by:

\( \text{Rate} = k \)
Here, \( k = 2.5 \times 10^{-4} \, \text{mol L}^{-1} \, \text{s}^{-1} \), which represents the rate of decomposition of \( NH_3 \).

2. Rate of Decomposition of \( NH_3 \):
For a zero-order reaction, the rate of reaction is:

\( -\frac{1}{2} \frac{d[NH_3]}{dt} = k = 2.5 \times 10^{-4} \, \text{mol L}^{-1} \, \text{s}^{-1} \)
The coefficient \( \frac{1}{2} \) accounts for the stoichiometry of the reaction (2 moles of \( NH_3 \)). Thus, the rate of consumption of \( NH_3 \) is:

\( \frac{d[NH_3]}{dt} = -2 \times 2.5 \times 10^{-4} = -5.0 \times 10^{-4} \, \text{mol L}^{-1} \, \text{s}^{-1} \)

3. Relating Rates to Stoichiometry:
From the balanced equation \( 2 NH_3 \rightarrow N_2 + 3 H_2 \), the rates of production of \( N_2 \) and \( H_2 \) are related to the rate of consumption of \( NH_3 \). The stoichiometric coefficients give:

\( -\frac{1}{2} \frac{d[NH_3]}{dt} = \frac{d[N_2]}{dt} = \frac{1}{3} \frac{d[H_2]}{dt} \)
Since the rate of reaction is \( k \), we have:

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

4. Calculating the Rate of Production of \( H_2 \):
Using the stoichiometric relationship:

\( \frac{d[H_2]}{dt} = 3 \times \frac{d[N_2]}{dt} = 3 \times k \)
Substitute \( k \):

\( \frac{d[H_2]}{dt} = 3 \times 2.5 \times 10^{-4} = 7.5 \times 10^{-4} \, \text{mol L}^{-1} \, \text{s}^{-1} \)

Final Answer:
The rate of production of \( N_2 \) is \( 2.5 \times 10^{-4} \, \text{mol L}^{-1} \, \text{s}^{-1} \), and the rate of production of \( H_2 \) is \( 7.5 \times 10^{-4} \, \text{mol L}^{-1} \, \text{s}^{-1} \).