5.9 Steady-State Approximation

Mastering the Steady-State Approximation in Chemical Kinetics

Chemical reactions often consist of multiple steps, each contributing to the overall rate of the reaction. When analyzing complex reaction mechanisms, the steady-state approximation can simplify calculations and help determine the reaction rate. This guide uses relatable analogies and examples to break down how the steady-state approximation works, making it a valuable tool for mastering kinetics.

The Bathtub Analogy

Imagine a bathtub being filled with water from a faucet while water drains out through a drain. Initially, the water level rises because the faucet adds water faster than it drains. However, over time, the water level reaches a steady state where the rate of water entering the tub equals the rate of water draining out.

In this analogy:

• The faucet represents the rate at which reactants are added.
• The drain represents the rate at which products are removed.
• The steady-state condition represents a situation in which the concentration of intermediates remains approximately constant because their rate of formation equals their rate of consumption.

Similarly, in chemical reactions, we can assume a steady state when the rate of formation of an intermediate equals its rate of consumption. This assumption simplifies calculations and helps determine the overall reaction rate without considering the detailed time-dependent changes in intermediate concentrations.

When to Use Steady-State Approximation

The steady-state approximation is especially useful when:

1. The first step of a mechanism is not the slow (rate-determining) step.
2. There is a fast equilibrium established before the slow step.

This method allows you to substitute terms from the fast step into the slow step, making complex kinetics more manageable.

Example: Decomposition of Nitrogen Dioxide with Hydrogen

Let’s illustrate how to apply the steady-state approximation using the reaction mechanism below:

1. Step 1 (Fast Equilibrium):
   $2 \text{NO} \rightleftharpoons \text{N}_2\text{O}_2$

2. Step 2 (Slow):
   $\text{N}_2\text{O}_2 + \text{H}_2 \rightarrow \text{H}_2\text{O} + \text{N}_2\text{O}$

The overall balanced equation for this reaction is:

$2 \text{NO} + \text{H}_2 \rightarrow \text{H}_2\text{O} + \text{N}_2\text{O}$

Identifying Intermediates and Catalysts

• Intermediate: $\text{N}_2\text{O}_2$ is formed in the first step and consumed in the second.
• Catalyst: If $\text{N}_2\text{O}_2$ were present initially and unchanged throughout, it would be considered a catalyst. In this case, it is an intermediate.

Writing the Rate Law Using Steady-State Approximation

1. Rate Law for the Slow Step:
   $\text{Rate} = k[\text{N}_2\text{O}_2][\text{H}_2]$
   However, $\text{N}_2\text{O}_2$ is an intermediate and does not appear in the overall balanced equation. Therefore, we need to eliminate it.

2. Using the Fast Equilibrium Step:
   Since Step 1 is a fast equilibrium, we can express the relationship between reactants and the intermediate:
   $\text{Rate}_{\text{forward}} = \text{Rate}_{\text{backward}}$
   $k_a[\text{NO}]^2 = k_e[\text{N}_2\text{O}_2]$
   Rearranging to solve for $[\text{N}_2\text{O}_2]$:
   $[\text{N}_2\text{O}_2] = \frac{k_a}{k_e}[\text{NO}]^2$

3. Substitute into the Rate Law:
   Substituting $[\text{N}_2\text{O}_2]$ into the rate law for the slow step:
   $\text{Rate} = k[\text{N}_2\text{O}_2][\text{H}_2]$
   $\text{Rate} = k \left( \frac{k_a}{k_e} \right) [\text{NO}]^2[\text{H}_2]$

   We can simplify by defining a new rate constant $k” = k \left( \frac{k_a}{k_e} \right)$:
   $\text{Rate} = k”[\text{NO}]^2[\text{H}_2]$

Key Takeaways

• Steady-State Approximation simplifies complex reaction mechanisms by assuming that the rate of formation of an intermediate equals its rate of consumption.
• Fast Equilibrium in an initial step allows substitution of intermediates using equilibrium expressions.
• The rate law derived must match the experimentally determined rate law to validate a proposed mechanism.

Practical Applications

Understanding the steady-state approximation helps explain and predict the rates of complex reactions in fields such as environmental chemistry, pharmacology, and industrial processes. Mastering this tool can enhance your ability to analyze mechanisms and derive accurate rate laws.