Question

Carbon monoxide reacts with \(\mathrm{O}_{2}\) to form \(\mathrm{CO}_{2}\) : $$2 \mathrm{CO}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{g}) \longrightarrow 2 \mathrm{CO}_{2}(\mathrm{g})$$ Information on this reaction is given in the table below. $$\begin{array}{lll}\hline[\mathrm{CO}](\mathrm{mol} / \mathrm{L}) & {\left[\mathrm{O}_{2}\right](\mathrm{mol} /\mathrm{L})} & \text { Rate }(\mathrm{mol} / \mathrm{L} \cdot \mathrm{min}) \\\\\hline 0.02 & 0.02 & 3.68 \times 10^{-5} \\\0.04 & 0.02 & 1.47 \times 10^{-4} \\\0.02 & 0.04 & 7.36 \times 10^{-5} \\\\\hline\end{array}$$ (a) What is the rate law for this reaction? (b) What is the order of the reaction with respect to CO? What is the order with respect \(\mathrm{O}_{2} ?\) What is the overall order of the reaction? (c) What is the value for the rate constant, \(k ?\)

Short answer

(a) Rate law: \( \text{Rate} = k [\text{CO}]^2 [\text{O}_2]^1 \). (b) Order with respect to CO: 2; O2: 1; Overall order: 3. (c) Rate constant \( k = 0.046 \text{ L}^2/\text{mol}^2/\text{min} \).

Step-by-step solution

Write the General Rate Law

The general form of a rate law is given by \( \text{Rate} = k [\text{CO}]^m [\text{O}_2]^n \), where \( k \) is the rate constant, \( m \) is the reaction order with respect to CO, and \( n \) is the reaction order with respect to \( \text{O}_2 \).

Analyze Initial Data for Order with Respect to CO

To find \( m \), compare experiments 1 and 2, where \([\text{O}_2]\) is constant. The concentration of \([\text{CO}]\) is doubled (from 0.02 to 0.04) and the rate quadruples (from \(3.68 \times 10^{-5}\) to \(1.47 \times 10^{-4}\)). Hence, \( 4^m = 2\), so \( m = 2 \).

Analyze Data for Order with Respect to O2

For \( n \), compare experiments 1 and 3, where \([\text{CO}]\) remains constant and \([\text{O}_2]\) is doubled, resulting in the rate doubling as well (from \(3.68 \times 10^{-5}\) to \(7.36 \times 10^{-5}\)). Thus, \( 2^n = 2\), giving \( n = 1 \).

Write the Complete Rate Law

Substituting the values of \(m\) and \(n\) into the rate law, we get \( \text{Rate} = k [\text{CO}]^2 [\text{O}_2]^{1} \).

Calculate the Rate Constant (k)

Use any set of experimental data to calculate \( k \). Using experiment 1: \(3.68 \times 10^{-5} = k (0.02)^2 (0.02) \). Solving gives \( k = \frac{3.68 \times 10^{-5}}{(0.02)^3} \). Calculating this results in \( k = 0.046 \text{ L}^2/\text{mol}^2/\text{min} \).

Determine the Overall Reaction Order

The overall reaction order is the sum of the orders with respect to each reactant: \( m + n = 2 + 1 = 3 \).

Key concepts

Rate Law Determination

Understanding how to determine the rate law for a chemical reaction is a crucial part of chemical kinetics. The rate law expresses the reaction rate as a function of the concentration of reactants. For the carbon monoxide and oxygen reaction forming carbon dioxide, the generalized form of the rate law is written as \( \text{Rate} = k [\text{CO}]^m [\text{O}_2]^n \). Here, \( k \) represents the rate constant, while \( m \) and \( n \) are the orders of the reaction with respect to CO and \( \text{O}_2 \), respectively.

To find these orders, the concentration of one reactant is held constant while varying the other. The change in rate helps determine the exponent values \( m \) and \( n \). This method reveals not just how sensitive the rate is to changes in reactant concentrations but also allows for the same conclusion to apply under varied experimental setups. By analyzing the experiments provided, both the \( \text{CO} \) and \( \text{O}_2 \) concentration changes and their correlating rate changes guide in determining the orders of components.

This systematic approach is applicable to any reaction, making it a pivotal skill for students analyzing kinetic data.

Reaction Order

The reaction order offers insight into how reactant concentration impacts the rate of reaction. In the context of the exercise, we focus separately on CO and \( \text{O}_2 \).

**Determining Order with Respect to CO**
By comparing experiments where \( \text{O}_2 \) remains constant, and \( \text{CO} \) is varied, the results show a pronounced effect. Doubling the concentration of CO causes the rate to quadruple. This indicates that the reaction order regarding CO, denoted as \( m \), is 2. This quadratic relationship reveals that the rate is heavily influenced by changes in the concentration of CO.

**Determining Order with Respect to \( \text{O}_2 \)**
Similarly, when analyzing \( \text{O}_2 \) while keeping the \( \text{CO} \) constant, doubling \( \text{O}_2 \) leads to a doubling of the rate. Such a change signifies a linear dependency, or an order of 1 for \( \text{O}_2 \). This first-order relationship means the rate is directly proportional to the concentration of \( \text{O}_2 \).

**Overall Reaction Order**
The total reaction order is the sum of the individual orders, \( m + n = 2 + 1 = 3 \). Thus, it's a third-order reaction, which is vital for predicting how the rate will change with different reactant concentrations.

Rate Constant Calculation

The rate constant, \( k \), forms the bridge between concentration, reaction order, and reaction rate in the rate law equation. To find \( k \), use experimental data and substitute it into the equation\[ \text{Rate} = k [\text{CO}]^2 [\text{O}_2]^1 \]. Using data from the experiments provided makes this step precise.

**Steps to Calculate \( k \):**

• Select an experiment, such as experiment 1. Here, the rate \( = 3.68 \times 10^{-5} \) mol/L·min, \([\text{CO}] = 0.02\) mol/L, and \([\text{O}_2] = 0.02\) mol/L.
• Substitute known values into the rate law equation, yielding \(3.68 \times 10^{-5} = k (0.02)^2 (0.02) \).
• Solve for \( k \), which gives \[ k = \frac{3.68 \times 10^{-5}}{(0.02)^3} \].
• Calculate to find \( k = 0.046 \text{ L}^2/\text{mol}^2/\text{min} \).

Knowing \( k \) is valuable because it allows for the prediction of reaction rates under various conditions, independent of concentration. This makes it a fundamental parameter in chemical kinetics.