Problem 64 - Chemistry and Chemical Reactivity

Question

Data in the table were collected at \(540 \mathrm{K}\) for the following reaction: $$\mathrm{CO}(\mathrm{g})+\mathrm{NO}_{2}(\mathrm{g}) \longrightarrow \mathrm{CO}_{2}(\mathrm{g})+\mathrm{NO}(\mathrm{g})$$ (a) Derive the rate equation. (b) Determine the reaction order with respect to each reactant. (c) Calculate the rate constant, giving the correct units for \(k.\) $$\begin{array}{lll}\hline \text { Initial Concentration }(\mathrm{mol} / \mathrm{L}) & {\text { Initial Rate }} \\\\\hline[\mathrm{C} 0] & {\left[\mathrm{NO}_{2}\right]} & (\mathrm{mol} / \mathrm{L} \cdot \mathrm{h}) \\\\\hline 5.1 \times 10^{-4} & 0.35 \times 10^{-4} & 3.4 \times 10^{-8} \\\5.1 \times 10^{-4} & 0.70 \times 10^{-4} & 6.8 \times 10^{-8} \\\5.1 \times 10^{-4} & 0.18 \times 10^{-4} & 1.7 \times 10^{-8} \\\1.0 \times 10^{-3} & 0.35 \times 10^{-4} & 6.8 \times 10^{-8} \\\1.5 \times 10^{-3} & 0.35 \times 10^{-4} & 10.2 \times 10^{-8} \\\\\hline\end{array}$$

Step-by-Step Solution

Answer

Rate equation: \( \text{Rate} = k[\mathrm{CO}][\mathrm{NO}_2] \) with \( k \approx 1.90 \times 10^4 \text{ L} \cdot \text{mol}^{-1} \cdot \text{h}^{-1} \). Order: first with respect to both reactants.

Step 1: Write the General Rate Equation

The general rate equation for the reaction \( \mathrm{CO}(\mathrm{g}) + \mathrm{NO}_{2}(\mathrm{g}) \to \mathrm{CO}_{2}(\mathrm{g}) + \mathrm{NO}(\mathrm{g}) \) is \( \text{Rate} = k [\mathrm{CO}]^m [\mathrm{NO}_2]^n \), where \( k \) is the rate constant, and \( m \) and \( n \) are the reaction orders with respect to \( \mathrm{CO} \) and \( \mathrm{NO}_2 \), respectively.

Step 2: Determine the Reaction Order with Respect to \(\mathrm{NO}_2\)

Compare experiments where the concentration of \( \mathrm{CO} \) is constant. Comparing the first and second experiments: \- \( \text{Initial Rate}_2 / \text{Initial Rate}_1 = (6.8 \times 10^{-8}) / (3.4 \times 10^{-8}) = 2 \) \- \( [\mathrm{NO}_2]_2 / [\mathrm{NO}_2]_1 = (0.70 \times 10^{-4}) / (0.35 \times 10^{-4}) = 2 \) \The rate doubles when \( [\mathrm{NO}_2] \) doubles, indicating a first-order reaction with respect to \( \mathrm{NO}_2 \) (\( n = 1 \)).

Step 3: Determine the Reaction Order with Respect to \(\mathrm{CO}\)

Compare experiments where the concentration of \( \mathrm{NO}_2 \) is constant. Comparing the first and fourth experiments: \- \( \text{Initial Rate}_4 / \text{Initial Rate}_1 = (6.8 \times 10^{-8}) / (3.4 \times 10^{-8}) = 2 \) \- \( [\mathrm{CO}]_4 / [\mathrm{CO}]_1 = (1.0 \times 10^{-3}) / (5.1 \times 10^{-4}) = 2 \) \The rate doubles when \( [\mathrm{CO}] \) doubles, indicating a first-order reaction with respect to \( \mathrm{CO} \) (\( m = 1 \)).

Step 4: Derive the Rate Equation

Since both orders are 1, the rate equation is: \( \text{Rate} = k [\mathrm{CO}] [\mathrm{NO}_2] \).

Step 5: Calculate the Rate Constant \(k\)

Use the data from any experiment to solve for \( k \). Using the first experiment: \- \( 3.4 \times 10^{-8} = k (5.1 \times 10^{-4}) (0.35 \times 10^{-4}) \) \- Solve for \( k \): \( k = \frac{3.4 \times 10^{-8}}{(5.1 \times 10^{-4}) (0.35 \times 10^{-4})} \approx 1.90 \times 10^{4} \text{ L} \cdot \text{mol}^{-1} \cdot \text{h}^{-1} \).

Key Concepts

Rate Law

A rate law is an equation that relates the rate of a chemical reaction to the concentration of its reactants. It provides crucial information about how the concentration of reactants affects the speed of the reaction. Typically, the rate law for a reaction can be expressed in the form:

• Rate = k [A]^m [B]^n

Here,

• k is the rate constant.
• [A] and [B] represent concentrations of reactants A and B.
• m and n are the reaction orders for A and B respectively.

This equation helps chemists predict the speed of reaction given the current reactant concentrations.

Reaction Order

Reaction order indicates how the rate of a reaction depends on the concentration of the reactants. It's a value that tells us by what factor the rate of reaction changes when the concentration of a reactant is altered.

• If the reaction is first-order with regard to a reactant, then doubling the concentration of that reactant will double the reaction rate.
• If it's second-order with respect to a reactant, doubling the concentration will quadruple the rate, and so on.

In the rate law, the reaction orders are the exponents, such as m and n from the equation:

• Rate = k [A]^m [B]^n

Reaction orders can be determined experimentally and are not necessarily related to the stoichiometric coefficients of the reactants in the balanced chemical equation.

Rate Constant

The rate constant, denoted as k, is a crucial parameter in the rate law equation that quantifies the speed of a reaction. It is a proportionality factor that helps adjust the units of concentration to those of the reaction rate. However, the rate constant is not influenced by the concentration of reactants.

• The units of the rate constant depend on the overall reaction order.
• For a first-order reaction, the units are s^-1.
• For a second-order reaction, k is expressed in L mol^-1 s^-1.

The value of k can be calculated using experimental data by rearranging the rate law to solve for k:
\[k = \frac{\text{Rate}}{[A]^m [B]^n}\]
where [A] and [B] are concentrations of reactants involved.

Kinetic Data Analysis

Kinetic data analysis involves analyzing experimental data to determine reaction rates and how they change with concentration. This analysis helps to identify the rate law and the order of the reaction.

• The initial rate method is often used, where the initial reaction rate is measured for different concentrations of reactants.
• By comparing how rates vary with different concentrations, reaction orders can be deduced.
• Once the rate law is established, it is possible to calculate the rate constant using data from one of the experiments and solve for k.

This process can also verify if the theoretical model is correct by checking if the calculated rate constant remains consistent regardless of which experimental data set is used.