Problem 134 - Chemistry: An Atoms-Focused Approach

Question

The following table contains kinetics data for the reaction $$2 \mathrm{NO}(g)+\mathrm{Cl}_{2}(g) \rightarrow 2 \mathrm{NOCl}(g)$$ $$\begin{array}{|c|c|c|c|}\hline \text { Experiment } & \text { [NO]o }(M) & \left[\mathrm{Cl}_{2}\right]_{0}(M) & \begin{array}{c}\text { Initial } \\\\\text { Rate }(M / \mathrm{s})\end{array} \\\\\hline 1 & 0.20 & 0.10 & 0.63 \\\\\hline 2 & 0.20 & 0.30 & 5.70 \\\\\hline 3 & 0.80 & 0.10 & 2.58 \\\\\hline 4 & 0.40 & 0.20 & ? \\\\\hline\end{array}$$ Predict the initial rate of reaction in experiment 4.

Step-by-Step Solution

Answer

Answer: The initial rate of reaction for Experiment 4 is 10.08 M/s.

Step 1: Determine the reaction order for each reactant

By analyzing the ratio of initial rates and the ratio of initial concentrations between different experiments, we can determine the reaction order for each reactant. The general rate law for the reaction is given by: $$\text{Rate} = k[\mathrm{NO}]^{m}[\mathrm{Cl}_{2}]^{n}$$ where m and n are the reaction orders for NO and Cl2, respectively. From the table, we have data for the following reaction experiments: - Experiment 1: $$\text{Rate}_1 = k[\mathrm{0.20}]^{m}[\mathrm{0.10}]^{n}$$ - Experiment 2: $$\text{Rate}_2 = k[\mathrm{0.20}]^{m}[\mathrm{0.30}]^{n}$$ - Experiment 3: $$\text{Rate}_3 = k[\mathrm{0.80}]^{m}[\mathrm{0.10}]^{n}$$

Step 2: Determine the order for NO

We need to examine two experiments where the concentration of Cl2 is kept constant. Experiments 1 and 3 meet this criterion, so we can divide their rate expressions to find the order with respect to NO: $$\frac{\text{Rate}_3}{\text{Rate}_1} = \frac{k[\mathrm{0.80}]^{m}[\mathrm{0.10}]^{n}}{k[\mathrm{0.20}]^{m}[\mathrm{0.10}]^{n}}$$ Using the given rate data, we have: $$\frac{2.58}{0.63} = \frac{[\mathrm{0.80}]^{m}}{[\mathrm{0.20}]^{m}}$$ After simplifying, we get: $$4.1 = 4^{m}$$ Solving for m, we find that the order of reaction for NO is m = 1.

Step 3: Determine the order for Cl2

To find the order for Cl2, we can examine two experiments where the concentration of NO is kept constant. Experiments 1 and 2 meet this criterion, so we can divide their rate expressions to find the order with respect to Cl2: $$\frac{\text{Rate}_2}{\text{Rate}_1} = \frac{k[\mathrm{0.20}]^{m}[\mathrm{0.30}]^{n}}{k[\mathrm{0.20}]^{m}[\mathrm{0.10}]^{n}}$$ Using the given rate data and the value of m we found in Step 2, we have: $$\frac{5.70}{0.63} = [\mathrm{0.30}]^{n}$$ After simplifying, we get: $$9 = 3^{n}$$ Solving for n, we find that the order of reaction for Cl2 is n = 2.

Step 4: Write the Rate Law

With the order of the reaction determined for both reactants, we can now write the rate law: $$\text{Rate} = k[\mathrm{NO}]^{1}[\mathrm{Cl}_{2}]^{2}$$

Step 5: Calculate the rate constant (k) using one of the previous experiments

We can use the rate law and the data from any of the earlier experiments to calculate the rate constant k. We will use Experiment 1 for this purpose: $$\text{Rate}_1 = k[\mathrm{0.20}]^{1}[\mathrm{0.10}]^{2}$$ $$0.63 = k[\mathrm{0.20}][\mathrm{0.01}]$$ Solving for k, we get: $$k = 315 \, M^{-2}s^{-1}$$

Step 6: Predict the initial rate for Experiment 4

Now that we have the rate law and the rate constant, we can predict the initial rate of the reaction for Experiment 4 by plugging in the initial concentrations of the reactants: $$\text{Rate}_4 = k[\mathrm{0.40}]^{1}[\mathrm{0.20}]^{2}$$ $$\text{Rate}_4 = 315 \, M^{-2}s^{-1}[\mathrm{0.40}][\mathrm{0.04}]$$ Calculating the rate, we find the initial rate of reaction for Experiment 4 is: $$\text{Rate}_4 = 10.08 \, M/s$$

Key Concepts

Rate Law

In chemical kinetics, the rate law is a mathematical expression that describes the speed of a chemical reaction based on the concentration of its reactants. For a reaction such as \[2 \mathrm{NO}(g)+\mathrm{Cl}_{2}(g) \rightarrow 2 \mathrm{NOCl}(g)\] the rate law can be represented as:\[\text{Rate} = k[\mathrm{NO}]^{m}[\mathrm{Cl}_{2}]^{n}\]Here,

• \(k\) is the rate constant, specific to the reaction at a given temperature.
• \([\mathrm{NO}]\) and \([\mathrm{Cl}_{2}]\) are the concentrations of the reactants.
• \(m\) and \(n\) are the reaction orders for \(\mathrm{NO}\) and \(\mathrm{Cl}_{2}\), respectively.

Understanding the rate law helps us to predict how changes in concentration will affect the rate. In our given example, finding the rate law allows us to determine the initial rate of the reaction under different conditions.

Reaction Order

Reaction order refers to the power dependence of the rate on the concentration of each reactant. It's the sum of the exponents in the rate law equation with regard to each reactant. For our reaction with NO and Cl\(_2\), the orders were determined by analyzing the experimental data:

• The order of reaction with respect to \(\mathrm{NO}\) was found by comparing experiments where \([\mathrm{Cl}_2]\) remained constant. It was determined to be \(m = 1\).
• The order for \(\mathrm{Cl}_2\) was analyzed where \([\mathrm{NO}]\) remained constant. It was found to be \(n = 2\).

This tells us that the reaction rate is directly proportional to the concentration of \(\mathrm{NO}\) and proportional to the square of the \(\mathrm{Cl}_2\) concentration. Understanding the reaction order is crucial for predicting how the reaction rate changes with different concentrations.

Rate Constant

The rate constant \(k\) in a rate law controls how fast a reaction proceeds under set conditions. It is unique to a specific reaction at a given temperature. In calculating \(k\) from experiment data, we substitute concentrations and the rate into the rate law.For the reaction \(2 \mathrm{NO}(g)+\mathrm{Cl}_{2}(g) \rightarrow 2 \mathrm{NOCl}(g)\), using Experiment 1:\[\text{Rate}_1 = k[0.20]^{1}[0.10]^{2} = 0.63 \, \text{M/s}\]Solving for \(k\):\[k = \frac{0.63}{0.20 \times 0.01} = 315 \, \text{M}^{-2}\text{s}^{-1}\]The value of \(k\) shows the rate of reaction per unit concentration and can be used to predict the rate under different conditions.

Chemical Kinetics

Chemical kinetics focuses on the speed or rate of chemical reactions and the factors affecting them. Understanding kinetics allows chemists to control reactions in industrial, biological, and environmental processes. The fundamental aspects of chemical kinetics include:

• The rate law, which relates reaction rate with reactant concentrations.
• Reaction order, describing the influence of each reactant's concentration on the rate.
• The rate constant, a specific value that determines reaction speed.

Through the study of chemical kinetics, we can optimize reaction conditions to enhance yields, control reaction pathways, and improve energy efficiency in chemical manufacturing.