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Rate Laws from Experimental Data: Initial Rates Method. The reaction $$2 \mathrm{NO}(g)+\mathrm{Cl}_{2}(g) \longrightarrow 2 \mathrm{NOCl}(g)$$ was studied at \(-10^{\circ} \mathrm{C}\). The following results were obtained where $$\text { Rate }=-\frac{\Delta\left[\mathrm{Cl}_{2}\right]}{\Delta t}$$ $$ \begin{array}{ccc} {[\mathrm{NO}]_{0}} & {\left[\mathrm{Cl}_{2}\right]_{0}} & \text { Initial Rate } \\ (\mathrm{mol} / \mathrm{L}) & (\mathrm{mol} / \mathrm{L}) & (\mathrm{mol} / \mathrm{L} \cdot \mathrm{min}) \\ 0.10 & 0.10 & 0.18 \\ 0.10 & 0.20 & 0.36 \\ 0.20 & 0.20 & 1.45 \end{array} $$ a. What is the rate law? b. What is the value of the rate constant?

Short Answer

Expert verified
a. The rate law for the given reaction is: \(\text { Rate } = k[\mathrm{NO}]^2[\mathrm{Cl}_2]\) b. The rate constant (k) is \(180 L^2/mol^2 min\).

Step by step solution

01

Determine the order of the reaction with respect to NO

Compare experiment 1 and 3, where the initial concentration of Cl2 is constant. This will help us determine the effect of changing the concentration of NO on the rate of the reaction. Experiment 1: [NO] = 0.10 M, [Cl2] = 0.10 M, Initial Rate = 0.18 M/min Experiment 3: [NO] = 0.20 M, [Cl2] = 0.20 M, Initial Rate = 1.45 M/min The concentration of NO doubles while the concentration of Cl2 and the rate of the reaction are kept constant. Observe how much the initial rate increases when [NO] doubles: $$\frac{1.45}{0.18} = 8.06 \approx 8$$ The rate increases by a factor of 8 when the concentration of NO doubles, which indicates that the reaction is second order with respect to NO.
02

Determine the order of the reaction with respect to Cl2

Compare experiment 1 and 2, where the initial concentration of NO is constant. This will help us determine the effect of changing the concentration of Cl2 on the rate of the reaction. Experiment 1: [NO] = 0.10 M, [Cl2] = 0.10 M, Initial Rate = 0.18 M/min Experiment 2: [NO] = 0.10 M, [Cl2] = 0.20 M, Initial Rate = 0.36 M/min The concentration of Cl2 doubles while the concentration of NO and the rate of the reaction are kept constant. Observe how much the initial rate increases when [Cl2] doubles: $$\frac{0.36}{0.18} = 2$$ The rate doubles when the concentration of Cl2 doubles, which indicates that the reaction is first order with respect to Cl2.
03

Write the rate law

Since the reaction is second order with respect to NO and first order with respect to Cl2, the rate law can be written as: $$\text { Rate } = k[\mathrm{NO}]^2[\mathrm{Cl}_2]$$ where k is the rate constant.
04

Calculate the rate constant, k

Use the rate law from step 3 and the experimental data from any of the experiments. In this case, let's use the data from experiment 1. Initial Rate = 0.18 M/min, [NO] = 0.10 M, [Cl2] = 0.10 M Substitute the values into the rate law: $$0.18 = k(0.10)^2(0.10)$$ Solve for k: $$k = \frac{0.18}{(0.10)^2(0.10)} = 180$$ The rate constant k is 180 L^2/mol^2 min. The final answers are: a. Rate law: \(\text { Rate } = k[\mathrm{NO}]^2[\mathrm{Cl}_2]\) b. Rate constant: k = \(180 L^2/mol^2 min\)

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Initial Rates Method
The Initial Rates Method is a crucial approach used in chemical kinetics to decipher how reactions proceed, specifically focusing on the reaction rate at the very beginning. When studying a chemical reaction, such as the one involving NO and Clâ‚‚, scientists measure the rate at which reactants are converted into products at the start of the reaction. The initial rate provides insights into the relationship between the concentration of reactants and the speed of the reaction.
In practice, several experiments are carried out with varying initial concentrations of the reactants. These experiments help determine how changes in concentrations affect the rate, thus guiding the formulation of a suitable rate law. For instance, by comparing different trials where only one concentration is changed at a time, it becomes possible to isolate the effect of each reactant on the reaction rate. This method helps in accurately pinpointing the reaction order for each reactant.

Overall, the Initial Rates Method simplifies complex chemical reactions by breaking down initial experimental data into manageable observations. It lays the foundation for understanding kinetics, helping to shape the mathematical expression governing the reaction's behavior.
Reaction Order
Understanding the Reaction Order is fundamental in predicting and controlling chemical reaction rates. Reaction order tells us how the rate of a reaction depends on the concentration of its reactants.
In the case of the reaction between NO and Clâ‚‚, the reaction was determined to be second order with respect to NO and first order with respect to Clâ‚‚. This means that the concentration of NO appears squared in the rate law, indicating that doubling [NO] results in an eightfold increase in the reaction rate (due to its square relation), while doubling [Clâ‚‚] leads to a doubling of the rate.
Reaction orders are determined experimentally, typically using the Initial Rates Method, and can be zero, positive, or even fractional. They are not always the same as the stoichiometric coefficients in the balanced chemical equation.

Knowing the order of a reaction is essential for understanding the mechanism and for effectively designing chemical processes. It provides valuable insights into the intricacies of molecular interactions during the reaction.
Rate Constant
The Rate Constant, denoted as "k" in the rate law equation, is a pivotal factor that helps quantify the reaction rate. It marks how quickly a reaction proceeds and is specific to a given reaction at a set temperature.
The rate constant for the reaction of NO with Cl₂ was determined to be 180 L²/mol² min, indicating how the speed of this reaction varies with changes in reactant concentrations. The dimensions of the rate constant will vary depending on the overall order of the reaction. For example, in a third-order reaction, like our example, the units are L²/mol² min.
  • Temperature Sensitivity: The rate constant is highly temperature-dependent; even slight changes can result in significant variations in k.
  • Role in Kinetics: It integrates both the intrinsic properties of the reactants and the effects of catalysts, showing the efficiency of converting reactants to products.
The rate constant represents more than just a number. It offers essential insights into the pace at which a chemical reaction will proceed under specific conditions.
Chemical Kinetics
Chemical Kinetics is the study of reaction rates and the mechanisms underlying them, providing both theoretical and practical insights into how chemical reactions unfold over time.
In essence, chemical kinetics allows us to comprehend the speed at which reactants transform into products and provides a framework to manipulate these rates for various applications such as manufacturing, pharmaceuticals, and environmental science. It involves analyzing factors like reactant concentration, temperature, and the presence of catalysts.

Focusing on reaction rates, chemical kinetics includes:
  • Determining Rate Laws: Establishing mathematical models that predict how reaction rates vary with different conditions and concentrations.
  • Studying Reaction Mechanisms: Exploring step-by-step pathways to unveil how molecules interact at a fundamental level.
Therefore, chemical kinetics is not just about solving mathematical equations. It's about connecting theory with real-world chemical behaviors to innovate, enhance efficiencies, and provide sustainable solutions.

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Most popular questions from this chapter

Write the rate laws for the following elementary reactions. a. \(\mathrm{CH}_{3} \mathrm{NC}(g) \rightarrow \mathrm{CH}_{3} \mathrm{CN}(g)\) b. \(\mathrm{O}_{3}(g)+\mathrm{NO}(g) \rightarrow \mathrm{O}_{2}(g)+\mathrm{NO}_{2}(g)\) c. \(\mathrm{O}_{3}(g) \rightarrow \mathrm{O}_{2}(g)+\mathrm{O}(g)\) d. \(\mathrm{O}_{3}(g)+\mathrm{O}(g) \rightarrow 2 \mathrm{O}_{2}(g)\)

Describe at least two experiments you could perform to determine a rate law.

The rate constant \((k)\) depends on which of the following (there may be more than one answer)? a. the concentration of the reactants b. the nature of the reactants c. the temperature d. the order of the reaction Explain.

The rate law for the reaction $$ \begin{array}{c}{\mathrm{Cl}_{2}(g)+\mathrm{CHCl}_{3}(g) \longrightarrow \mathrm{HCl}(g)+\mathrm{CCl}_{4}(g)} \\ {\text { Rate }=k\left[\mathrm{Cl}_{2}\right]^{1 / 2}\left[\mathrm{CHCl}_{3}\right]}\end{array} $$ What are the units for \(k,\) assuming time in seconds and concentration in mol/L?

Consider the reaction $$ 3 \mathrm{A}+\mathrm{B}+\mathrm{C} \longrightarrow \mathrm{D}+\mathrm{E} $$ where the rate law is defined as $$ -\frac{\Delta[\mathrm{A}]}{\Delta t}=k[\mathrm{A}]^{2}[\mathrm{B}][\mathrm{C}] $$ An experiment is carried out where \([\mathrm{B}]_{0}=[\mathrm{C}]_{0}=1.00 \mathrm{M}\) and \([\mathrm{A}]_{0}=1.00 \times 10^{-4} \mathrm{M}\) a. If after \(3.00 \min ,[\mathrm{A}]=3.26 \times 10^{-5} M,\) calculate the value of \(k .\) b. Calculate the half-life for this experiment. c. Calculate the concentration of \(B\) and the concentration of A after 10.0 min.

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