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Chapter 12: Problem 2

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

Short Answer

Expert verified
To determine a rate law, perform two experiments: 1. Vary the concentration of one reactant (A) while keeping the other reactant(s) (B) constant. Prepare a series of reaction mixtures, measure initial reaction rates, plot rates against the concentration of A, and analyze the plot to determine the order of the reaction with respect to A. 2. Vary the concentrations of both reactants (A and B) simultaneously. Prepare new reaction mixtures, measure initial rates, create a plot with rates against both reactant concentrations, compare it with the plot from the first experiment, and modify the rate law as necessary to account for the dependence of both reactant concentrations, determining the overall order of the reaction.

Step by step solution

01

Experiment 1: Vary the concentration of one reactant while keeping the concentration of the other reactant(s) constant.

In this experiment, we will choose a reaction with two reactants, A and B, and take the following steps: 1. Prepare a series of reaction mixtures with varying concentrations of reactant A, while keeping the concentration of reactant B constant. 2. Measure the initial rate of reaction for each mixture. 3. Plot the initial reaction rate against the concentration of reactant A. 4. Analyze the plot to determine the order of the reaction with respect to reactant A (i.e., how the rate depends on the concentration of A). This experiment will provide us with information about how the rate law is influenced by reactant A.
02

Experiment 2: Vary the concentration of both reactants simultaneously and comparing the results.

In this experiment, we will build upon the results obtained from Experiment 1 by considering both reactants A and B and taking the following steps: 1. Prepare a new series of reaction mixtures where the concentrations of both reactants A and B are varied. 2. Measure the initial rate of reaction for each new mixture. 3. Create a plot with the initial reaction rates against the concentrations of reactant A and reactant B. 4. Compare this new plot with the one obtained from Experiment 1, analyzing how the rate law depends on both reactant A and reactant B concentrations together. 5. Modify the rate law as necessary to account for the dependence of both reactant concentrations, determining the overall order of the reaction. By comparing the results from Experiment 1 and Experiment 2, we can determine the rate law for the given reaction.

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Most popular questions from this chapter

Consider the following initial rate data for the decomposition of compound AB to give A and B: Determine the half-life for the decomposition reaction initially having 1.00\(M \mathrm{AB}\) present.

Table 12.2 illustrates how the average rate of a reaction decreases with time. Why does the average rate of a reaction generally decrease with time? How does the instantaneous rate of a reaction depend on time? Why are initial rates of a reaction primarily used by convention?

Chemists commonly use a rule of thumb that an increase of 10 \(\mathrm{K}\) in temperature doubles the rate of a reaction. What must the activation energy be for this statement to be true for a temperature increase from 25 to \(35^{\circ} \mathrm{C} ?\)

Consider the hypothetical reaction $$ \mathrm{A}+\mathrm{B}+2 \mathrm{C} \longrightarrow 2 \mathrm{D}+3 \mathrm{E} $$ where the rate law is $$ \text {Rate} =-\frac{\Delta[\mathrm{A}]}{\Delta t}=k[\mathrm{A}][\mathrm{B}]^{2} $$ An experiment is carried out where \([\mathrm{A}]_{0}=1.0 \times 10^{-2} M\) \([\mathrm{B}]_{0}=3.0 M,\) and \([\mathrm{C}]_{0}=2.0 M .\) The reaction is started, and after 8.0 seconds, the concentration of \(\mathrm{A}\) is \(3.8 \times 10^{-3} \mathrm{M}\) a. Calculate the value of k for this reaction. b. Calculate the half-life for this experiment. c. Calculate the concentration of A after 13.0 seconds. d. Calculate the concentration of C after 13.0 seconds.

The activation energy for the reaction $$ \mathrm{NO}_{2}(g)+\mathrm{CO}(g) \longrightarrow \mathrm{NO}(g)+\mathrm{CO}_{2}(g) $$ is 125 \(\mathrm{kJ} / \mathrm{mol}\) , and \(\Delta E\) for the reaction is \(-216 \mathrm{kJ} / \mathrm{mol}\) . What is the activation energy for the reverse reaction \(\left[\mathrm{NO}(g)+\mathrm{CO}_{2}(g) \longrightarrow \mathrm{NO}_{2}(g)+\mathrm{CO}(g)\right] ?\)
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