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Chapter 14: Problem 34

The reaction \(2 \mathrm{ClO}_{2}(a q)+2 \mathrm{OH}^{-}(a q) \longrightarrow \mathrm{ClO}_{3}^{-}(a q)+\) \(\mathrm{ClO}_{2}^{-}(a q)+\mathrm{H}_{2} \mathrm{O}(l)\) was studied with the following results: $$ \begin{array}{lccc} \hline \text { Experiment } & {\left[\mathrm{CIO}_{2}\right](M)} & {\left[\mathrm{OH}^{-}\right](M)} & \text { Initial Rate }(M / s) \\ \hline 1 & 0.060 & 0.030 & 0.0248 \\ 2 & 0.020 & 0.030 & 0.00276 \\ 3 & 0.020 & 0.090 & 0.00828 \\ \hline \end{array} $$ (a) Determine the rate law for the reaction. (b) Calculate the rate constant with proper units. (c) Calculate the rate when \(\left[\mathrm{ClO}_{2}\right]=0.100 \mathrm{M}\) and \(\left[\mathrm{OH}^{-}\right]=0.050 \mathrm{M}\)

Short Answer

Expert verified
(a) The rate law for the reaction is Rate = k[ClO2]^2[OH-]^1. (b) The rate constant (k) is 9.2 M^(-2)s^(-1). (c) The rate of the reaction when [ClO2] = 0.100 M and [OH-] = 0.050 M is 0.046 M/s.

Step by step solution

01

Determine the order of the reaction with respect to each reactant

From the given table, we can compare the rates of the different experiments to determine the order of the reaction with respect to each reactant. Consider experiments 1 and 2. The [ClO2] in experiment 1 is 3 times that of experiment 2, while [OH-] remains the same. Comparing the rates of these two experiments, we get: \(\frac{0.0248}{0.00276} = 9\) Since the [ClO2] increased by a factor of 3, and the rate increased by a factor of 9, we can deduce that the order of the reaction with respect to ClO2 is 2 (first order: rate increase would be 3; second order: rate increase would be \(3^2 = 9\); etc.) Now, let's consider experiments 2 and 3. The [OH-] in experiment 3 is 3 times that of experiment 2 while the [ClO2] remains the same. Comparing the rates of these two experiments, we get: \(\frac{0.00828}{0.00276} = 3\) Since the [OH-] increased by a factor of 3, and the rate increased by a factor of 3 as well, we can deduce that the order of the reaction with respect to OH- is 1 (first-order reaction since rate increase is same as concentration increase). Thus, the rate law for the reaction can be written as: Rate = k[ClO2]^2[OH-]^1
02

Calculate the rate constant

Now that we have determined the rate law, we can calculate the rate constant 'k' using the concentration and rate data from any of the experiments. We will use the data from experiment 1. Rate = k[ClO2]^2[OH-]^1 0.0248 = k(0.060)^2(0.030) Solving for k: k = \(\frac{0.0248}{(0.060)^2(0.030)}\) = 9.2 M^(-2)s^(-1) The rate constant (k) is 9.2 M^(-2)s^(-1).
03

Calculate the rate for given concentrations

Now that we have the rate law and the rate constant, we can calculate the rate of the reaction when [ClO2] = 0.100 M and [OH-] = 0.050 M. Rate = k[ClO2]^2[OH-]^1 Rate = (9.2 M^(-2)s^(-1))(0.100 M)^2(0.050 M) Rate = 0.046 M/s The rate of the reaction when [ClO2] = 0.100 M and [OH-] = 0.050 M is 0.046 M/s.

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Key Concepts

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

Reaction Order
The reaction order tells us how a reaction rate is affected by the concentration of each reactant. In this exercise, we determine the reaction order by comparing the initial rates of different experiments while altering reactant concentrations.

By examining how changes in the concentration of reactants affect the rate, we deduce whether the reaction is first order, second order, or another type.

For instance, holding the concentration of OH- constant while varying ClO2 helps us observe the direct impact of ClO2 on the reaction rate.
  • If the rate changes in direct proportion to a reactant's concentration change, it indicates a first-order reaction concerning that reactant.
  • If the rate changes as the square (or some other power) of the concentration change, it indicates a second (or higher) order reaction.
In this exercise, the data shows that the reaction is second order with respect to ClO2 (since the rate triples when the concentration triples) and first order with respect to OH-.
Rate Law
The rate law is an equation that relates the reaction rate to the concentration of reactants. This model allows us to predict the rate under different conditions.

In our example, the rate law was determined based on the reaction orders derived. The general form is:
  • Rate = k[A]^m[B]^n
where m and n are the orders of the reaction with respect to reactants A and B, respectively, and k is the rate constant. Using the reaction orders determined earlier, our rate law is expressed as:
  • Rate = k[ClO2]^2[OH-]^1
The exponents reflect the reaction order for ClO2 and OH-. This expression tells us how the rate is affected by each reactant. Knowing the rate law is crucial for calculating the reaction rate under different concentration scenarios.
Rate Constant
The rate constant, denoted as k, is a crucial part of the rate law that combines with the reactant concentrations to predict the reaction rate. It is notably affected by factors such as temperature and the presence of a catalyst.

Its units depend on the overall order of the reaction. For example, in our second-order reaction with respect to ClO2 and first-order with respect to OH-, the rate constant has units of M^(-2)s^(-1).
  • Rate constants help in comparing the speeds of different reactions under similar conditions.
  • They're determined experimentally, as shown through calculations using data from one of the experiments.
In this exercise, using the given concentration data from Experiment 1, the rate constant is calculated to be 9.2 M^(-2)s^(-1). Understanding the rate constant is key for accurately modeling the kinetics of chemical reactions.

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Most popular questions from this chapter

(a) Consider the combustion of ethylene, \(\mathrm{C}_{2} \mathrm{H}_{4}(g)+\) \(3 \mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{CO}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(g) .\) If the concentration of \(\mathrm{C}_{2} \mathrm{H}_{4}\) is decreasing at the rate of \(0.025 \mathrm{M} / \mathrm{s}\), what are the rates of change in the concentrations of \(\mathrm{CO}_{2}\) and \(\mathrm{H}_{2} \mathrm{O}\) ? (b) The rate of decrease in \(\mathrm{N}_{2} \mathrm{H}_{4}\) partial pressure in a closed reaction vessel from the reaction \(\mathrm{N}_{2} \mathrm{H}_{4}(g)+\mathrm{H}_{2}(g) \longrightarrow 2 \mathrm{NH}_{3}(g)\) is \(10 \mathrm{kPa}\) per hour. What are the rates of change of \(\mathrm{NH}_{3}\) partial pressure and total pressure in the vessel?

The dimerization of \(\mathrm{C}_{2} \mathrm{~F}_{4}\) to \(\mathrm{C}_{4} \mathrm{~F}_{8}\) has a rate constant \(k=0.045 \mathrm{M}^{-1} \mathrm{~s}^{-1}\) at \(450 \mathrm{~K} .\) (a) Based on the unit of \(k\) what is the reaction order in \(\mathrm{C}_{2} \mathrm{~F}_{4} ?(\mathbf{b})\) If the initial concentration of \(\mathrm{C}_{2} \mathrm{~F}_{4}\) is \(0.100 \mathrm{M}\), how long would it take for the concentration to decrease to \(0.020 \mathrm{M}\) at \(450 \mathrm{~K}\) ?

(a) Most commercial heterogeneous catalysts are extremely finely divided solid materials. Why is particle size important? (b) What role does adsorption play in the action of a heterogeneous catalyst?

Consider two reactions. Reaction (1) has a half-life that gets longer as the reaction proceeds. Reaction (2) has a half-life that gets shorter as the reaction proceeds. What can you conclude about the rate laws of these reactions from these observations?

The reaction between ethyl iodide and hydroxide ion in ethanol \(\left(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\right)\) solution, \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{I}(a l c)+\mathrm{OH}^{-}(a l c) \longrightarrow\) \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(l)+\mathrm{I}^{-}(\) alc \()\) has an activation energy of \(86.8 \mathrm{~kJ} / \mathrm{mol}\) and a frequency factor of \(2.1 \times 10^{11} \mathrm{M}^{-1} \mathrm{~s}^{-1}\) (a) Predict the rate constant for the reaction at \(30^{\circ} \mathrm{C}\). (b) A solution of KOH in ethanol is made up by dissolving \(0.500 \mathrm{~g} \mathrm{KOH}\) in ethanol to form \(500 \mathrm{~mL}\) of solution. Similarly, \(1.500 \mathrm{~g}\) of \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{I}\) is dissolved in ethanol to form \(500 \mathrm{~mL}\) of solution. Equal volumes of the two solutions are mixed. Assuming the reaction is first order in each reactant, what is the initial rate at \(30^{\circ} \mathrm{C} ?(\mathbf{c})\) Which reagent in the reaction is limiting, assuming the reaction proceeds to completion? ((d) Assuming the frequency factor and activation energy do not change as a function of temperature, calculate the rate constant for the reaction at \(40^{\circ} \mathrm{C} .\)
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