/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 30 The reaction \(2 \mathrm{ClO}_{2... [FREE SOLUTION] | 91影视

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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{tabular}{llll} \hline Experiment & {\(\left[\mathrm{ClO}_{2}\right](M)\)} & {\(\left[\mathrm{OH}^{-}\right](M)\)} & Rate \((M / \mathrm{s})\) \\ \hline 1 & \(0.060\) & \(0.030\) & \(0.0248\) \\ 2 & \(0.020\) & \(0.030\) & \(0.00276\) \\ 3 & \(0.020\) & \(0.090\) & \(0.00828\) \\ \hline \end{tabular} (a) Determine the rate law for the reaction. (b) Calculate the rate constant. (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
The rate law for the reaction is Rate = k [ClO_2]^2 [OH^-], with a rate constant (k) of approximately 2.76 M鈦宦瞫鈦宦. When \([\text{ClO}_2] = 0.100\text{ M}\) and \([\text{OH}^-] = 0.050\text{ M}\), the reaction rate is about 0.069 M/s.

Step by step solution

01

Determine the order of the reaction with respect to ClO_2 and OH^-

First, let's compare experiments 1 and 2. In these experiments, the concentration of OH^- is the same, while the concentration of ClO_2 changes by a factor of 3 (0.060 M to 0.020 M). We can see that when the concentration of ClO_2 is reduced, the rate drops from 0.0248 M/s to 0.00276 M/s, which is approximately reduced by a factor of 9. So, with respect to ClO_2, the reaction is second order. To find the order with respect to OH^-, let's compare experiments 2 and 3. Now, let's compare Experiments 2 and 3. In these experiments, the concentration of ClO_2 is the same, while the concentration of OH^- changes by a factor of 3 (0.030 M to 0.090 M). When the concentration of OH^- is increased by this factor, the rate rises from 0.00276 M/s to 0.00828 M/s, which is approximately an increase by a factor of 3. So, with respect to OH^-, the reaction is first-order.
02

Write down the rate law for the reaction

Now that we know the order with respect to each reactant, we can write down the rate law for the reaction. The rate of the reaction is given by the following expression: Rate = k [ClO_2]^2 [OH^-] Where k is the rate constant for the reaction.
03

Calculate the rate constant

Let's calculate the rate constant (k) by using the first experiment. Based on the experimental data: Rate = 0.0248 M/s [ClO_2] = 0.060 M [OH^-] = 0.030 M Substitute these values into the rate law we found in Step 2: 0.0248 = k (0.060)^2 (0.030) Now, solve for the rate constant k: k = 0.0248 / ((0.060)^2 * (0.030)) k 鈮 2.76 M鈦宦瞫鈦宦
04

Calculate the reaction rate when given concentrations of ClO_2 and OH^-

Finally, we will calculate the reaction rate when [ClO_2] = 0.100 M and [OH^-] = 0.050 M. Substitute these values, along with the calculated rate constant (k 鈮 2.76 M鈦宦瞫鈦宦), into the rate law: Rate = 2.76 (0.100)^2 (0.050) Rate 鈮 0.069 M/s So, when [ClO_2] = 0.100 M and [OH^-] = 0.050 M, the reaction rate is about 0.069 M/s.

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Key Concepts

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

Reaction Order
Understanding the reaction order is crucial in chemical kinetics, as it helps us know how the concentration of reactants affects the rate of a reaction. In this case, we are examining the reaction \(2 \text{ClO}_2(aq) + 2 \text{OH}^-(aq) \rightarrow \text{ClO}_3^-(aq) + \text{ClO}_2^-(aq) + \text{H}_2\text{O}(l)\). For such reactions, determining the reaction order with respect to each reactant is essential. To find out, we compare experimental data where only one reactant concentration changes, while the other remains constant. In comparing experiments 1 and 2: The concentration of \(\text{ClO}_2\) decreases by a factor of 3, but the rate drops by a factor of 9. This indicates that the reaction is
  • Second order with respect to \(\text{ClO}_2\)
Similarly, by comparing experiments 2 and 3 where the concentration of \(\text{OH}^-\) changes by a factor of 3, the rate also changes by the same factor. This means:
  • First order with respect to \(\text{OH}^-\)
Thus, the overall order of the reaction is based on the sum of these individual orders.
Rate Law
The rate law is a mathematical expression that relates the reaction rate to the concentrations of reactants, each raised to a power corresponding to their respective reaction orders.For the given reaction, the rate law is determined once the reaction orders have been established: \[\text{Rate} = k [\text{ClO}_2]^2 [\text{OH}^-]^1\]Here,
  • \(k\) is the rate constant, a unique value for a given reaction at a specified temperature.
  • \([\text{ClO}_2]^2\) denotes that the reaction is second order in \(\text{ClO}_2\).
  • \([\text{OH}^-]^1\) denotes first order in \(\text{OH}^-\).
The rate law is fundamental because it allows us to predict how changes in concentration affect reaction rates. It provides insights into the stoichiometric relationship between reactants and the kinetics of the chemical process.
Rate Constant Calculation
The rate constant, \(k\), is a crucial parameter in the rate law that is determined through experimental data and calculations. It provides insights into the speed of a reaction under certain conditions.To find \(k\), we use experimental results, inserting values into the rate law. Let's take Experiment 1 as an example:
  • Rate = 0.0248 M/s
  • \([\text{ClO}_2] = 0.060\) M
  • \([\text{OH}^-] = 0.030\) M
Inserting these into the rate law, \[0.0248 = k (0.060)^2 (0.030)\]Solving for \(k\), we find: \[k = \frac{0.0248}{(0.060)^2 \times 0.030} \approx 2.76 \, \text{M}^{-2}\text{s}^{-1}\]This constant is crucial for calculating the reaction rate at different concentrations and understanding the reaction's nature and speed more deeply.
Reaction Rate
The reaction rate describes how quickly a reaction proceeds over time. It depends on the concentration of reactants, which influences how frequently molecules collide to react.To determine the reaction rate with given concentrations, we use the rate law together with the rate constant. For instance, let's calculate the rate when \([\text{ClO}_2] = 0.100\) M and \([\text{OH}^-] = 0.050\) M:Using the rate law: \[\text{Rate} = k [\text{ClO}_2]^2 [\text{OH}^-]\]Where \(k \approx 2.76 \, \text{M}^{-2}\text{s}^{-1}\), we substitute the values:\[\text{Rate} = 2.76 \times (0.100)^2 \times 0.050\]Calculating this, we find:\[\text{Rate} \approx 0.069 \, \text{M/s}\]This rate indicates how fast the reaction progresses at these specific concentrations. Understanding reaction rates helps in controlling reaction conditions effectively in both laboratory and industrial settings.

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Most popular questions from this chapter

The reaction between ethyl iodide and hydroxide ion in ethanol \(\left(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\right)\) solution, \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{I}(a l c)+\mathrm{OH}^{-}(a l c)\) \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(l)+\mathrm{I}^{-}(a l c)\), has an activation energy of \(86.8 \mathrm{~kJ} / \mathrm{mol}\) and a frequency factor of \(2.10 \times 10^{11} \mathrm{M}^{-1} \mathrm{~s}^{-1}\). (a) Predict the rate constant for the reaction at \(35^{\circ} \mathrm{C} .\) (b) \(\mathrm{A}\) solution of KOH in ethanol is made up by dissolving \(0.335\) g KOH in ethanol to form \(250.0 \mathrm{~mL}\) of solution. Similarly, \(1.453 \mathrm{~g}\) of \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{I}\) is dissolved in ethanol to form \(250.0 \mathrm{~mL}\) 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 \(35^{\circ} \mathrm{C} ?(\mathrm{c})\) Which reagent in the reaction is limiting, assuming the reaction proceeds to completion?

The enzyme invertase catalyzes the conversion of sucrose, a disaccharide, to invert sugar, a mixture of glucose and fructose. When the concentration of invertase is \(4.2 \times 10^{-7} \mathrm{M}\) and the concentration of sucrose is \(0.0077 \mathrm{M}\) invert sugar is formed at the rate of \(1.5 \times 10^{-4} \mathrm{M} / \mathrm{s}\). When the sucrose concentration is doubled, the rate of formation of invert sugar is doubled also. (a) Assuming that the enzyme-substrate model is operative, is the fraction of enzyme tied up as a complex large or small? Explain. (b) Addition of inositol, another sugar, decreases the rate of formation of invert sugar. Suggest a mechanism by which this occurs.

Explain why rate laws generally cannot be written from balanced equations. Under what circumstance is the rate law related directly to the balanced equation for a reaction?

NO catalyzes the decomposition of \(\mathrm{N}_{2} \mathrm{O}\), possibly by the following mechanism: $$ \begin{array}{r} \mathrm{NO}(g)+\mathrm{N}_{2} \mathrm{O}(g) \longrightarrow \mathrm{N}_{2}(g)+\mathrm{NO}_{2}(g) \\ 2 \mathrm{NO}_{2}(g) \longrightarrow 2 \mathrm{NO}(g)+\mathrm{O}_{2}(g) \end{array} $$ (a) What is the chemical equation for the overall reaction? Show how the two steps can be added to give the overall equation. (b) Why is NO considered a catalyst and not an intermediate? (c) If experiments show that during the decomposition of \(\mathrm{N}_{2} \mathrm{O}, \mathrm{NO}_{2}\) does not accumulate in measurable quantities, does this rule out the proposed mechanism? If you think not, suggest what might be going on.

(a) What factors determine whether a collision between two molecules will lead to a chemical reaction? (b) According to the collision model, why does temperature affect the value of the rate constant?

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