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Chapter 9: Problem 11

The rate constant for the decomposition of a compound in solution is \(2.0 \times 10^{-4} \mathrm{s}^{-1}\). If the initial concentration is \(0.02 \mathrm{moldm}^{-3},\) what will the concentration of the compound be after \(10 \text { min? (Section } 9.4)\)

Short Answer

Expert verified
The concentration after 10 minutes is approximately 0.0177 mol/dm鲁.

Step by step solution

01

Identify the Reaction Order and Relevant Equation

The given rate constant has units of \(s^{-1}\), indicating a first-order reaction. For a first-order reaction, the concentration at time \(t\) is given by the equation \([A]_t = [A]_0 e^{-kt}\), where \([A]_0\) is the initial concentration, \(k\) is the rate constant, and \(t\) is the time.
02

Convert Time to Appropriate Units

The time given is 10 minutes, but the rate constant is given in seconds. Therefore, we must convert 10 minutes into seconds: \(10 \text{ minutes} = 10 \times 60 = 600 \text{ seconds}\).
03

Apply the First-Order Kinetics Formula

Substitute the values into the first-order kinetics formula: \([A]_t = 0.02 \times e^{-2.0 \times 10^{-4} \times 600}\).
04

Calculate the Exponential Term

Calculate the value of the exponential term: \(e^{-2.0 \times 10^{-4} \times 600} = e^{-0.12}\).
05

Evaluate the Exponential Term

Using a calculator, evaluate \(e^{-0.12} \approx 0.8869\).
06

Calculate Final Concentration

Substitute the evaluated exponential value back into the equation to find the concentration after 600 seconds: \([A]_t = 0.02 \times 0.8869 \).
07

Compute the Final Concentration

Compute the result \([A]_t = 0.017738 \text{ mol/dm}^3\).

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Key Concepts

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

Decomposition Reaction
Decomposition reactions are chemical reactions where a single compound breaks down into two or more simpler substances. This type of reaction is quite common and occurs in many natural and industrial processes. In the context of a first-order reaction, these reactions have a specific characteristic: the rate of reaction depends solely on the concentration of one reactant. Decomposition reactions can be spontaneous or require energy input, like heat or light, to occur. During a decomposition reaction, the compound undergoes a transformation, and the rate at which it decomposes can be determined by observing changes in the concentration over time.
Rate Constant
The rate constant, denoted as \(k\), is a crucial parameter in chemical kinetics. It provides the speed at which a reaction proceeds. For first-order reactions, the rate constant has units of \(s^{-1}\), indicating that it鈥檚 related to the time it takes for the reactant concentration to decrease. A higher rate constant means a faster reaction, while a lower rate constant suggests a slower reaction. The value of \(k\) is often determined experimentally and can vary with factors such as temperature and the presence of a catalyst. Understanding the rate constant helps in predicting how a reaction will progress under different conditions.
Kinetics Formula
In first-order reactions, the kinetics formula plays a vital role in determining the concentration of a reactant at any given time. The formula is \([A]_t = [A]_0 e^{-kt}\), where:
  • \([A]_t\) is the concentration at time \(t\)
  • \([A]_0\) is the initial concentration
  • \(k\) is the rate constant
  • \(t\) is the time
This formula illustrates how the concentration decreases exponentially over time. It is widely used to model reactions in various fields, including biology and environmental science. By applying the formula, one can calculate how much of a substance remains after a certain period, allowing scientists to make predictions and design experiments accordingly.
Exponential Decay
Exponential decay describes the process where a quantity decreases at a rate proportional to its current value. In the context of a first-order reaction, the concentration of a reactant decreases exponentially over time. This means that the rate of decomposition is faster when more of the reactant is present and slows down as the concentration decreases. In mathematical terms, exponential decay is represented by the equation \([A]_t = [A]_0 e^{-kt}\). This kind of decay is often seen in radioactive decay, population decline, and other natural processes. Understanding exponential decay is essential for predicting how reactants behave over time in a controlled setting, such as a laboratory experiment.

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Most popular questions from this chapter

For a particular first order reaction, half of the reactant is used up after 15 s. What fraction of the reactant will remain after 1 \(\min ?(\text { Section } 9.4)\)

At low substrate concentrations, the initial rate of an enzyme catalysed reaction was found to be directly proportional to the initial substrate concentration, [S]o, and directly proportional to the initial enzyme concentration, \([\mathrm{E}]_{0} .\) (Section 9.9 ) (a) Outline the series of experiments that led to these results. (b) Write a rate equation for the reaction under these conditions and explain the observed kinetics in terms of the mechanism for the reaction. What is the rate-determining step under these conditions? At much higher substrate concentrations, the initial rate was found to be constant and independent of the initial substrate concentrations. The initial enzyme concentration was the same in each experiment. (c) Write a rate equation for the reaction under these conditions and explain the observed kinetics in terms of the mechanism for the reaction. What is the rate-determining step under these conditions?

The mechanism for the formation of a DNA double helix from two strands \(A \text { and } B \text { is as follows. (Section } 9.6)\) (a) Experiments show that the overall reaction is first order with respect to strand A and first order with respect to strand B. Write the equation for the overall reaction. (b) Write the rate equation for the overall reaction. (c) Assuming the processes involved are elementary reactions, derive an expression for the rate constant for the overall reaction in terms of the rate constants for the individual steps.

From the data provided in the table, deduce the rate equation and the value of the rate constant for the following reaction. (Section \(9.4)\) \\[ \begin{aligned} \mathrm{CH}_{3} \mathrm{COCH}_{3}(\mathrm{aq})+\mathrm{Br}_{2}(\mathrm{aq})+\mathrm{H}^{+}(\mathrm{aq}) \rightarrow & \mathrm{CH}_{3} \mathrm{COCH}_{2} \mathrm{Br}(\mathrm{aq}) \\ &+2 \mathrm{H}^{+}(\mathrm{aq})+\mathrm{Br}(\mathrm{aq}) \end{aligned} \\] $$\begin{array}{llll} \hline & \text { Initial } & \text { Initial } & \text { Initial } & \text { Initial } \\ & \text { concentration } & \text { concentration } & \text { concentration } & \text { rate of } \\ & \begin{array}{l} \text { of } \mathrm{CH}_{3} \mathrm{COCH}_{3} \\ \text { /moldm }^{-3} \end{array} & \begin{array}{l} \text { of } \mathrm{Br}_{2} \\ \text { /moldm }^{-3} \end{array} & \begin{array}{l} \text { of } \mathrm{H}^{+} \\ \text {/moldm }^{-3} \end{array} & \begin{array}{l} \text { reaction } \\ \text { /moldm }^{-3} \mathrm{s}^{-1} \end{array} \\ \hline 1 & 1.00 & 1.00 & 1.00 & 4.0 \times 10^{-3} \\ 2 & 2.00 & 1.00 & 1.00 & 8.0 \times 10^{3} \\ 3 & 2.00 & 2.00 & 1.00 & 8.0 \times 10^{3} \\ 4 & 1.00 & 1.00 & 2.00 & 8.0 \times 10^{-3} \end{array}$$

The dissociation of propane in which a C-C bond breaks to form a methyl radical and an ethyl radical is a unimolecular reaction. \\[ \mathrm{C}_{3} \mathrm{H}_{8} \rightarrow \mathrm{CH}_{3}^{*}+\mathrm{C}_{2} \mathrm{H}_{5}^{*} \\] The rate of formation of \(\mathrm{CH}_{3}^{*}\) (and of \(\mathrm{C}_{2} \mathrm{H}_{5}\) ') is given by \(\frac{\mathrm{d}\left[\mathrm{CH} \mathrm{b}^{*}\right]}{\mathrm{d} t}=\mathrm{k}_{\text {overd }}\left[\mathrm{C}_{3} \mathrm{H}_{8}\right],\) where \(k_{\text {overd }}\) is the unimolecular rate constant for the reaction. (Sections 9.6 and 9.8 ) (a) Explain what is meant by the term 'unimolecular? The propane molecules obtain sufficient energy to dissociate by colliding with other molecules, \(M\), where \(M\) may be an unreactive gas such as nitrogen. The mechanism for this process can be written as where \(\mathrm{C}_{3} \mathrm{H}_{\mathrm{g}}^{*}\) is a propane molecule in a high energy state, which has sufficient energy to dissociate. (b) By applying the steady state approximation to \(\mathrm{C}_{3} \mathrm{H}_{8}^{*}\), derive an expression for \(\left[\mathrm{C}_{3} \mathrm{H}_{4}^{\prime}\right] .\) since the rate of formation of \(\mathrm{CH}_{3}^{*}\) is equal to \(k_{2}\left[\mathrm{C}_{3} \mathrm{H}_{8}^{*}\right],\) show that \\[ k_{\text {overal }}=\frac{k_{1} k_{2}[M]}{k_{-1}[M]+k_{2}} \\] (c) Show that: (i) when \([\mathrm{M}]\) is very large, \(k_{\text {overal }}=\frac{k_{1} k_{2}}{k_{-1}}\) (ii) when \([\mathrm{M}]\) is very small, \(k_{\text {oved }}=k_{1}[\mathrm{M}]\) (d) What are the rate-determining stages in the mechanism under each of the conditions in (c)(1) and (c)(ii)? Sketch a graph to show the dependence of \(k_{\text {overall }}\) on \([\mathrm{M}]\) (e) It is always better to find a linear expression to analyse experimental data. Show that \\[ \frac{1}{k_{\text {overal }}}=\frac{k_{-1}}{k_{\sqrt{2}}}+\frac{1}{k_{1}[M]} \\] A plot of \(1 / k_{\text {werall }}(y-\text { axis ) against } 1 /[\mathrm{M}] \text { ( } x\) -axis) is a straight line. What are the gradient and the intercept on the \(y\) -axis at \(1 /[\mathrm{M}]=0 ?\)
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