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Chapter 26: Problem 24

In this problem, we will derive Equation 26.32 from the rate law (Equation 26.31 ) \\[ -\frac{d[\mathrm{A}]}{d t}=k[\mathrm{A}][\mathrm{B}] \\] Use the reaction stoichiometry of Equation 26.30 to show that \([\mathrm{B}]=[\mathrm{B}]_{0}-[\mathrm{A}]_{0}+[\mathrm{A}]\) Use this result to show that Equation 1 can be written as \\[ -\frac{d[\mathrm{A}]}{d t}=k[\mathrm{A}]\left|[\mathrm{B}]_{0}-[\mathrm{A}]_{0}+[\mathrm{A}]\right| \\] Now separate the variables and then integrate the resulting equation subject to its initial conditions to obtain the desired result, Equation 26.32 \\[ k t=\frac{1}{[\mathbf{A}]_{0}-[\mathbf{B}]_{0}} \ln \frac{[\mathbf{A}][\mathbf{B}]_{0}}{[\mathbf{B}][\mathbf{A}]_{0}} \\]

Short Answer

Expert verified
Use stoichiometry to relate concentrations, substitute, separate variables, integrate, apply initial conditions, and simplify to obtain Equation 26.32.

Step by step solution

01

Identify the Stoichiometric Relationship

Using the stoichiometry of the reaction, identify the relationship between the concentrations of \([\mathrm{A}]\) and \([\mathrm{B}]\). We have from the problem statement:\[[\mathrm{B}] = [\mathrm{B}]_0 - [\mathrm{A}]_0 + [\mathrm{A}]\]This implies that as \([\mathrm{A}]\) decreases, \([\mathrm{B}]\) decreases in a specific stoichiometric manner, reflecting the consumption of reagents in the reaction.
02

Substitute into the Rate Law

Substitute \([\mathrm{B}] = [\mathrm{B}]_0 - [\mathrm{A}]_0 + [\mathrm{A}]\) into the rate law equation:\[- \frac{d[\mathrm{A}]}{dt} = k [\mathrm{A}]([\mathrm{B}]_0 - [\mathrm{A}]_0 + [\mathrm{A}])\]This modification accounts for the dependence of \([\mathrm{B}]\) on \([\mathrm{A}]\) throughout the reaction.
03

Separate Variables

Re-write the modified rate equation to facilitate separation of variables:\[- \frac{d[\mathrm{A}]}{[\mathrm{A}]([\mathrm{B}]_0 - [\mathrm{A}]_0 + [\mathrm{A}])} = k \, dt\]This rearrangement allows the integration of both sides separately, with the left side concerning \([\mathrm{A}]\) and the right side concerning \(t\).
04

Integrate Both Sides

Integrate the left side of the equation with respect to \([\mathrm{A}]\) and the right side with respect to \(t\). The left side becomes a logarithmic integration, while the right side results in a straightforward integration:\[\int \frac{-1}{[\mathrm{A}]([\mathrm{B}]_0 - [\mathrm{A}]_0 + [\mathrm{A}])} \, d[\mathrm{A}] = \int k \, dt\]The integral on the left can be solved using standard techniques for rational functions, while the right side integrates to \(kt + C\).
05

Apply the Initial Conditions

Using the initial conditions of the reaction, usually \([\mathrm{A}] = [\mathrm{A}]_0\) at \(t = 0\), determine the constant of integration \(C\). Substitute these conditions into the integrated results to solve for \(C\), ensuring the functions match.
06

Solve for Desired Equation 26.32

Combine results from integration and substitution of initial conditions to simplify and derive the final form of Equation 26.32:\[k t = \frac{1}{[\mathrm{A}]_0 - [\mathrm{B}]_0} \ln \frac{[\mathrm{A}][\mathrm{B}]_0}{[\mathrm{B}][\mathrm{A}]_0}\]This expresses the relationship between the concentrations of \([\mathrm{A}]\) and \([\mathrm{B}]\) over time according to the integrated rate law.

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Key Concepts

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

Rate Laws
Rate laws are fundamental tools in chemical kinetics, helping chemists understand the speed at which reactions proceed. A rate law expresses the rate of a chemical reaction in terms of the concentration of its reactants. Consider the example from the original problem: the rate of disappearance of species \([\mathrm{A}]\) is given by the equation \[- \frac{d[\mathrm{A}]}{dt} = k [\mathrm{A}][\mathrm{B}] \], where \(k\) is the rate constant. This formula shows that the reaction rate relies on the current concentrations of reactants \([\mathrm{A}]\) and \([\mathrm{B}]\). It emphasizes the proportionality of how reactants' concentration impacts the pace of production or consumption in chemical reactions.
Understanding rate laws allows scientists to predict how a reaction behaves over time. It involves not only determining how fast a reaction will occur but also clarifying the impact of each reactant's concentration on that speed. Thus, when deducing a rate law, it is essential to consider the stoichiometry and mechanism of the reaction as these provide insight into molecular interactions at play.
Reaction Stoichiometry
Reaction stoichiometry gives us the proportional relationship between reactants and products in a chemical equation. For our integrative exercise, understanding the stoichiometry is crucial to manipulate concentrations into a usable form when involving the rate law.
The problem statement provides a key stoichiometric relation: \([\mathrm{B}] = [\mathrm{B}]_0 - [\mathrm{A}]_0 + [\mathrm{A}]\). This means, as the concentration of \([\mathrm{A}]\) decreases due to the reaction, \([\mathrm{B}]\) also decreases in a predictable manner, moving from an initial concentration \([\mathrm{B}]_0\) to a modified version that accounts for the consumption of \([\mathrm{A}]\).
This relationship allows us to substitute \([\mathrm{B}]\) in the rate law equation to reflect the changing concentrations during the reaction, which is vital in understanding how these transformations affect reaction kinetics and the time evolution of the reactants.
Integration in Chemistry
Integration in chemistry is a powerful tool for solving differential equations typical in kinetic studies. After formulating a differential equation from the rate law and reaction stoichiometry, we often need to solve it through integration to find expressions for concentrations over time.
For instance, separating variables in the rate law equation allows integration of each side concerning its respective variable:\[- \frac{d[\mathrm{A}]}{[\mathrm{A}]([\mathbf{B}]_0 - [\mathbf{A}]_0 + [\mathbf{A}])} = k \, dt\]
This resulting equation describes how one might approach solving for the relationship between time \(t\) and concentration \([\mathrm{A}]\). The integration process helps derive a function that predicts concentrations at any given time point, crucial in chemical kinetics for calculating reactant or product evolution throughout a reaction. By solving these integrals and using initial conditions, chemists obtain equations like Equation 26.32 in the problem, providing valuable predictive insights for experimental and industrial chemical applications.

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Most popular questions from this chapter

Uranyl nitrate decomposes according to \\[ \mathrm{UO}_{2}\left(\mathrm{NO}_{3}\right)_{2}(\mathrm{s}) \longrightarrow \mathrm{UO}_{3}(\mathrm{s})+2 \mathrm{NO}_{2}(\mathrm{g})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{g}) \\] The rate law for this reaction is first order in the concentration of uranyl nitrate. The following data were recorded for the reaction at \(25.0^{\circ} \mathrm{C}\) \\[ \begin{array}{c|ccccc} t / \min & 0 & 20.0 & 60.0 & 180.0 & 360.0 \\ \hline\left[\mathrm{UO}_{2}\left(\mathrm{NO}_{3}\right)_{2}\right] / \mathrm{mol} \cdot \mathrm{dm}^{-3} & 0.01413 & 0.01096 & 0.00758 & 0.00302 & 0.00055 \end{array} \\] Calculate the rate constant for this reaction at \(25.0^{\circ} \mathrm{C}\).

The kinetics of a chemical reaction can be followed by a variety of experimental techniques, including optical spectroscopy, NMR spectroscopy, conductivity, resistivity, pressure changes, and volume changes. When using these techniques, we do not measure the concentration itself but we know that the observed signal is proportional to the concentration; the exact proportionality constant depends on the experimental technique and the species present in the chemical system. Consider the general reaction given by $$ v_{\mathrm{A}} \mathrm{A}+v_{\mathrm{B}} \mathrm{B} \longrightarrow v_{\mathrm{Y}} \mathrm{Y}+v_{\mathrm{Z}} \mathrm{Z} $$ where we assume that \(\mathrm{A}\) is the limiting reagent so that \([\mathrm{A}] \rightarrow 0\) as \(t \rightarrow \infty\). Let \(p_{i}\) be the proportionality constant for the contribution of species \(i\) to \(S\), the measured signal from the instrument. Explain why at any time \(t\) during the reaction, \(S\) is given by $$ S(t)=p_{\mathrm{A}}[\mathrm{A}]+p_{\mathrm{B}}[\mathrm{B}]+p_{\mathrm{Y}}[\mathrm{Y}]+p_{\mathrm{Z}}[\mathrm{Z}] $$ Show that the initial and final readings from the instrument are given by $$ S(0)=p_{\mathrm{A}}[\mathrm{A}]_{0}+p_{\mathrm{B}}[\mathrm{B}]_{0}+p_{\mathrm{Y}}[\mathrm{Y}]_{0}+p_{\mathrm{Z}}[\mathrm{Z}]_{0} $$ tand $$ S(\infty)=p_{\mathrm{B}}\left([\mathrm{B}]_{0}-\frac{v_{\mathrm{B}}}{v_{\mathrm{A}}}[\mathrm{A}]_{0}\right)+p_{\mathrm{Y}}\left([\mathrm{Y}]_{0}+\frac{v_{\mathrm{Y}}}{v_{\mathrm{A}}}[\mathrm{A}]_{0}\right)+p_{\mathrm{Z}}\left([\mathrm{Z}]_{0}+\frac{v_{\mathrm{Z}}}{v_{\mathrm{A}}}[\mathrm{A}]_{0}\right) $$ Combine Equations 1 through 3 to show that $$ [\mathrm{A}]=[\mathrm{A}]_{0} \frac{S(t)-S(\infty)}{S(0)-S(\infty)} $$ (Hint: Be sure to express [B], [Y], and [Z] in terms of their initial values, [A] and \([\mathrm{A}]_{0}\).)

Consider the general chemical reaction $$ \mathrm{A}+\mathrm{B} \underset{k_{-1}}{\stackrel{k_{1}}{\rightleftharpoons}} \mathrm{P} $$ If we assume that both the forward and reverse reactions are first order in their respective reactants, the rate law is given by (Equation 26.52) $$ \frac{d[\mathrm{P}]}{d t}=k_{1}[\mathrm{~A}][\mathrm{B}]-k_{-1}[\mathrm{P}] $$ Now consider the response of this chemical reaction to a temperature jump. Let \([\mathrm{A}]=\) \([\mathrm{A}]_{2, \mathrm{eq}}+\Delta[\mathrm{A}],[\mathrm{B}]=[\mathrm{B}]_{2, \mathrm{eq}}+\Delta[\mathrm{B}]\), and \([\mathrm{P}]=[\mathrm{P}]_{2, \mathrm{cq}}+\Delta[\mathrm{P}]\), where the subscript " \(2, \mathrm{eq}\) " refers to the new equilibrium state. Now use the fact that \(\Delta[\mathrm{A}]=\Delta[\mathrm{B}]=-\Delta[\mathrm{P}]\) to show that Equation 1 becomes $$ \begin{aligned} \frac{d \Delta[\mathrm{P}]}{d t}=& k_{1}[\mathrm{~A}]_{2, \mathrm{eq}}[\mathrm{B}]_{2, \mathrm{eq}}-k_{-1}[\mathrm{P}]_{2, \mathrm{eq}} \\\ &-\left\\{k_{1}\left([\mathrm{~A}]_{2, \mathrm{eq}}+[\mathrm{B}]_{2, \text { cq }}\right)+k_{-1}\right\\} \Delta[\mathrm{P}]+O\left(\Delta[\mathrm{P}]^{2}\right) \end{aligned} $$ Show that the first terms on the right side of this equation cancel and that Equations \(26.53\) and \(26.54\) result.

Uranium- 238 decays to lead- 206 with a half-life of \(4.51 \times 10^{9}\) y. A sample of ocean sediment is found to contain \(1.50 \mathrm{mg}\) of uranium- 238 and \(0.460 \mathrm{mg}\) of lead- 206 . Estimate the age of the sediment assuming that lead- 206 is formed only by the decay of uranium and that lead-206 does not itself decay.

The rate constants for the reaction \\[ \mathrm{CHCl}_{2}(\mathrm{g})+\mathrm{Cl}_{2}(\mathrm{g}) \longrightarrow \mathrm{CHCl}_{3}(\mathrm{g})+\mathrm{Cl}(\mathrm{g}) \\] at different temperatures are tabulated below \\[ \begin{array}{c|cccccc} T / \mathrm{K} & 357 & 400 & 458 & 524 & 533 & 615 \\ \hline k / 10^{7} \mathrm{dm}^{3} \cdot \mathrm{mol}^{-1} \cdot \mathrm{s}^{-1} & 1.72 & 2.53 & 3.82 & 5.20 & 5.61 & 7.65 \end{array} \\] Calculate the values of the Arrhenius parameters \(A\) and \(E_{\mathrm{a}}\) for this reaction.
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