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Chapter 14: Problem 89

The reaction \(A+B \longrightarrow\) products is first order in \(A\) first order in \(\mathrm{B},\) and second order overall. Consider that the starting concentrations of the reactants are \([\mathrm{A}]_{0}\) and [ \(\mathrm{B}]_{0},\) and that \(x\) represents the decrease in these concentrations at the time \(t .\) That is, \([\mathrm{A}]_{t}=[\mathrm{A}]_{0}-x\) and \([\mathrm{B}]_{t}=[\mathrm{B}]_{0}-x .\) Show that the integrated rate law for this reaction can be expressed as shown below. $$\ln \frac{[\mathrm{A}]_{0} \times[\mathrm{B}]_{t}}{[\mathrm{B}]_{0} \times[\mathrm{A}]_{t}}=\left([\mathrm{B}]_{0}-[\mathrm{A}]_{0}\right) \times k t$$

Short Answer

Expert verified
The integrated rate law for the given reaction can be obtained by expressing the rate of reaction in terms of concentrations of the reacting species, substituting for concentrations at time \(t\), and integrating the resulting differential equation. The result can be manipulated to match the given integrated rate law expression.

Step by step solution

01

Understanding the Given Reaction Order and Composition of Reagents

The given reaction involves two reagents A and B, and is first order in both, which makes it an overall second order reaction. The decrease in concentrations of both reactants over time \(t\) is represented by \(x\). Therefore, at time \(t\), the concentrations of A and B will be \([A]_t = [A]_0 - x\) and \([B]_t = [B]_0 - x\), respectively.
02

Express the Rate of Reaction

Given that the reaction is first order with respect to both A and B, the rate of reaction can be expressed as follows: \(-\frac{d[A]}{dt} = -\frac{d[B]}{dt} = k[A][B]\) . Here, \(k\) is the rate constant.
03

Substitute for Concentrations at Time t

Substitute \([A]_t = [A]_0 - x\) and \([B]_t = [B]_0 - x\) into the rate equation to express the rates in terms of \(x\): \(-\frac{dx}{dt} = k([A]_0 - x)([B]_0 - x)\).
04

Integrate the Rate Equation

The equation obtained in Step 3 is a differential equation that can be integrated to get the time evolution of the concentrations. This is a non-linear differential equation which can be solved by separation of variables and integration. Before the integration, rearrange the equation to isolate terms involving \(x\) on one side: \(\frac{dx}{([A]_0 - x)([B]_0 - x)} = -k dt\). Integrate both sides of the equation integrating lhs from \(0\) to \(x\) and rhs from \(0\) to \(t\).
05

Simplify the Integrated Equation

Simplify the result obtained from the integration and rearrange into the required form to get: \(\ln \frac{[A]_0 [B]_t}{[B]_0 [A]_t} = ([B]_0 - [A]_0) k t \).

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Most popular questions from this chapter

A reaction is \(50 \%\) complete in 30.0 min. How long after its start will the reaction be \(75 \%\) complete if it is (a) first order; (b) zero order?

The rate equation for the reaction \(2 \mathrm{A}+\mathrm{B} \longrightarrow \mathrm{C}\) is found to be rate \(=k[\mathrm{A}][\mathrm{B}] .\) For this reaction, we can conclude that (a) the unit of \(k=s^{-1} ;\) (b) \(t_{1 / 2}\) is constant; (c) the value of \(k\) is independent of the values of \([\mathrm{A}]\) and \([\mathrm{B}] ;\) (d) the rate of formation of \(\mathrm{C}\) is twice the rate of disappearance of A.

For the reaction \(A+2 B \longrightarrow 2 C\), the rate of reaction is \(1.76 \times 10^{-5} \mathrm{M} \mathrm{s}^{-1}\) at the time when \([\mathrm{A}]=0.3580 \mathrm{M}.\) (a) What is the rate of formation of \(\mathrm{C}\) ? (b) What will \([\mathrm{A}]\) be 1.00 min later? (c) Assume the rate remains at \(1.76 \times 10^{-5} \mathrm{M} \mathrm{s}^{-1}\) How long would it take for \([\mathrm{A}]\) to change from 0.3580 to \(0.3500 \mathrm{M} ?\)

The initial rate of the reaction \(A+B \longrightarrow C+D\) is determined for different initial conditions, with the results listed in the table. (a) What is the order of reaction with respect to A and to B? (b) What is the overall reaction order? (c) What is the value of the rate constant, \(k ?\) $$\begin{array}{llll} \hline \text { Expt } & \text { [A], M } & \text { [B], M } & \text { Initial Rate, M s }^{-1} \\ \hline 1 & 0.185 & 0.133 & 3.35 \times 10^{-4} \\ 2 & 0.185 & 0.266 & 1.35 \times 10^{-3} \\ 3 & 0.370 & 0.133 & 6.75 \times 10^{-4} \\ 4 & 0.370 & 0.266 & 2.70 \times 10^{-3} \\ \hline \end{array}$$

The rate of a chemical reaction generally increases rapidly, even for small increases in temperature, because of a rapid increase in (a) collision frequency; (b) fraction of reactant molecules with very high kinetic energies; (c) activation energy; (d) average kinetic energy of the reactant molecules.
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