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Chapter 14: Problem 83

The following data are for the reaction \(2 \mathrm{A}+\mathrm{B} \longrightarrow\) products. Establish the order of this reaction with respect to A and to B. $$\begin{array}{cccc} \hline \text { Expt 1, }[\mathrm{B}]=1.00 \mathrm{M} & & {\text { Expt 2, }[\mathrm{B}]=0.50 \mathrm{M}} \\ \hline \begin{array}{cccc} \text { Time, } \\ \text { min } \end{array} & \begin{array}{c} \text { [A], M } \\ \end{array} & \text { Time, } \text { min } &\text { [A], M } \\ \hline 0 & 1.000 \times 10^{-3} & 0 & 1.000 \times 10^{-3} \\ 1 & 0.951 \times 10^{-3} & 1 & 0.975 \times 10^{-3} \\ 5 & 0.779 \times 10^{-3} & 5 & 0.883 \times 10^{-3} \\ 10 & 0.607 \times 10^{-3} & 10 & 0.779 \times 10^{-3} \\ 20 & 0.368 \times 10^{-3} & 20 & 0.607 \times 10^{-3} \\ \hline \end{array}$$

Short Answer

Expert verified
The order of the reaction is first order with respect to B and zeroth order with respect to A.

Step by step solution

01

Reactant Concentration Comparison

Examine the change in concentration of A after 1 minute in both experiments. In experiment 1 when B is 1M, A goes from 1.000 * 10-3 M to 0.951 * 10-3 M. In experiment 2, when B is .5M, A goes from 1.000 * 10-3 M to 0.975 * 10-3 M.
02

Rate Calculation

Calculate the ratio of the rates of reaction for the two experiments. Remember that the rate of reaction is essentially the change in concentration per unit time. For experiment 1, the rate is (1.000 - 0.951) x 10-3 M/min = 0.049 * 10-3 M/min. For experiment 2, the rate is (1.000 - 0.975) x 10-3 M/min = 0.025 * 10-3 M/min. The ratio of the rates is thus 0.049/0.025 = 1.96, which is approximately 2.
03

Order of Reaction Determination

Because the rate approximately doubled when we halved the concentration of B, we can deduce that the order of the reaction with respect to B is first order. Moving on to A, since the concentration of A is constant in both the experiments, the order with respect to A does not affect the result and is therefore 0. Thus, the reaction is first order with respect to B and zeroth order with respect to A.

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Most popular questions from this chapter

For the first-order reaction $$\mathrm{N}_{2} \mathrm{O}_{5}(\mathrm{g}) \longrightarrow 2 \mathrm{NO}_{2}(\mathrm{g})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{g})$$ \(t_{1 / 2}=22.5 \mathrm{h}\) at \(20^{\circ} \mathrm{C}\) and \(1.5 \mathrm{h}\) at \(40^{\circ} \mathrm{C}.\) (a) Calculate the activation energy of this reaction. (b) If the Arrhenius constant \(A=2.05 \times 10^{13} \mathrm{s}^{-1}\) determine the value of \(k\) at \(30^{\circ} \mathrm{C}\).

Explain the important distinctions between each pair of terms: (a) first-order and second-order reactions; (b) rate law and integrated rate law; (c) activation energy and enthalpy of reaction; (d) elementary process and overall reaction; (e) enzyme and substrate.

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. $$\ln \frac{[\mathrm{A}]_{0} \times[\mathrm{B}]_{t}}{[\mathrm{B}]_{0} \times[\mathrm{A}]_{t}}=\left([\mathrm{B}]_{0}-[\mathrm{A}]_{0}\right) \times k t$$

The reaction \(A \longrightarrow\) products is second order. The initial rate of decomposition of \(A\) when \([\mathrm{A}]_{0}=0.50 \mathrm{M}\) is \((\mathrm{a})\) the same as the initial rate for any other value of \([\mathrm{A}]_{0} ;\) (b) half as great as when \([\mathrm{A}]_{0}=1.00 \mathrm{M} ;(\mathrm{c})\) five times as great as when \([\mathrm{A}]_{0}=[\mathrm{A}]_{0}=0.25 \mathrm{M}.\)

The following data are obtained for the initial rates of reaction in the reaction \(A+2B+C \longrightarrow 2 D+E.\) $$\begin{array}{lllll} \hline & \text { Initial } & \text { Initial } & & \\ \text { Expt } & \text { [A], M } & \text { [B],M } & \text { [C], M } & \text { Initial Rate } \\ \hline 1 & 1.40 & 1.40 & 1.00 & R_{1} \\ 2 & 0.70 & 1.40 & 1.00 & R_{2}=\frac{1}{2} \times R_{1} \\ 3 & 0.70 & 0.70 & 1.00 & R_{3}=\frac{1}{4} \times R_{2} \\ 4 & 1.40 & 1.40 & 0.50 & R_{4}=16 \times R_{3} \\ 5 & 0.70 & 0.70 & 0.50 & R_{5}=? \\ \hline \end{array}$$ (a) What are the reaction orders with respect to A, B, and C? (b) What is the value of \(R_{5}\) in terms of \(R_{1} ?\)
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