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Chapter 13: Problem 47

For each of the following rate laws, determine the order with respect to each reactant and the overall reaction order. a. Rate \(=k[\mathrm{A}][\mathrm{B}]\) b. Rate \(=k[\mathrm{A}]^{2}[\mathrm{B}]\) c. Rate \(=k[\mathrm{A}][\mathrm{B}]^{3}\)

Short Answer

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Question: Determine the reaction order with respect to each reactant and the overall reaction order for the following rate laws: a. Rate \(= k[\mathrm{A}][\mathrm{B}]\) b. Rate \(= k[\mathrm{A}]^{2}[\mathrm{B}]\) c. Rate \(= k[\mathrm{A}][\mathrm{B}]^{3}\) Answer: a. The reaction is first order with respect to A and first order with respect to B. The overall reaction order is 2. b. The reaction is second order with respect to A and first order with respect to B. The overall reaction order is 3. c. The reaction is first order with respect to A and third order with respect to B. The overall reaction order is 4.

Step by step solution

01

Finding orders related to reactants

The rate law is given as Rate \(= k[\mathrm{A}][\mathrm{B}]\). This means that the reaction is first order with respect to A and first order with respect to B, since the exponents of the concentrations are both equal to 1.
02

Finding overall reaction order

To find the overall reaction order, we add the individual orders with respect to each reactant: \(1 + 1 = 2\). Thus, the overall reaction order is 2. b. Rate \(=k[\mathrm{A}]^{2}[\mathrm{B}]\)
03

Finding orders related to reactants

The rate law is given as Rate \(= k[\mathrm{A}]^{2}[\mathrm{B}]\). This means that the reaction is second order with respect to A and first order with respect to B, as shown by the exponents of the concentrations.
04

Finding overall reaction order

To find the overall reaction order, we add the individual orders with respect to each reactant: \(2 + 1 = 3\). Thus, the overall reaction order is 3. c. Rate \(=k[\mathrm{A}][\mathrm{B}]^{3}\)
05

Finding orders related to reactants

The rate law is given as Rate \(= k[\mathrm{A}][\mathrm{B}]^{3}\). This means that the reaction is first order with respect to A and third order with respect to B, as shown by the exponents of the concentrations.
06

Finding overall reaction order

To find the overall reaction order, we add the individual orders with respect to each reactant: \(1 + 3 = 4\). Thus, the overall reaction order is 4.

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Most popular questions from this chapter

Hydrogen gas reduces NO to \(\mathrm{N}_{2}\) in the following reaction: $$2 \mathrm{H}_{2}(g)+2 \mathrm{NO}(g) \rightarrow 2 \mathrm{H}_{2} \mathrm{O}(g)+\mathrm{N}_{2}(g)$$ The initial reaction rates of four mixtures of \(\mathrm{H}_{2}\) and \(\mathrm{NO}\) were measured at \(900^{\circ} \mathrm{C}\) with the following results: $$\begin{array}{cccc}\text { Experiment } & \left[\mathrm{H}_{2}\right]_{0}(\mathrm{M}) & [\mathrm{NO}]_{0}(\mathrm{M}) & \begin{array}{c}\text { Initial } \\\\\text { Rate }(M / \mathrm{s})\end{array} \\\\\hline 1 & 0.212 & 0.136 & 0.0248 \\\\\hline 2 & 0.212 & 0.272 & 0.0991 \\\\\hline 3 & 0.424 & 0.544 & 0.793 \\\\\hline 4 & 0.848 & 0.544 & 1.59 \\\\\hline\end{array}$$ Determine the rate law and the rate constant for the reaction at \(900^{\circ} \mathrm{C}\).

Why doesn't a quadrupling of the rate correspond to a reaction order of \(4-\) for example, Rate \(\propto[\mathrm{NO}]^{4} ?\)

Can the half-life of a second-order reaction have the same units as the half- life of a first-order reaction?

A student inserts a glowing wood splint into a test tube filled with \(\mathrm{O}_{2}\). The splint quickly catches on fire (Figure P13.121). Why does the splint burn so much faster in pure \(\mathrm{O}_{2}\) than in air? (IMAGE NOT COPY)

The rate constant for the reaction of ozone with oxygen atoms was determined at four temperatures. Calculate the activation energy and frequency factor \(A\) for the reaction $$\mathrm{O}(g)+\mathrm{O}_{3}(g) \rightarrow 2 \mathrm{O}_{2}(g)$$ given the following data: $$\begin{array}{cc}T(\mathrm{K}) & k\left[\mathrm{cm}^{3} /(\text { molecule } \cdot \mathrm{s})\right] \\\250 & 2.64 \times 10^{-4} \\\\\hline 275 & 5.58 \times 10^{-4} \\\\\hline 300 & 1.04 \times 10^{-3} \\\\\hline 325 & 1.77 \times 10^{-3} \\\\\hline\end{array}$$
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