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In the reaction \(\mathrm{CH}_{3} \mathrm{Br}(\mathrm{aq})+\mathrm{OH}^{-}(\mathrm{aq}) \rightarrow \mathrm{CH}_{3} \mathrm{OH}(\mathrm{aq})+\) \(\mathrm{Br}^{-}(\mathrm{aq})\), when the \(\mathrm{OH}^{-}\)concentration alone was doubled, the rate doubled; when the \(\mathrm{CH}_{3} \mathrm{Br}\) concentration alone was increased by a factor of \(1.2\), the rate increased by a factor of \(1.2\). Write the rate law for the reaction.

Short Answer

Expert verified
The rate law for the reaction is Rate = k [CH3Br][OH-].

Step by step solution

01

Analyze the effect of doubling OH- concentration

The reaction's rate doubles when the concentration of OH- is doubled, keeping the concentration of CH3Br constant. This indicates that the rate is directly proportional to the concentration of OH-. Hence, the reaction is first order with respect to OH-.
02

Analyze the effect of increasing CH3Br concentration

When the concentration of CH3Br is increased by a factor of 1.2, the reaction's rate also increases by a factor of 1.2. This implies that the rate is directly proportional to the concentration of CH3Br. Thus, the reaction is first order with respect to CH3Br.
03

Write the rate law

Based on the given information, the rate law can be expressed by combining the dependency on both reactants. The rate law is: Rate = k [CH3Br][OH-].

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Key Concepts

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

Chemical Kinetics
Chemical kinetics plays a pivotal role in understanding how chemical reactions occur. It is the branch of physical chemistry that studies the rates of chemical processes and the factors that affect them. The kinetic analysis of a reaction involves examining how the rate changes in response to changes in variables such as concentration, temperature, and the presence of catalysts.

Through experiments, chemists can determine the speed at which reactants are transformed into products, which is crucial for various applications from industrial synthesis to pharmaceuticals. Kinetics also aids in elucidating reaction mechanisms, offering a microscopic glimpse of the steps that lead from reactants to products. By analyzing the effects of modifying reactant concentrations, scientists can derive a rate law that sums up the quantitative relationship between the rate and the concentrations of reactants.
Reaction Rates
The rate of a chemical reaction quantifies the speed at which reactants turn into products. It can be measured by the change in concentration of reactants or products per unit time. In the studied reaction, \( \mathrm{CH}_{3}\mathrm{Br}(\mathrm{aq})+\mathrm{OH}^{-}(\mathrm{aq}) \rightarrow \mathrm{CH}_{3} \mathrm{OH}(\mathrm{aq})+ \mathrm{Br}^{-}(\mathrm{aq}) \), understanding the reaction rate is fundamental to formulating the rate law.

For instance, observing that doubling the concentration of \( \mathrm{OH}^{-} \) leads to a doubling of the rate shows a direct relationship, which is linear and proportional. Similarly, increasing \( \mathrm{CH}_{3} \mathrm{Br} \) concentration by a factor of 1.2 causing the rate to rise by the same factor supports the inference about its proportional effect on the rate. These experimental observations are integral to defining the rate law and establishing the stoichiometry of the rate-determining step.
Order of Reaction
The term 'order of reaction' refers to the exponent of the concentration of a reactant in the rate law equation, reflecting how the rate is affected by the concentration of that reactant. For the given reaction, we've deduced the order with respect to each reactant based on experimental data.

The reaction is first-order with respect to \( \mathrm{OH}^{-} \) because when its concentration is doubled, the rate also doubles. Similarly, the reaction is first-order with respect to \( \mathrm{CH}_{3} \mathrm{Br} \) since the rate is directly proportional to its concentration. In conclusion, this reaction is overall second-order, which is the sum of the individual orders (first-order with respect to both reactants). The rate law, thus, takes the form \( \text{Rate} = k [\mathrm{CH}_{3}\mathrm{Br}][\mathrm{OH}^{-}] \) where \( k \) is the rate constant. Understanding the order of reaction is crucial for predicting how changes in concentrations will alter the reaction rate and for figuring out the precise mechanism by which the reaction proceeds.

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Most popular questions from this chapter

Determine which of the following statements about catalysts are true. If the statement is false, explain why. (a) A heterogeneous catalyst works by binding one or more of the molecules undergoing reaction to the surface of the catalyst. (b) Enzymes are naturally occurring proteins that serve as catalysts in biological systems. (c) The equilibrium constant for a reaction is greater in the presence of a catalyst. (d) A catalyst changes the pathway of a reaction in such a way that the reaction becomes more exothermic.

(a) Using a graphing calculator or standard graphing software, such as that on the Web site for this book, make an appropriate Arrhenius plot of the data shown here for the conversion of cyclopropane into propene and calculate the activation energy for the reaction. (b) What is the value of the rate constant at \(600^{\circ} \mathrm{C}\) ? $$ \begin{array}{lcccc} T(\mathrm{~K}) & 750 . & 800 . & 850 . & 900 . \\ k\left(\mathrm{~s}^{-1}\right) & 1.8 \times 10^{-4} & 2.7 \times 10^{-3} & 3.0 \times 10^{-2} & 0.26 \end{array} $$

Manganate ions, \(\mathrm{MnO}_{4}^{2-}\), react at \(2.0 \mathrm{~mol} \cdot \mathrm{L}^{-1} \cdot \mathrm{min}^{-1}\) in acidic solution to form permanganate ions and manganese(IV) oxide: \(3 \mathrm{MnO}_{4}{ }^{2-}(\mathrm{aq})+4 \mathrm{H}^{+}(\mathrm{aq}) \rightarrow 2 \mathrm{MnO}_{4}{ }^{-}(\mathrm{aq})+\mathrm{MnO}_{2}(\mathrm{~s})+\) \(2 \mathrm{H}_{2} \mathrm{O}(\mathrm{l})\). (a) What is the rate of formation of permanganate ions? (b) What is the rate of reaction of \(\mathrm{H}^{+}(\mathrm{aq})\) ? (c) What is the unique rate of the reaction?

The half-life for the second-order reaction of a substance A is \(50.5 \mathrm{~s}\) when \([\mathrm{A}]_{0}=0.84 \mathrm{~mol} \cdot \mathrm{L}^{-1}\). Calculate the time needed for the concentration of A to decrease to (a) one- sixteenth; (b) onefourth; (c) one-fifth of its original value.

(a) Using a graphing calculator or graphing software, such as that on the Web site for this book, calculate the activation energy for the acid hydrolysis of sucrose to give glucose and fructose from an Arrhenius plot of the data shown here. (b) Calculate the rate constant at \(37^{\circ} \mathrm{C}\) (body temperature). (c) From data in Appendix 2A, calculate the enthalpy change for this reaction, assuming that the solvation enthalpies of the sugars are negligible. Draw an energy profile for the overall process. $$ \begin{array}{cc} \text { Temperature }\left({ }^{\circ} \mathrm{C}\right) & k\left(\mathrm{~s}^{-1}\right) \\ \hline 24 & 4.8 \times 10^{-3} \\ 28 & 7.8 \times 10^{-3} \\ 32 & 13 \times 10^{-3} \\ 36 & 20 . \times 10^{-3} \\ 40 . & 32 \times 10^{-3} \\ \hline \end{array} $$

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