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

Models of population growth are analogous to chemical reaction rate equations. In the model developed by Malthus in 1798 , the rate of change of the population \(N\) of Earth is \(\mathrm{d} N / \mathrm{d} t=\) births - deaths. The numbers of births and deaths are proportional to the population, with proportionality constants \(b\) and \(d\). Derive the integrated rate law for population change. How well does it fit the approximate data for the population of Earth over time given below? $$ \begin{array}{lccccccc} \text { Year } & 1750 & 1825 & 1922 & 1960 & 1974 & 1987 & 2000 \\ N / 10^{9} & 0.5 & 1 & 2 & 3 & 4 & 5 & 6 \end{array} $$

The half-life of a substance taking part in a third-order reaction \(\mathrm{A} \rightarrow\) products is inversely proportional to the square of the initial concentration of A. How can this half-life be used to predict the time needed for the concentration to fall to (a) onehalf; (b) one-fourth; (c) one- sixteenth of its initial value?

For the reversible, one-step reaction \(\mathrm{A}+\mathrm{B} \rightleftarrows \mathrm{C}+\mathrm{D}\), the forward rate constant is \(52.4 \mathrm{~L} \cdot \mathrm{mol}^{-1} \cdot \mathrm{h}^{-1}\) and the rate constant for the reverse reaction is \(32.1 \mathrm{~L} \cdot \mathrm{mol}^{-1} \cdot \mathrm{h}^{-1}\). The activation energy was found to be \(35.2 \mathrm{~kJ} \cdot \mathrm{mol}^{-1}\) for the forward reaction and \(44.0 \mathrm{~kJ} \cdot \mathrm{mol}^{-1}\) for the reverse reaction. (a) What is the equilibrium constant for the reaction? (b) Is the reaction exothermic or endothermic? (c) What will be the effect of raising the temperature on the rate constants and the equilibrium constant?

Determine whether each of the following statements is true or false. If a statement is false, explain why. (a) For a reaction with a very large equilibrium constant, the rate constant of the forward reaction is much larger than the rate constant of the reverse reaction. (b) At equilibrium, the rate constants of the forward and reverse reactions are equal. (c) Increasing the concentration of a reactant increases the rate of a reaction by increasing the rate constant in the forward direction.

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?

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