A reaction in which \(\mathrm{A}, \mathrm{B},\) and \(\mathrm{C}\) react to form products is zero order in \(\mathrm{A},\) one-half order in \(\mathrm{B},\) and second order in \(\mathrm{C} .\) \begin{equation} \begin{array}{l}{\text { a. Write a rate law for the reaction. }} \\ {\text { b. What is the overall order of the reaction? }} \\ {\text { c. By what factor does the reaction rate change if }[\mathrm{A}] \text { is doubled (and the }} \\ {\text { other reactant concentrations are held constant)? }}\end{array} \end{equation} \begin{equation} \begin{array}{l}{\text { d. By what factor does the reaction rate change if }[\mathrm{B}] \text { is doubled (and the }} \\ {\text { other reactant concentrations are held constant)? }} \\ {\text { e. By what factor does the reaction rate change if }[\mathrm{C}] \text { is doubled (and the }} \\\ {\text { other reactant concentrations are held constant)? }}\\\\{\text { f. By what factor does the reaction rate change if the concentrations of }} \\\ {\text { all three reactants are doubled? }}\end{array} \end{equation}

Short Answer

Expert verified
a. Rate = k[A]^{0}[B]^{1/2}[C]^{2}, b. Overall order is 2.5, c. No change in rate, d. Rate increases by √2, e. Rate increases by 4, f. Rate increases by 4√2.

Step by step solution

01

Write the Rate Law

The rate law is determined by the orders of reaction with respect to each reactant. Given that the reaction is zero order in A, one-half order in B, and second order in C, the rate law can be written as: Rate = k[A]^{0}[B]^{1/2}[C]^{2}. The exponents correspond to the reaction order of each reactant.
02

Determine the Overall Order of the Reaction

The overall order of reaction is the sum of the orders of each reactant: (0) + (1/2) + (2) = 5/2 or 2.5. The reaction is 2.5 order overall.
03

Effect of Doubling [A] on the Reaction Rate

Since the reaction is zero order in A, changing its concentration has no effect on the reaction rate. Therefore, doubling [A] would result in no change in the reaction rate: Rate Change Factor = [A]^{0} = 1^{0} = 1.
04

Effect of Doubling [B] on the Reaction Rate

The reaction is one-half order in B. If [B] is doubled, the rate would increase by a factor of 2^{1/2} or √2. Rate Change Factor when [B] is doubled: (2)^{1/2} = √2.
05

Effect of Doubling [C] on the Reaction Rate

The reaction is second order in C. If [C] is doubled, the rate would increase by a factor of 2^{2} = 4. Rate Change Factor when [C] is doubled: (2)^{2} = 4.
06

Effect of Doubling All Reactants on the Reaction Rate

When all the reactants are doubled, the rate change factor will be the product of the individual change factors. For [A], the factor is 1; for [B], it is √2; and for [C], it is 4. Hence, the total factor is: 1 * √2 * 4 = 4√2. Note that increasing [A] has no effect as the reaction is zero order with respect to A.

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Key Concepts

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

Rate Law
The rate law is a mathematical equation that describes the relationship between the rate of a chemical reaction and the concentration of reactants. It allows us to understand how different factors, such as changes in concentration or the presence of a catalyst, affect the speed at which a reaction takes place. For instance, if we have a reaction where substances A, B, and C combine to form products, the rate law might look something like this:
\( \text{Rate} = k[A]^x[B]^y[C]^z \).
Here, \( k \) is the rate constant, a unique value that depends on the reaction and conditions, while \( x \), \( y \), and \( z \) represent the reaction orders with respect to A, B, and C respectively. These exponents tell us how the rate is affected by changes in concentration of each reactant. If any of the exponents are zero, it indicates that changing the concentration of that reactant has no effect on the rate of reaction.
Overall Reaction Order
Understanding overall reaction order is essential for a comprehensive study of reaction kinetics. The overall reaction order is simply the sum of the exponents in the rate law equation. Taking our previous example, with exponents as the reaction orders of each reactant in the rate law, the overall reaction order is the addition of these values: \( x + y + z \). If we have a reaction where A is zero order, B is one-half order, and C is second order, by adding up these orders (0 for A, 1/2 for B, and 2 for C), we determine that the reaction has an overall order of 2.5. This gives us insight into how the reaction rate will respond to changes in the concentration of all the reactants combined. A reaction's overall order is crucial when comparing different reactions and predicting the impact of concentration changes on the rate of reaction.
Reactant Concentration Effects
Reactant concentration has a pivotal role in chemical kinetics. Each reactant's impact is governed by its reaction order, as depicted in the rate law. A zero-order reaction means that changes in the reactant's concentration don't influence the rate. A reaction with a higher reaction order, like second order, is more sensitive to changes in reactant concentration. For a reaction that's second order in one reactant, doubling the concentration of that particular reactant will quadruple the reaction rate, since the rate change is proportional to the square of its concentration change (\( 2^2 = 4 \)). Understanding how individual and combined changes in reactant concentrations affect the reaction rate can be highly useful in industrial processes and experimental design, allowing chemists to control reaction speeds for desired outcomes.

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Most popular questions from this chapter

This reaction is first order in \(\mathrm{N}_{2} \mathrm{O}_{5}:\) $$\mathrm{N}_{2} \mathrm{O}_{5}(g) \longrightarrow \mathrm{NO}_{3}(g)+\mathrm{NO}_{2}(g)$$ The rate constant for the reaction at a certain temperature is 0.053\(/ \mathrm{s}.\) \begin{equation} \begin{array}{l}{\text { a. Calculate the rate of the reaction when }\left[\mathrm{N}_{2} \mathrm{O}_{5}\right]=0.055 \mathrm{M} \text { . }} \\\ {\text { b. What is the rate of the reaction at the concentration indicated in part }} \\ {\text { a if the reaction is second order? Zero order? (Assume the same } n u-} \\ {\text { merical value for the rate constant with the appropriate units.) }}\end{array} \end{equation}

\(\begin{aligned} \text { Consider the reaction. } \\ & 2 \mathrm{N}_{2} \mathrm{O}(g) \longrightarrow 2 \mathrm{N}_{2}(g)+\mathrm{O}_{2}(g) \end{aligned}\) \begin{equation} \begin{array}{l}{\text { a. Express the rate of the reaction in terms of the change in }} \\ {\text { concentration of each of the reactants and products. }} \\ {\text { b. In the first } 15.0 \text { s of the reaction, 0.015 mol of } \mathrm{O}_{2} \text { is produced in a }} \\ {\text { reaction vessel with a volume of } 0.500 \text { L. What is the average rate }} \\ {\text { of the reaction during this time interval? }}\end{array} \end{equation} \begin{equation} \begin{array}{l}{\text { c. Predict the rate of change in the concentration of } \mathrm{N}_{2} \mathrm{O} \text { during this }} \\ {\text { time interval. In other words, what is } \Delta\left[\mathrm{N}_{2} \mathrm{O}\right] / \Delta t?}\end{array} \end{equation}

Anthropologists can estimate the age of a bone or other sample of or- ganic matter by its carbon-14 4 content. The carbon-14 in a living organ- ism is constant until the organism dies, after which carbon-14 decays with first-order kinetics and a half-life of 5730 years. Suppose a bone from an ancient human contains 19.5\(\%\) of the \(\mathrm{C}-14\) found in living or- ganisms. How old is the bone?

In this chapter we have seen a number of reactions in which a single reactant forms products. For example, consider the following first- order reaction: $$\mathrm{CH}_{3} \mathrm{NC}(g) \longrightarrow \mathrm{CH}_{3} \mathrm{CN}(g)$$ However, we also learned that gas-phase reactions occur through collisions. \begin{equation} \begin{array}{l}{\text { a. One possible explanation is that two molecules of } \mathrm{CH}_{3} \mathrm{NC} \text { collide }} \\ {\text { with each other and form two molecules of the product in a single ele- }} \\ {\text { mentary step. If that is the case, what reaction order would you expect? }}\\\\{\text { b. Another possibility is that the reaction occurs through more than }} \\\ {\text { one step. For example, a possible mechanism involves one step in }} \\\ {\text { which the two CH}_{3} \mathrm{NC} \text { molecules collide, resulting in the activation" }} \\ {\text { of one of them. In a second step, the activated molecule goes on to }}\\\\{l}{\text { form the product. Write down this mechanism and determine which }} \\ {\text { step must be rate determining in order for the kinetics of the reaction }} \\ {\text { to be first order. Show explicitly how the mechanism predicts first- }} \\ {\text { order kinetics. }}\end{array} \end{equation}

A particular reaction, \(A \longrightarrow\) products, has a rate that slows down as the reaction proceeds. The half-life of the reaction is found to depend on the initial concentration of A. Determine whether each statement is like- ly to be true or false for this reaction. \begin{equation} \begin{array}{l}{\text { a. A doubling of the concentration of A doubles the rate of the reaction. }} \\ {\text { b. A plot of } 1 /[\mathrm{A}] \text { versus time is linear. }} \\ {\text { c. The half-life of the reaction gets longer as the initial concentration of }} \\ {\text { A increases. }} \\\ {\text { d. A plot of the concentration of A versus time has a constant slope. }}\end{array} \end{equation}

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