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Chapter 14: Problem 53

A reaction has the following experimental rate equation: Rate \(=k[\mathrm{A}]^{2}[\mathrm{B}] .\) If the concentration of \(\mathrm{A}\) is doubled and the concentration of B is halved, what happens to the reaction rate?

Short Answer

Expert verified
The reaction rate doubles.

Step by step solution

01

Understand the Rate Law

The rate law for the reaction is given by: \( \text{Rate} = k[\mathrm{A}]^{2}[\mathrm{B}] \). This means the reaction rate is proportional to the square of the concentration of \( \mathrm{A} \) and directly proportional to the concentration of \( \mathrm{B} \). \( k \) is the rate constant.
02

Determine the Effect of Doubling \( \mathrm{A} \)

If the concentration of \( \mathrm{A} \) is doubled, it goes from \( \mathrm{A} \) to \( 2\mathrm{A} \). Since the rate law involves \( [\mathrm{A}]^2 \), you substitute \( 2\mathrm{A} \) for \( \mathrm{A} \): \( (2[\mathrm{A}])^2 = 4[\mathrm{A}]^2 \). Thus, this change increases the rate by a factor of 4.
03

Determine the Effect of Halving \( \mathrm{B} \)

If \( \mathrm{B} \) concentration is halved, it changes from \( [\mathrm{B}] \) to \( \frac{1}{2}[\mathrm{B}] \). Substituting this into the rate equation, the effect of \([\mathrm{B}] \) becomes \( \frac{1}{2}[\mathrm{B}] \). Thus, this change decreases the rate by a factor of \( \frac{1}{2} \).
04

Combine Effects of Both Changes

Now, combine both effects: \( 4[\mathrm{A}]^2 \) from doubling \( \mathrm{A} \) and \( \frac{1}{2}[\mathrm{B}] \) from halving \( \mathrm{B} \). The overall rate change is \( 4 \times \frac{1}{2} = 2 \).
05

Conclude the Result

The final reaction rate increases by a factor of 2, which means it doubles when \( \mathrm{A} \) is doubled and \( \mathrm{B} \) is halved.

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Key Concepts

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

Rate Constant
The rate constant, often denoted as \( k \), is a fundamental part of the rate equation. It links the reaction rate with the concentrations of reactants raised to specific powers. Despite changes in concentration, \( k \) remains unchanged. This property makes it essential in determining how quickly a reaction progresses under given conditions.
Every reaction has its own specific rate constant at a particular temperature, and it's determined experimentally. In the reaction \( ext{Rate} = k[ ext{A}]^2[ ext{B}] \), \( k \) demonstrates the inherent speed of the reaction. Remember:
  • \( k \) is determined by characteristics of the reaction itself, such as the types of reactants and the temperature.
  • It does not change with concentration changes of reactants.
  • Units of \( k \) vary depending on the overall order of the reaction, making it crucial to understand dimensionality in kinetics.
Understanding the rate constant is key to grasping the overall kinetics of a chemical reaction.
Concentration Changes
Changes in concentration of reactants directly affect the reaction rate in accordance with the reaction's rate law. In our example rate law \( ext{Rate} = k[ ext{A}]^2[ ext{B}] \), the reaction rate is influenced by changes in concentrations of \( ext{A} \) and \( ext{B} \). This showcases how different orders of reactions affect the contribution of each reactant to the reaction rate:
  • Doubling \( ext{A} \) has a squared effect because it is raised to the power of 2, showing a greater sensitivity to its concentration change.
  • Halving \( ext{B} \) results in the rate being multiplied by \( \frac{1}{2} \), exhibiting a direct and proportional effect since it is raised to the first power.
Different reactions will have rate laws that specify how concentrations influence rates, helping to predict how varying amounts of reactants impact the progression of reactions. This principle is fundamental in the study of kinetics and understanding how reactions can be controlled and manipulated.
Proportional Relationships
The principle of proportional relationships in reaction kinetics is a cornerstone concept. When examining the rate law \( ext{Rate} = k[ ext{A}]^2[ ext{B}] \), it becomes clear how the relationship between concentration and rate directly influences how the reaction behaves.
Key aspects include:
  • The reaction rate is proportional to \( [ ext{A}]^2 \), meaning every change in the concentration of \( ext{A} \) drastically affects the reaction since the change is squared.
  • For \( [ ext{B}] \), the reaction rate is linearly proportional. A simple change in its concentration results in a directly proportional change in the rate.
  • Understanding these relationships enables prediction of outcomes when reactants are added or removed in varying quantities.
The interplay between different proportional relationships helps in tailoring reactions to achieve desired speeds, important in industrial and laboratory settings where efficiency and rate are critical.

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Most popular questions from this chapter

Isotopes are often used as "tracers" to follow an atom through a chemical reaction, and the following is an example. Acetic acid reacts with methanol. Explain how you could use the isotope \(^{18} \mathrm{O}\) to show whether the oxygen atom in the water comes from the \(-\) OH of \(\mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{H}\) or the \(-\mathrm{OH}\) of \(\mathrm{CH}_{3} \mathrm{OH}\).

The following statements relate to the reaction for the formation of HI: $$ \mathrm{H}_{2}(\mathrm{g})+\mathrm{I}_{2}(\mathrm{g}) \rightarrow 2 \mathrm{HI}(\mathrm{g}) \quad \text { Rate }=k\left[\mathrm{H}_{2}\right]\left|\mathrm{I}_{2}\right| $$ Determine which of the following statements are true. If a statement is false, indicate why it is incorrect. (a) The reaction must occur in a single step. (b) This is a second-order reaction overall. (c) Raising the temperature will cause the value of \(k\) to decrease. (d) Raising the temperature lowers the activation energy for this reaction. (e) If the concentrations of both reactants are doubled, the rate will double. (f) Adding a catalyst in the reaction will cause the initial rate to increase.

Nitramide, \(\mathrm{NO}_{2} \mathrm{NH}_{2}\), decomposes slowly in aqueous solution according to the following reaction: $$ \mathrm{NO}_{2} \mathrm{NH}_{2}(\mathrm{aq}) \rightarrow \mathrm{N}_{2} \mathrm{O}(\mathrm{g})+\mathrm{H}_{2} \mathrm{O}(\ell) $$ The reaction follows the experimental rate law $$ \text { Rate }=\frac{k\left[\mathrm{NO}_{2} \mathrm{NH}_{2}\right]}{\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]} $$ (a) What is the apparent order of the reaction in a pH buffered solution? (In a pH buffered solution, the concentration of \(\mathrm{H}_{3} \mathrm{O}^{+}\) is a constant.) (b) Which of the following mechanisms is the most appropriate for the interpretation of this rate law? Explain. (Note that when writing the expression for \(K,\) the equilibrium constant, \(\left.\left[\mathrm{H}_{2} \mathrm{O}\right] \text { is not involved. See Chapter } 15 .\right)\) Mechanism 1 \(\mathrm{NO}_{2} \mathrm{NH}_{2} \stackrel{k_{1}}{\longrightarrow} \mathrm{N}_{2} \mathrm{O}+\mathrm{H}_{2} \mathrm{O}\) Mechanism 2 $$ \mathrm{NO}_{2} \mathrm{NH}_{2}+\mathrm{H}_{3} \mathrm{O}^{+} \frac{k_{2}}{\overleftarrow{k_{2}^{\prime}}} \mathrm{NO}_{2} \mathrm{NH}_{3}^{+}+\mathrm{H}_{2} \mathrm{O} $$ \(\mathrm{NO}_{2} \mathrm{NH}_{3}^{+} \stackrel{k_{3}}{\longrightarrow} \mathrm{N}_{2} \mathrm{O}+\mathrm{H}_{3} \mathrm{O}^{+} \quad\) (rate-limiting step) Mechanism 3 \(\mathrm{NO}_{2} \mathrm{NH}_{2}+\mathrm{H}_{2} \mathrm{O} \frac{k_{4}}{k_{4}^{\prime}} \mathrm{NO}_{2} \mathrm{NH}^{-}+\underset{(\text { rapid equilibrium })}{\mathrm{H}_{3} \mathrm{O}^{+}}\) \(\mathrm{NO}_{2} \mathrm{NH}^{-} \stackrel{k_{5}}{\longrightarrow} \mathrm{N}_{2} \mathrm{O}+\mathrm{OH}^{-} \quad\) (rate-limiting step) \(\mathrm{H}_{3} \mathrm{O}^{+}+\mathrm{OH}^{-} \stackrel{k_{6}}{\longrightarrow} 2 \mathrm{H}_{2} \mathrm{O} \quad\) (very fast reaction) (c) Show the relationship between the experimentally observed rate constant, \(k\), and the rate constants in the selected mechanism. (d) Based on the experimental rate law, will the reaction rate increase or decrease if the pH of the solution is increased?

Radioactive gold- 198 is used in the diagnosis of liver problems. The half- life of this isotope is 2.7 days. If you begin with a 5.6 -mg sample of the isotope, how much of this sample remains after 1.0 day?

The decomposition of \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\) to \(\mathrm{SO}_{2}\) and \(\mathrm{Cl}_{2}\) is first-order in \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\) $$ \begin{array}{c} \mathrm{SO}_{2} \mathrm{Cl}_{2}(\mathrm{g}) \rightarrow \mathrm{SO}_{2}(\mathrm{g})+\mathrm{Cl}_{2}(\mathrm{g}) \\ \text { Rate }=k\left[\mathrm{SO}_{2} \mathrm{Cl}_{2}\right] \text { where } k=0.17 / \mathrm{hr} \end{array} $$ (a) What is the rate of decomposition when \(\left[\mathrm{SO}_{2} \mathrm{Cl}_{2}\right]=0.010 \mathrm{M} ?\) (b) What is the half-life of the reaction? (c) If the initial pressure of \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\) in a flask is \(0.050 \mathrm{atm},\) what is the pressure of all gases (i.e., the total pressure) in the flask after the reaction has proceeded for one half-life?
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