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Chapter 11: Problem 20

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

Short Answer

Expert verified
The reaction rate is doubled.

Step by step solution

01

Understanding the Rate Law

The rate law for the reaction is given by Rate = \( k[A]^2[B] \). This means the rate depends on the concentration of \( A \) squared and \( B \) linearly.
02

Identify Initial Rate Expression

Initial rate can be expressed as \( ext{Rate}_1 = k[A]^2[B] \). This represents the original conditions of the reaction.
03

Adjust Concentrations

The concentration of \( A \) is doubled, so \( [A] \to 2[A] \). The concentration of \( B \) is halved, so \( [B] \to \frac{1}{2}[B] \).
04

Substitute Adjusted Concentrations in Rate Law

Substitute the changed concentrations into the rate law to find the new rate: \( ext{Rate}_2 = k(2[A])^2(\frac{1}{2}[B]) \).
05

Simplify the Expression

Calculate the new rate: \( ext{Rate}_2 = k(4[A]^2)(\frac{1}{2}[B]) = 4[A]^2 \times \frac{1}{2}[B] = 2k[A]^2[B] \).
06

Compare Old and New Rates

The original rate was \( k[A]^2[B] \) and the new rate is \( 2k[A]^2[B] \). Therefore, the rate is doubled.

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Key Concepts

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

Rate Law Expression
The rate law expression is a mathematical relationship that describes the rate of a chemical reaction based on the concentration of reactants. It is represented in the general form, \( ext{Rate} = k[A]^m[B]^n\), where \(k\) is the rate constant, and \(m\) and \(n\) are the orders of the reaction concerning reactants \(A\) and \(B\) respectively. These orders indicate how the concentration of each reactant affects the rate. In our example, Rate = \(k[A]^2[B]\), the reaction order is 2 with respect to \(A\) and 1 with respect to \(B\). Understanding the exponents helps predict how changes in concentrations will affect the rate.
Concentration Effects on Reaction Rate
The concentration of reactants greatly influences the reaction rate, as described by the rate law. If you increase the concentration of a reactant, you generally increase the rate at which the reaction occurs. In our given problem, the concentration of \([A]\) is doubled, and \([B]\) is halved. This change directly affects the rate, which can be calculated by plugging the new concentrations into the rate law expression.
  • Doubling \([A]\) implies substituting \(2[A]\) for \([A]\) in the rate law, leading to \((2[A])^2 = 4[A]^2\).
  • Halving \([B]\) means replacing \([B]\) with \(\frac{1}{2}[B]\), modifying the terms that include \([B]\).
By calculating, we see that the new rate is twice the original rate, demonstrating the marked effect of concentration changes on reaction kinetics.
Chemical Kinetics
Chemical kinetics is the branch of chemistry that studies the speed or rate of chemical reactions. It helps predict how different conditions influence the speed and mechanism of the reaction. Through understanding kinetics, we can comprehend why some reactions are instantaneous while others take years to reach completion. Rate laws, including both concentration effects and temperature dependencies, are fundamental components of this field.
In practice, chemical kinetics allows chemists to optimize reactions to be faster, more efficient, and economically feasible. By manipulating variables such as concentration, which we analyzed using our specific reaction rate law, scientists can determine the best conditions for industrial or laboratory chemical processes. This ability to control reactions is crucial in a wide range of fields, including pharmaceuticals, agriculture, and materials science.

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Most popular questions from this chapter

Assume that each gas-phase reaction occurs via a single bimolecular step. For which reaction would you expect the steric factor to be more important? Why? $$ \begin{aligned} \mathrm{H}_{2} \mathrm{C} &=\mathrm{CH}_{2}+\mathrm{H}_{2} \longrightarrow \mathrm{H}_{3} \mathrm{C}-\mathrm{CH}_{3} \text { or } \\ \left(\mathrm{CH}_{3}\right)_{2} \mathrm{C}=\mathrm{CH}_{2}+\mathrm{HBr} \longrightarrow &\left(\mathrm{CH}_{3}\right)_{2} \mathrm{CBr}-\mathrm{CH}_{3} \end{aligned} $$

Using the rate law, rate \(=k[\mathrm{~A}]^{2}[\mathrm{~B}],\) define the order of the reaction with respect to \(\mathrm{A}\) and \(\mathrm{B}\) and the overall reaction order.

In a time-resolved picosecond spectroscopy experiment, Sheps, Crowther, Carrier, and Crim (Journal of Physical Chemistry \(A,\) Vol. 110,\(2006 ;\) pp. \(3087-3092\) ) generated chlorine atoms in the presence of pentane. The pentane was dissolved in dichloromethane, \(\mathrm{CH}_{2} \mathrm{Cl}_{2}\). The chlorine atoms are free radicals and are very reactive. After a nanosecond the chlorine atoms have reacted with pentane molecules, removing a hydrogen atom to form \(\mathrm{HCl}\) and leaving behind a pentane radical with a single unpaired electron. The equation is \(\mathrm{Cl} \cdot(\mathrm{dcm})+\mathrm{C}_{5} \mathrm{H}_{12}(\mathrm{dcm}) \longrightarrow \mathrm{HCl}(\mathrm{dcm})+\mathrm{C}_{5} \mathrm{H}_{11} \cdot(\mathrm{dcm})\) where \((\mathrm{dcm})\) indicates that a substance is dissolved in dichloromethane. Measurements of the concentration of chlorine atoms were made as a function of time at three different concentrations of pentane in the dichloromethane. These results are shown in the table. of Chlorine Atoms (M) \begin{tabular}{clcc} & \multicolumn{3}{c} { Concentration of Chlorine Atoms (m) } \\ Time (ps) & \multicolumn{4}{c} { for Different \(\mathrm{C}_{5} \mathrm{H}_{12}\) Concentrations } \\ \cline { 3 - 5 } & \(0.15-\mathrm{M} \mathrm{C}_{5} \mathrm{H}_{12}\) & \(0.30-\mathrm{M} \mathrm{C}_{5} \mathrm{H}_{12}\) & \(0.60-\mathrm{M} \mathrm{C}_{5} \mathrm{H}_{12}\) \\ \hline 100.0 & \(4.42 \times 10^{-5}\) & \(3.11 \times 10^{-5}\) & \(1.77 \times 10^{-5}\) \\ 140.0 & \(3.823 \times 10^{-5}\) & \(2.39 \times 10^{-5}\) & \(1.15 \times 10^{-5}\) \\\ 180.0 & \(3.38 \times 10^{-5}\) & \(1.94 \times 10^{-5}\) & \(8.48 \times 10^{-6}\) \\\ 220.0 & \(3.03 \times 10^{-5}\) & \(1.49 \times 10^{-5}\) & \(6.04 \times 10^{-6}\) \\\ 260.0 & \(2.68 \times 10^{-5}\) & \(1.19 \times 10^{-5}\) & \(4.12 \times 10^{-6}\) \\\ 300.0 & \(2.42 \times 10^{-5}\) & \(9.45 \times 10^{-6}\) & \(3.14 \times 10^{-6}\) \\\ 340.0 & \(2.08 \times 10^{-5}\) & \(7.75 \times 10^{-6}\) & \(2.38 \times 10^{-6}\) \\\ 380.0 & \(1.91 \times 10^{-5}\) & \(6.35 \times 10^{-6}\) & \(1.75 \times 10^{-6}\) \\\ 420.0 & \(1.71 \times 10^{-5}\) & \(4.58 \times 10^{-6}\) & \(1.61 \times 10^{-6}\) \\\ 460.0 & \(1.53 \times 10^{-5}\) & \(3.77 \times 10^{-6}\) & \(9.98 \times 10^{-7}\) \\\ \hline \end{tabular} (a) Determine the order of the reaction with respect to chlorine. (b) Determine whether the reaction rate depends on the concentration of pentane in dichloromethane. If so, determine the order of the reaction with respect to pentane. (c) Explain why the concentration of pentane in dichloromethane does not affect the data analysis that you performed in part (a). (d) Write the rate law for the reaction and calculate the rate of reaction for a concentration of chlorine atoms equal to \(1.0 \mu \mathrm{M}\) and a pentane concentration of \(0.23 \mathrm{M}\). (e) Sheps, Crowther, Carrier, and Crim found that the rate of formation of \(\mathrm{HCl}\) matched the rate of disappearance of Cl. From this they concluded that there were no intermediates and side reactions were not important. Explain the ic

The decomposition of ammonia to nitrogen and hydrogen on tungsten at \(1100^{\circ} \mathrm{C}\) is zeroth-order with a rate constant of \(2.5 \times 10^{-4} \mathrm{~mol} \mathrm{~L}^{-1} \min ^{-1}\). (a) Write the rate expression. (b) Calculate the rate when \(\left[\mathrm{NH}_{3}\right]=0.075 \mathrm{M}\).

If a reaction has the experimental rate law, Rate \(=k[\mathrm{~A}]^{2}\), explain what happens to the rate when (a) the concentration of A is tripled. (b) the concentration of A is halved.
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