/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 49 A reaction has the following exp... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 15: Problem 49

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 \(\mathrm{B}\) is halved, what happens to the reaction rate?

Short Answer

Expert verified
The reaction rate doubles.

Step by step solution

01

Identify the Initial Rate Equation

The initial rate equation is given as: Rate \(= k[A]^2[B]\), where \([A]\) and \([B]\) are the concentrations of A and B, respectively, and \(k\) is the rate constant.
02

Modify the Concentrations

The problem states that the concentration of \(A\) is doubled and \(B\) is halved. Therefore, the new concentrations are \([A]' = 2[A]\) and \([B]' = \frac{1}{2}[B]\).
03

Substitute New Concentrations into the Rate Equation

Insert the new concentrations into the rate equation: Rate' \(= k(2[A])^2\left(\frac{1}{2}[B]\right)\).
04

Simplify the Expression

Calculate the expression: \((2[A])^2 = 4[A]^2\), and then multiply by \(\frac{1}{2}[B]\): Rate' \(= k \cdot 4[A]^2 \cdot \frac{1}{2}[B] = 2k[A]^2[B]\).
05

Compare New Rate with Original Rate

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

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Rate Equation
A rate equation is an expression that links the rate of a chemical reaction to the concentration of its reactants. In the given exercise, the rate equation is: Rate \(= k[A]^2[B] \). This tells us how the rate of reaction is affected by the concentrations of substances \( A \) and \( B \). The \( k \) in the equation is the rate constant, a number that helps scale the relationship between the concentrations of reactants and the rate. The exponents on \([A]\) and \([B]\) indicate how the concentration changes affect the rate. For example, in this equation, since \([A]\) is squared, any change in its concentration has a significant impact on the rate. Understanding how modifications in reactant concentrations translate into changes in the reaction rate is crucial in predicting how the reaction proceeds under varying conditions.
Chemical Kinetics
Chemical kinetics is the branch of chemistry that studies the speed or rate at which chemical reactions occur. It helps determine how different factors like concentration, temperature, and catalysts impact reaction rates. In the provided exercise, we see a practical application of chemical kinetics. By altering the concentrations of \( A \) and \( B \), we're able to observe the effect on the reaction rate. Kinetics gives insights into the mechanism of the reaction, explaining the step-by-step process at the molecular level that leads to product formation. Kinetics is essential for both academic research and industrial applications, as it can optimize reactions to make them faster and more efficient.
Concentration
Concentration refers to the amount of a substance present in a certain volume of solution. It plays a pivotal role in reactions as it influences the rate at which they occur. The rate equation in our exercise demonstrates this by showing that the rate is proportional to \([A]^2[B] \), meaning that an increase or decrease in these concentrations will directly affect the speed of the reaction. For instance, doubling the concentration of \( A \) results in a quadrupled effect since it is squared in the rate equation, illustrating a relatively larger impact on the reaction rate compared to changes in \( B \). Understanding how concentration impacts reaction rates can help in designing better experimental conditions and achieving desired reaction outcomes more efficiently.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

The mechanism for the reaction of \(\mathrm{CH}_{3} \mathrm{OH}\) and \(\mathrm{HBr}\) is believed to involve two steps. The overall reaction is exothermic. Step 1: Fast, endothermic $$\mathrm{CH}_{3} \mathrm{OH}+\mathrm{H}^{+} \rightleftarrows \mathrm{CH}_{3} \mathrm{OH}_{2}^{+}$$ Step 2: Slow $$\mathrm{CH}_{3} \mathrm{OH}_{2}^{+}+\mathrm{Br}^{-} \rightarrow \mathrm{CH}_{3} \mathrm{Br}+\mathrm{H}_{2} \mathrm{O}$$ (a) Write an equation for the overall reaction. (b) Draw a reaction coordinate diagram for this reaction. (c) Show that the rate law for this reaction is Rate \(=k\left[\mathrm{CH}_{3} \mathrm{OH}\right]\left[\mathrm{H}^{+}\right]\left[\mathrm{Br}^{-}\right]\)

Gaseous azomethane, \(\mathrm{CH}_{3} \mathrm{N}=\mathrm{NCH}_{3},\) decomposes in a first-order reaction when heated: $$\mathrm{CH}_{3} \mathrm{N}=\mathrm{NCH}_{3}(\mathrm{g}) \rightarrow \mathrm{N}_{2}(\mathrm{g})+\mathrm{C}_{2} \mathrm{H}_{6}(\mathrm{g})$$ The rate constant for this reaction at \(600 \mathrm{K}\) is 0.0216 \(\min ^{-1} .\) If the initial quantity of azomethane in the flask is \(2.00 \mathrm{g},\) how much remains after 0.0500 hour? What quantity of \(\mathrm{N}_{2}\) is formed in this time?

Experimental data are listed here for the reaction \(A \rightarrow 2 B.\) $$\begin{array}{cc}\text { Time (s) } & {[\mathrm{B}](\mathrm{mol} / \mathrm{L})} \\\\\hline 0.00 & 0.000 \\\10.0 & 0.326 \\\20.0 & 0.572 \\\30.0 & 0.750 \\\40.0 & 0.890 \\\\\hline\end{array}$$ (a) Prepare a graph from these data; connect the points with a smooth line; and calculate the rate of change of [B] for each 10 -second interval from 0.0 to 40.0 seconds. Does the rate of change decrease from one time interval to the next? Suggest a reason for this result. (b) How is the rate of change of [A] related to the rate of change of \([\mathrm{B}]\) in each time interval? Calculate the rate of change of \([\mathrm{A}]\) for the time interval from 10.0 to 20.0 seconds. (c) What is the instantaneous rate, \(\Delta[\mathrm{B}] / \Delta \mathrm{t},\) when \([\mathrm{B}]=0.750 \mathrm{mol} / \mathrm{L} ?\)

Using the rate equation "Rate \(=k[\mathrm{A}]^{2}[\mathrm{B}],\) " define the order of the reaction with respect to \(\mathrm{A}\) and \(\mathrm{B}\). What is the total order of the reaction?

The substitution of \(\mathrm{CO}\) in \(\mathrm{Ni}(\mathrm{CO})_{4}\) by another molecule L [where \(\left.L \text { is an electron-pair donor such as } P\left(C H_{3}\right)_{3}\right]\) was studied some years ago and led to an understanding of some of the general principles that govern the chemistry of compounds having metal-CO bonds. (See J. P. Day, F. Basolo, and R. G. Pearson: Journal of the American Chemical Society, Vol. \(90,\) p. \(6927,1968 .\) ) A detailed study of the kinetics of the reaction led to the following mechanism: Slow \(\quad \mathrm{Ni}(\mathrm{CO})_{4} \rightarrow \mathrm{Ni}(\mathrm{CO})_{3}+\mathrm{CO}\) Fast \(\quad \mathrm{Ni}(\mathrm{CO})_{3}+\mathrm{L} \rightarrow \mathrm{Ni}(\mathrm{CO})_{3} \mathrm{L}\) (a) What is the molecularity of each of the elementary reactions? (b) Doubling the concentration of \(\mathrm{Ni}(\mathrm{CO})_{4}\) increased the reaction rate by a factor of \(2 .\) Doubling the concentration of L. had no effect on the reaction rate. Based on this information, write the rate equation for the reaction. Does this agree with the mechanism described? (c) The experimental rate constant for the reaction, when \(\mathrm{L}=\mathrm{P}\left(\mathrm{C}_{6} \mathrm{H}_{5}\right)_{3},\) is \(9.3 \times 10^{-3} \mathrm{s}^{-1}\) at \(20^{\circ} \mathrm{C}\) If the initial concentration of \(\mathrm{Ni}(\mathrm{CO})_{4}\) is \(0.025 \mathrm{M}\) what is the concentration of the product after 5.0 minutes?
See all solutions

Recommended explanations on Chemistry Textbooks

Kinetics

Read Explanation

Nuclear Chemistry

Read Explanation

Inorganic Chemistry

Read Explanation

Making Measurements

Read Explanation

Ionic and Molecular Compounds

Read Explanation

The Earths Atmosphere

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.