Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 12: Problem 89

Consider two reaction vessels, one containing \(\mathrm{A}\) and the other containing \(\mathrm{B}\) , with equal concentrations at \(t=0 .\) If both substances decompose by first-order kinetics, where $$ \begin{array}{l}{k_{\mathrm{A}}=4.50 \times 10^{-4} \mathrm{s}^{-1}} \\\ {k_{\mathrm{B}}=3.70 \times 10^{-3} \mathrm{s}^{-1}}\end{array} $$ how much time must pass to reach a condition such that \([\mathrm{A}]=\) 4.00\([\mathrm{B}]\) ?

Short Answer

Expert verified
To find the time when the concentration of A is 4 times the concentration of B, we can use the first-order reaction equation and the given rate constants. The equation is \(e^{-k_At} = 4 \times e^{-k_Bt}\). Solving for \(t\) gives \(t = \frac{\ln{4}}{k_B - k_A}\). Substituting the given values of \(k_A\) and \(k_B\) and calculating, we find that it takes approximately \(538.8\) seconds for the condition to be met.

Step by step solution

01

Identify the formula for the first-order reaction

The general equation to describe first-order reactions is given by: $$ [A]_t = [A]_0 \times e^{-k_At} $$ Similarly, for substance B, it would be: $$ [B]_t = [B]_0 \times e^{-k_Bt} $$ Where: - \([A]_t\) and \([B]_t\) are the concentrations of substances A and B, respectively, at a given time \(t\), - \([A]_0\) and \([B]_0\) are the initial concentrations at time \(t=0\), - \(k_A\) and \(k_B\) are the rate constants of substances A and B, respectively, - \(t\) is the time elapsed.
02

Express the given condition \([A] = 4[B]\) using the formulas

To find the time when the condition \([A] = 4[B]\) is met, we can substitute the concentration expressions from the previous step into this equation. It would look like this: $$ [A]_0 \times e^{-k_At} = 4 \times ([B]_0 \times e^{-k_Bt}) $$Since \([A]_0 = [B]_0\) (given that both substances have equal concentrations initially), we can simplify the equation to: $$ e^{-k_At} = 4 \times e^{-k_Bt} $$
03

Solve for time t

In order to solve for time \(t\), we can first take the natural logarithm of both sides to get rid of the exponentials: $$ \ln{e^{-k_At}} = \ln{(4 \times e^{-k_Bt})} $$This simplifies to: $$ -k_At = \ln{4} - k_Bt $$Now we can solve for \(t\): $$ t = \frac{\ln{4}}{k_B - k_A} $$Plug in the given values of \(k_A\) and \(k_B\): $$ t = \frac{\ln{4}}{3.70 \times 10^{-3} s^{-1} - 4.50 \times 10^{-4} s^{-1}} $$
04

Calculate the time

Finally, evaluate the expression to find the time: $$ t = \frac{\ln{4}}{3.25 \times 10^{-3} s^{-1}} \approx 538.8 \,\text{seconds} $$ So, it takes approximately 538.8 seconds for the concentration of A to be 4 times the concentration of B.

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.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

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

A certain substance, initially at 0.10\(M\) in solution, decomposes by second- order kinetics. If the rate constant for this process is 0.40 $\mathrm{L} / \mathrm{mol} \cdot \min$ , how much time is required for the concentration to reach 0.020 \(\mathrm{M}\) ?

In the Haber process for the production of ammonia, $$ \mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \longrightarrow 2 \mathrm{NH}_{3}(g) $$ what is the relationship between the rate of production of ammonia and the rate of consumption of hydrogen?

Consider the hypothetical reaction $$ \mathrm{A}+\mathrm{B}+2 \mathrm{C} \longrightarrow 2 \mathrm{D}+3 \mathrm{E} $$ where the rate law is $$ \text {Rate} =-\frac{\Delta[\mathrm{A}]}{\Delta t}=k[\mathrm{A}][\mathrm{B}]^{2} $$ An experiment is carried out where \([\mathrm{A}]_{0}=1.0 \times 10^{-2} M\) \([\mathrm{B}]_{0}=3.0 M,\) and \([\mathrm{C}]_{0}=2.0 M .\) The reaction is started, and after 8.0 seconds, the concentration of \(\mathrm{A}\) is $3.8 \times 10^{-3} \mathrm{M}$ a. Calculate the value of k for this reaction. b. Calculate the half-life for this experiment. c. Calculate the concentration of A after 13.0 seconds. d. Calculate the concentration of C after 13.0 seconds.

Individuals who suffer from lactose intolerance do not produce enough of the enzyme lactase. Explain why this is an issue.

Draw a rough sketch of the energy profile for each of the following cases: a. $\Delta E=+10 \mathrm{kJ} / \mathrm{mol}, E_{\mathrm{a}}=25 \mathrm{kJ} / \mathrm{mol}$ b. $\Delta E=-10 \mathrm{kJ} / \mathrm{mol}, E_{\mathrm{a}}=50 \mathrm{kJ} / \mathrm{mol}$ c. $\Delta E=-50 \mathrm{kJ} / \mathrm{mol}, E_{\mathrm{a}}=50 \mathrm{kJ} / \mathrm{mol}$
See all solutions

Recommended explanations on Chemistry Textbooks

Inorganic Chemistry

Read Explanation

The Earths Atmosphere

Read Explanation

Chemical Reactions

Read Explanation

Kinetics

Read Explanation

Organic Chemistry

Read Explanation

Chemical Analysis

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.

Sign-up for free