/*! 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 51 A certain first-order reaction i... [FREE SOLUTION] | 91影视

91影视

A certain first-order reaction is 45.0\(\%\) complete in 65 s. What are the values of the rate constant and the half-life for this process?

Short Answer

Expert verified
The rate constant (k) for this first-order reaction is approximately 0.00899 s鈦宦, and its half-life (t_half) is approximately 77.1 seconds.

Step by step solution

01

Write down the given information

We are given that the reaction is 45.0% complete in 65 seconds. Let's represent the initial concentration of the reactant as [A]_0 and the remaining concentration after 65 seconds as [A]. Since the reaction is 45.0% complete, 55.0% of the reactant is remaining. So, we have the following relationship between [A]_0 and [A]: \[ A = 0.55 * A_0 \] Additionally, we know that the time taken for this change is 65 seconds.
02

Use the first-order reaction formula

For a first-order reaction, the rate law is given by: \[ \ln \frac{[A]}{[A]_0} = -kt \] Where: k is the rate constant, t is the time taken (in seconds), [A] is the remaining concentration of the reactant, [A]_0 is the initial concentration of the reactant. We'll now plug in our known values into the formula to find the rate constant k: \[ \ln \frac{0.55 * [A]_0}{[A]_0} = -k * 65 \]
03

Calculate the rate constant (k)

Now, let's simplify our equation and solve for the rate constant k: \[ \ln (0.55) = -k * 65 \] \[ k = -\frac{\ln (0.55)}{65} \] Using a calculator, we find: \[ k = 0.00899 s^{-1} \] Hence, the rate constant (k) for this reaction is approximately 0.00899 s鈦宦.
04

Calculate the half-life (t_half) using the rate constant

For a first-order reaction, the half-life formula is given by: \[ t_{1/2} = \frac{\ln 2}{k} \] Now, we can plug in the value of the rate constant (k) that we calculated in step 3 to find the half-life (t_half): \[ t_{1/2} = \frac{\ln 2}{0.00899} \] Using a calculator, we find: \[ t_{1/2} 鈮 77.1 s \] Hence, the half-life (t_half) for this reaction is approximately 77.1 seconds. In conclusion, the rate constant (k) for this first-order reaction is approximately 0.00899 s鈦宦, and its half-life (t_half) is approximately 77.1 seconds.

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 Constant
The rate constant, often symbolized as \( k \), is a crucial part of understanding the speed of a chemical reaction. In simple terms, it defines how fast the reactants transform into products. For first-order reactions, the rate of the reaction is directly proportional to the concentration of one reactant. This means if the concentration doubles, the rate of reaction also doubles.

To find the rate constant in a first-order reaction such as the one given in the exercise, we use the formula\[ \ln \frac{[A]}{[A]_0} = -kt \]where:
  • \([A]\) is the concentration of the reactant at time \( t \),
  • \([A]_0\) is the initial concentration,
  • \( k \) is the rate constant,
  • \( t \) is time.
By using the given values, you can solve for \( k \) and determine its value, indicating how quickly the reaction proceeds.

This provides invaluable insight into the reaction's behavior under various conditions.
Half-Life
The half-life, emblematic of first-order reactions, is the time needed for half of the reactants to convert into products. Understanding half-life not only helps in predicting how long a reaction will take to reach completion but also in assessing its speed and practicality in chemical processes.

A unique feature of first-order reactions is that the half-life remains constant regardless of the initial concentration. The mathematical formula to calculate half-life \( t_{1/2} \) for a first-order reaction is:\[ t_{1/2} = \frac{\ln 2}{k} \] where:
  • \( \ln 2 \approx 0.693 \),
  • \( k \) is the rate constant you previously calculated.
This property shows that each equal interval of time results in the reduction of half of the existing reactant concentration, offering a simple way to predict how long a reaction will persist.
Reaction Kinetics
Reaction kinetics is the study of rates at which chemical processes occur and the factors that affect these rates. For a first-order reaction, it's essential to comprehend how the concentration of reactants and the time influence the reaction speed.

In a first-order kinetics scenario, the rate at which a reaction proceeds is directly dependent on the concentration of a single reactant, which offers a straightforward approach to predicting outcome behaviors. Kinetics involves understanding and using equations that relate time, reactant concentration, and the rate constant. Experimentally, it provides chemists with tools to control rates of reactions, optimize conditions, and develop safer and more efficient processes.

By mastering the fundamentals of reaction kinetics, students can discern how various elements like temperature, pressure, and presence of catalysts modify how chemical reactions unfold. This leads to deeper insights into the chemistry and practical applications in fields such as pharmacology, environmental science, and material engineering.

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

At \(40^{\circ} \mathrm{C}, \mathrm{H}_{2} \mathrm{O}_{2}(a q)\) will decompose according to the following reaction: $$ 2 \mathrm{H}_{2} \mathrm{O}_{2}(a q) \longrightarrow 2 \mathrm{H}_{2} \mathrm{O}(l)+\mathrm{O}_{2}(\mathrm{~g}) $$ The following data were collected for the concentration of \(\mathrm{H}_{2} \mathrm{O}_{2}\) at various times. $$ \begin{array}{|cc|} \hline \begin{array}{c} \text { Time } \\ (\mathbf{s}) \end{array} & \begin{array}{c} {\left[\mathrm{H}_{2} \mathrm{O}_{2}\right]} \\ (\mathrm{mol} / \mathrm{L}) \end{array} \\ \hline 0 & 1.000 \\ \hline 2.16 \times 10^{4} & 0.500 \\ \hline 4.32 \times 10^{4} & 0.250 \\ \hline \end{array} $$ a. Calculate the average rate of decomposition of \(\mathrm{H}_{2} \mathrm{O}_{2}\) between 0 and \(2.16 \times 10^{4} \mathrm{~s}\). Use this rate to calculate the average rate of production of \(\mathrm{O}_{2}(g)\) over the same time period. b. What are these rates for the time period \(2.16 \times 10^{4} \mathrm{~s}\) to \(4.32 \times 10^{4} \mathrm{~s} ?\)

Consider the following initial rate data for the decomposition of compound AB to give A and B: Determine the half-life for the decomposition reaction initially having 1.00\(M \mathrm{AB}\) present.

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\([\mathrm{B}]\) ?

Would the slope of a \(\ln (k)\) versus 1\(/ T\) plot (with temperature in kelvin) for a catalyzed reaction be more or less negative than the slope of the \(\ln (k)\) versus 1\(/ T\) plot for the uncatalyzed reaction? Explain. Assume both rate laws are first-order overall.

The type of rate law for a reaction, either the differential rate law or the integrated rate law, is usually determined by which data is easiest to collect. Explain.

See all solutions

Recommended explanations on Chemistry Textbooks

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.