This reaction was monitored as a function of time: A → B + C A plot of ln[A] versus time yields a straight line with slope -0.0045/s. c. What is the half-life?

Verified step by step guidance

Identify that the reaction is first-order because a plot of \( \ln[A] \) versus time yields a straight line.

Recall the formula for the half-life of a first-order reaction: \( t_{1/2} = \frac{0.693}{k} \), where \( k \) is the rate constant.

Recognize that the slope of the line in a plot of \( \ln[A] \) versus time for a first-order reaction is equal to \(-k\).

Given that the slope is \(-0.0045/s\), determine that \( k = 0.0045/s \).

Substitute \( k = 0.0045/s \) into the half-life formula \( t_{1/2} = \frac{0.693}{0.0045/s} \) to find the half-life.

Verified video answer for a similar problem:

This video solution was recommended by our tutors as helpful for the problem above.

Video duration:

Was this helpful?

Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

First-Order Reactions

In chemistry, a first-order reaction is one where the rate of reaction is directly proportional to the concentration of one reactant. This means that as the concentration of the reactant decreases, the rate of the reaction also decreases. The relationship can be expressed mathematically, and for first-order reactions, the natural logarithm of the concentration of the reactant versus time yields a straight line.

Half-Life

The half-life of a reaction is the time required for the concentration of a reactant to decrease to half of its initial value. For first-order reactions, the half-life is constant and does not depend on the initial concentration. It can be calculated using the formula t1/2 = 0.693/k, where k is the rate constant derived from the slope of the ln[A] versus time plot.

Rate Constant (k)

The rate constant (k) is a proportionality factor in the rate equation that relates the rate of a reaction to the concentration of reactants. For first-order reactions, k can be determined from the slope of the ln[A] versus time plot, which is negative. In this case, the slope is -0.0045/s, indicating the rate at which the concentration of A decreases over time.

Recommended video:

Guided course

01:14

Equilibrium Constant K

Related Practice

Textbook Question

The reaction A¡products was monitored as a function of time. The results are shown here. Time (s) [A] (M) 0 1.000 25 0.914 50 0.829 75 0.744 100 0.659 125 0.573 150 0.488 175 0.403 200 0.318 Determine the order of the reaction and the value of the rate constant. What is the rate of reaction when [A] = 0.10 M?

Textbook Question

This reaction was monitored as a function of time: A → B + C A plot of ln[A] versus time yields a straight line with slope -0.0045/s. a. What is the value of the rate constant (k) for this reaction at this temperature?

Textbook Question

This reaction was monitored as a function of time: A → B + C A plot of ln[A] versus time yields a straight line with slope -0.0045/s. b. Write the rate law for the reaction.

Textbook Question

This reaction was monitored as a function of time: A → B + C A plot of ln[A] versus time yields a straight line with slope -0.0045/s. d. If the initial concentration of A is 0.250 M, what is the concentration after 225 s?

Textbook Question

This reaction was monitored as a function of time: AB → A + B A plot of 1/[AB] versus time yields a straight line with a slope of +0.55/Ms.

a. What is the value of the rate constant (k) for this reaction at this temperature?

Textbook Question

This reaction was monitored as a function of time: AB → A + B A plot of 1/[AB] versus time yields a straight line with a slope of +0.55/Ms. b. Write the rate law for the reaction.