   Chapter 12, Problem 4RQ

Chapter
Section
Textbook Problem

# The initial rate for a reaction is equal to the slope of the tangent line at t ≈ 0 in a plot of [A] versus time. From calculus, initial rate = − d [ A ] d t . Therefore. the differential rate law for a reaction is Rate = − d [ A ] d t = k [ A ] n .Assuming you have some calculus in your background, derive the zero-, first-, and second-order integrated rate laws using the differential rate law.

Interpretation Introduction

Interpretation: It is stated that the initial; rate for a reaction is equal to the slope of the tangent line at t0 in a plot of [A] vs time. The zero-, first- and the second order integrated rate laws are to be determined.

Concept introduction: The dependence of the reaction rate on the concentration of the reactants is represented by a differential rate law, whereas the dependence on time is represented by an integrated rate law.

To determine: The zero-, first- and the second order integrated rate laws.

Explanation

Given

The differential rate law for a reaction is,

Rate=d[A]dt=k[A]n

Where,

• [A] is the molar concentrations of the reactant ‘A’.
• k’ is the rate constant.
• n is the order of the reaction.

For a zero order reaction,

Rate=d[A]dt=k[A]0

Simplify the above expression.

d[A]=kdtd[A]=kdt

Integrate the above expression.

[A]=kt+C

At t=0, [A]=[A]ο

The above expression simplifies to,

[A]=kt+[A]ο

This is the integrated rate law for a zero order reaction.

To determine: The first order integrated rate law.

Given

The differential rate law for a reaction is,

Rate=d[A]dt=k[A]n

Where,

• [A] is the molar concentrations of the reactant ‘A’.
• k’ is the rate constant.
• n is the order of the reaction.

For a zero order reaction,

Rate=d[A]dt=k[A]1

Simplify the above expression.

d[A][A]=kdtd[A][A]=kdt

Integrate the above expression

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