   Chapter 12, Problem 3ALQ

Chapter
Section
Textbook Problem

# Make a graph of [A] versus time for zero-, first-, and second-order reactions.  From these graphs, compare successive half-lives.

Interpretation Introduction

Interpretation: The [A] versus time graph for a zero-, first- and a second order reaction is to be stated and value of half-life for each is to be calculated and compared.

Concept introduction: The order of a reaction is determined by adding the powers of the concentration of the reactants in the rate equation. Order of a reaction can be a whole number like 0,1,2 or 3 or can even be a fractional value.

Explanation

To determine: The concentration versus time graph for a zero order reaction and the expression for half life for the zero order reaction.

Solution: The half life for a zero order reaction is, t12=[A]ο2k

The rate constant (k) value is given by the expression,

k=xt

Where,

• x is the concentration.
• t is time.

Simplify the above expression.

x=ktx=kt+0

This is similar to the equation for a straight line, that is, y=mx+c .

Where,

• y is the y-intercept.
• x is the x-intercept.
• m is the slope.
• c is a constant.

The [A] versus time graph for a zero order reaction,

Figure 1

The half life for a zero order reaction is, t12=[A]ο2k

Where,

• [A]ο is the initial concentration of the reactant A .

To determine: The concentration versus time graph for a first order reaction and the expression for half life for the second order reaction.

Solution: The half life for a first order reaction is, t12=0.693k

The rate law equation for a first order reaction is,

k=1tln[A][A]ο

Where,

• [A]ο is the initial concentration of the reactant A .
• [A] is the final concentration of the reactant A .
• t is the time.

Simplify the above expression.

k=2.303tlog[A][A]οkt2.303=log[A]log[A]οlog[A]=kt2.303+log[A]ο

This is similar to the equation for a straight line, that is y=mx+C .

Where,

• y is the y-intercept.
• x is the x-intercept.
• m is the slope

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