The Foundations of Chemistry

(f ) We use the straight-line plot in Figure 16-8c to find the value of the rate constant for
this first-order reaction from the relationship

slopeak or k

To determine the slope of the straight line, we pick any two points, such as Pand Q,on
the line. From their coordinates, we calculate

slope0.300 min^1

k0.300 min^1

You should now work Exercises 44 and 45.

The graphical interpretations of concentration-versus-time data for some common
reaction orders are summarized in Table 16-3.

0.300 min^1



1

slope



a

(1.83)(0.27)



(8.501.50) min

change in ordinate



change in abscissa

slope



a

16-4 Concentration versus Time: The Integrated Rate Equation 675

Problem-Solving Tip:Some Warnings About the Graphical Method

for Determining Reaction Order

1.If we were dealing with real experimental data, there would be some error in each

of the data points on the plot. For this reason, we should notuse experimental data

points to determine the slope. (Random experimental errors of only 10% can intro-

duce errors of more than 100% in slopes based on only two points.) Rather we

should draw the best straight line and then use points on that line to find its slope.

Errors are further minimized by choosing points that are widely separated.

2.Remember that the ordinate is the vertical axis and the abscissa is the horizontal

one. If you are not careful to keep the points in the same order in the numerator

and denominator, you will get the wrong sign for the slope.

TABLE 16-3 Graphical Interpretations for Various Orders of the Reaction

aAnProducts

Order

Zero First Second

Plot that gives straight line [A] vs. t ln [A] vs. t vs. t

Direction of straight-line slope down with time down with time up with time

Interpretation of slope ak ak ak

Interpretation of intercept [A] 0 ln [A] 0

1



[A] 0

1



[A]