Concise Physical Chemistry

c10 JWBS043-Rogers September 13, 2010 11:26 Printer Name: Yet to Come

160 CHEMICAL KINETICS

PROBLEMS AND EXAMPLES

Example 10.1 The Enthalpy of Activation

If the half-time of a first-order reaction is 20 minutes at 298 K and 4 minutes at 313 K,

what is the activation enthalpy?

Solution 10.1 The half-times of the reaction at the two temperatures are inversely

related to the rate constantsk= 0. 693 /t 1 / 2 ; therefore the rate constants are in the

ratio of 5:1 with temperature ratioT 2 /T 1 = 313 /298. These values are then loaded

into the rate equation, which has been integrated between the appropriate limits:

ln

k 2

k 1

=−

aH

R

(

1

T 2

−

1

T 1

)

The ratio of the rate constants is 5.0 and the difference of the reciprocal temperatures

is− 1. 60 × 10 −^4 , thereforeaH= 83 .2kJmol−^1 :

ln 5=−

aH

R

(

1

313

−

1

298

)

1. 609 =−

aH

R

( 0. 00319 − 0. 00336 )=

aH

R

( 0. 0001608 )

aH

R

=

1. 609

0. 0001608

= 1. 00 × 104

aH= 1. 00 × 104 R= 8. 314 × 104 Jmol−^1 = 83 .1kJmol−^1

Example 10.2

The relative intensityIof fluorescence from electronically excited iodine drops off

according to the time sequence below, where the time is in milliseconds.

Time, ms 0 50 100 150 200 250 300

I, relative 1 0.61 0.42 0.30 0.21 0.08 0.02

What is the rate constant and half-time for this radiative decay? Give units. Plot

relative intensity as a function of time in milliseconds (Fig. 10.7).

Solution 10.2 The sequence, when sketched out, may be a first-order decay. If it

is, drawing a horizontal fromI= 0 .5 (relative) cuts the curve at aboutt=70 ms, a

first estimate of the half-time. Let’s try to convert to lnIvs. time to see if the plot is

linear.

The reaction is first order, as demonstrated by the linear plot of lnIvs. time in

Fig. 10.8 (for more detail see Houston, 2001).