The Foundations of Chemistry

EXAMPLE 16-11 Arrhenius Equation

The specific rate constant, k,for the following first-order reaction is 9.16 10 ^3 s^1 at 0.0°C.

The activation energy of this reaction is 88.0 kJ/mol. Determine the value of kat 2.0°C.

N 2 O 5 88nNO 2 
NO 3

Plan

First we tabulate the values, remembering to convert temperature to the Kelvin scale.

Ea88,000 J/mol R8.314 J/molK

k 1 9.16 10 ^3 s^1 at T 1 0.0°C
 273 273 K

k 2 __?atT 2 2.0°C
 273 275 K

We use these values in the “two-temperature” form of the Arrhenius equation.

Solution

ln  


T

1

1



T

1

2



ln
 


273

1

K



275

1

K

0.282

Taking inverse (natural) logarithms of both sides,

1.32

k 2 1.32(9.16 10 ^3 s^1 ) 1.21 10 ^2 s^1

We see that a very small temperature difference, only 2°C, causes an increase in the rate constant

(and hence in the reaction rate for the same concentrations) of about 32%. Such sensitivity of

rate to temperature change makes the control and measurement of temperature extremely

important in chemical reactions.

You should now work Exercise 53.

EXAMPLE 16-12 Activation Energy

The gas-phase decomposition of ethyl iodide to give ethylene and hydrogen iodide is a first-

order reaction.

C 2 H 5 I88nC 2 H 4 
HI

At 600. K, the value of kis 1.60 10 ^5 s^1. When the temperature is raised to 700. K, the

value of kincreases to 6.36 10 ^3 s^1. What is the activation energy for this reaction?

Plan

We know kat two different temperatures. We solve the two-temperature forms of the Arrhe-

nius equation for Eaand evaluate.

Solution

k 1 1.60^10 ^5 s^1 at T 1 600. K k 2 6.36^10 ^3 s^1 at T 2 700. K

R8.314 J/molK Ea_?_

k 2



9.16 10 ^3 s^1

88,000 J/mol



8.314

mo

J

lK



k 2



9.16 10 ^3 s^1

Ea



R

k 2



k 1

686 CHAPTER 16: Chemical Kinetics