Lecture 29 : Temperature Dependence of Reaction Rates
Example 1
The rate constant for the aqueous transformation of thiourea to ammonium thiocyanate is measured at 90oC, 120oC and 130oC. Obtain the Arrhenius parameters for this reaction.
Solution :
The values of k and temperature are shown in the first two rows of Table 29.1. The first step is converting oC into Kelvins. Since ln k = ln A -
Ea / RT ( ln = loge ), lnk plotted against 1 / T should yield a straight line with slope =
- E a / R and intercept = ln A. If there are departures from Arrhenius like behaviour, the graph will depart from a straight line. Plot the graph and verify that you get a straight line. Assuming that you have verified that, let us get values of Ea and A in a simpler way.
Table 29.1 Temperature dependence of k for the reaction thiourea
ammonium thiocyamate.
Temp / oC
k / s –1
ln ( k / s –1 )
Temp / K
{ 1/ T} K –1
103 K / T
90
4.53 * 10 - 8
-16.910
363.15
0.00275
2.754
110
59.3 * 10 - 8
- 14.338
383.15
0.00261
2.610
120
197 * 10 -8
-13.137
393.15
0.00254
2.544
130
613 * 10 -8
-12.002
403.15
0.00248
2.480
Let k1 correspond to T 1 = 363.15 K and k2 correspond to T2 = 403.15 K.
ln k2 - ln k1 = +12.002 - 16.910 = - 4.902
(29.12)
= (- Ea / R) ( 1 / T1 - 1 / T2) = -( Ea/ R) (0.002754 - 0.002480) = -0.000274 Ea / R
Substituting the values of Ea in Arrhenius equation, we get the value of A
ln A = ln k + Ea / RT = - 13.137 + 148 / (8.314 * 393.15)
= 8.8 * 1013 s -1
(29.15)
Note that when ln of k was taken, k was taken dimensionless (as we are interested in the difference between ln k values). Ea and RT have to be in the same units in the calculations.
ammonium thiocyamate. 
