Find the activation energy and the pre-exponential factor for the reaction:
2HI(g) --> H2(g) + I2(g)
| T/K | 556 | 629 | 700 | 781 |
| k/dm3mol-1s-1 | 7.04*10-7 | 6.04*10-5 | 2.32*10-3 | 7.90*10-2 |
| 103*1/T | 1.80 | 1.59 | 1.43 | 1.28 |
| log k | -6.15 | -4.21 | -2.63 | -1.10 |
Answer:
The plot of log k against 1/T gives a straight line with the
gradient = -5.40*104K
gradient = -5.40 K =-E/2.3R
activation energy E = 195kJmol-1
Using any point on the graph when logk =...... 1/T = .....
Substitute into logA =logk +E/2.3RT to show:
A = 7.9*1011dm3mol-1s-1