Page 155 - [B._MURPHY,_C._MURPHY,_B._HATHAWAY]_A_working_meth

Chemical Kinetics II 139

7. Slope of graph: (xl, y1) = (1.1 1,O); (x2, y2) = (1.90, - 15.52)
m = Ay/Ax = 02 - yl)/(xz - x1) = (-15.52 - 0)/(1.90 -
1.11) = -19.646 103~.
m = -E,t/R = -E,t/(8.314) = -18.354 x lo3 K * Eact M
163.34 kJ mol-'.
8. Intercept of the graph: y = mx + c =$ c = y, - mx,. Choosing
(xC, yc) = (1.3, - 3.76)
JC = (-3.76) - (-19.646)(1.3) = 21.78 = 1nA
+A = 2.88 x 109M-'s-'.

RATE CONSTANTS AND TEMPERATURE
The Arrhenius equation states that:
Ink = -E&RT+ 1nA
If kl is the rate constant at temperature T1, and k2 is the rate constant
at temperature T2, an expression can be derived relating these four
parameters:

In kl - In k2 = (-E,,/RT1) + In A + (E,,/RTz) - In A
= (4Ct/R)(l/T2 - 1/Tl)
But since In kl - In k2 = log A - log B = log (A/B), then:
In (W2) = (E,t/Jo(Tl - T2)/(TlT2)

I y=mx+c I

i.e. if a graph is drawn of In (kl/k2) versus (Tl - T')/(T17'2), a straight
line of positive gradient m = (EacJR) would be obtained, from which
the activation energy Eact could subsequently be determined. The
graph would pass through the origin, (O,O), as the intercept c = 0. The
major value of the above expression is that given any four of the five
variables, kl, k2, Tl, T2 or Eact, the fourth can be determined from the
equation.