Page 156 - [B._MURPHY,_C._MURPHY,_B._HATHAWAY]_A_working_meth

140 Chapter 9

Examples

Exumple No. 1: Determine the rate constant at 42 "C for the
hydrolysis of a sugar, S, given that the activation energy of the
reaction is 132 kJ mol-I and the rate constant at 53 "C is 1.12 x
dm3 mol-' s-' and R = 8.314 J K-' mol-'.

1. Identify the data in the question:
Ti = 42°C
T2 = 53°C
k2 = 1.12 x 10-3dm3mol-1 s-'
Eact = 132 kJ mol-'
R = 8.314JK-'mol-'
2. Convert all units to SI:
Ti = (42 + 273)K = 315K
T2 = (53 + 273)K = 326K
k2 = 1.12 x 10-3dm3mol-' s-l
Eact = 132kJmol-' = 132000Jmol-'
R = 8.314JK-'mol-'
3. Identify the unknown@) in the question to be determined: the
rate constant kl at temperature TI.
4. What expression is involved?
In (k 1 /k2) = [(Eact/R)( TI - T2)1/( TI T2)
5. Rearrange the equation for kl before substituting the numerical
values:
In k~ - In k2 = [(Eact/MTI - ~'z)I/(TITz)
In kl = [(Eact/R)(Tl - T2)11(TlT2) + In k2
6. Substitute the values into the equation, taking care of signs and
brackets:
In kl = (132000/8.314)(315-326)/[(315)(326)]+ln (1.12 x loF3)
= -8.4951295
kl = -8.4951295 = 2.045 10-4
7. What are the units of kl? kl = 2.045 x dm3 mol-' s-'
Answer: kl = 2.045 x dm3 mol-'s-'