Page 157 - [B._MURPHY,_C._MURPHY,_B._HATHAWAY]_A_working_meth

Chemical Kinetics 11 141
Example No. 2: A first-order reaction, A + B, takes 5.8 minutes at
32°C to complete a 15% loss of A. If the activation energy of the
reaction is 65.2 kJ mol-', determine the half-life of the reaction at
45 "C, given that R = 8.314 J K-' mol-'.

1. Identify the data in the question:

7'1 = 32°C
t 1 = 5.8 min
T2 = 45°C
Eact = 65.2 kJ mol-
R = 8.314JK-'mol-'
2. Convert all units to SI:
7'1 = (32 + 273)K = 305K
tl = 5.8 min = 348s
7'2 = (45 + 273)K = 318K
Eact = 65200 Jmol-'
R = 8.314JK-'mol-'
3. Identify the unknown, to be evaluated: the half-life, fl12 at
temperature 45 "C (7'2).
4. What expressions are involved?
(a) ln(kllk2) = (Eact/MTl - 7'2)/(7'17'2)
(b) ln[A] = -kt + In[&], for a first-order reaction.
(c) tl12 = (In 2)/k for a first-order reaction.
5. To determine t1/2, k2 must first be evaluated from equation (a).
Rearrange the equation for k2 before substituting the numerical
values:

In kl - In k2 = (&Ct/R)(7'1 - 7'2)/(T1 T2)
In k2 = - (EWt/R)(G - 7'2)/(7'1 7'2) + In kl
6. Substitute the values into the equation, taking care of signs and
brackets:
In k2 = -(65200/8.314)(305 - 318)/[(305)(318)] + In kl
= 1.051124 + Ink1 (t)
However, no explicit value is given in the question for kl. Hence,
this must be determined from another expression, before sub-
stitution into the above equation.
1" reaction: ln[A] = - kt + ln[Ao]. Rearrange this equation for
k, before substituting the numerical values:
kt = In[&] - ln[A] + k = (l/t)(ln [Ao] - In [A]) = (l/t)(ln
[&]/[A]), (since log (A/B) = log A - log B).