Reaction Half -life
Note:
[A]o = CAo , [A] = CA
1- Zero Order Reaction:
The t1/2 formula for a zero order reaction suggests the half-life depends on the amount of initial concentration and rate constant.
2- First Order Reaction:
The formula for t1/2 shows that for first order reactions, the half-life depends solely on the reaction rate constant, k. We can visually see this on the graph for first order reactions when we note that the amount of time between one half life and the next are the same. Another way to see it is that the half life of a first order reaction is independent of its initial concentration.
3- Second Order Reaction:
The formula for t1/2 shows that for second order reactions, the half-life only depends on the initial concentration and the rate constant.
4 - nth Order Reaction:
Fractional Life Method tFThe half-life method can be extended to any fractionallife method in which the concentration of reactant drops to any fractionalvalue F = CA/CAoin time tF. The derivation is a direct extension of the half-lifemethod giving:
Thus, a plot of log tF versus log CAo, as shown in Figure bellow, will give the reactionorder.
Summary:
Problem: Calculate half-life for first-order reaction if 68% of a substance is reacted within 66 s.
Solution:
1) 68% reacted means 32% remains:
ln A = -kt + lnAo
ln 0.32 = - k (66 s) + ln 1
k = 0.0172642 s-1
2) for the half-life:
ln 0.5 = - (0.0172642 s-1) (t) + ln 1
t = 40. s
or: t1/2 = (ln 2) / k
t1/2 = (ln 2) / 0.0172642 s-1t1/2 = 40. S
Problem: The decomposition of hydrogen peroxide is a first-order reaction. The half-life of the reaction is 17.0 minutes.
(a) What is the rate constant of the reaction?
(b) If you had a bottle of H2O2, how long would it take for 86% todecompose?
(c) If you started the reaction with [H2O2] = 0.1 M, what would be the hydrogen peroxide concentration after 15.0 minutes?
Solution:
Part (a)
k = (ln 2) / t1/2
k = (ln 2) / 17.0 min = 0.04077 min-1
Part (b)
86% decomposed means 14% remains.
ln A = -kt + lnAo
ln 0.14 = - (0.04077 min-1) (t) + ln 1
t = 48.2 min
Part (c)
0.1 M is Ao
ln A = - (0.04077 min-1) (15.0 min) + ln 0.1
ln A = -2.914135
A = 0.0542 M
Comment: one half-life is 17.0 min, so the [H2O2] would be 0.05 M at the end of 17.0 min. Note that part c involves a time frame slightly less than one half-life, so the ending concentration is slightly more than 0.05 M.
Q:
Liquid A decomposes by second-order kinetics, and in a batch reactor 50%of A is converted in a 5-minute run. How much longer would it take to reach 75% conversion?
Solution:
Problem: The initial reactant concentration in a first-order reaction was 7.30 x 10-2 M and 8.70 x 10-3 M after 20 s . What is the rate constant for this reaction?
Solution:
ln A = -kt + lnAo
ln 8.70 x 10-3 = - (k) (20. s) + ln 7.30 x 10-2
-4.74443 = - (k) (20. s) + (-2.61730)
2.12713 = (k) (20. s)
k = 0.106 s-1
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