Half life equation for first order reactions t1/2 = 0.693/k where t1/2 is the ha
ID: 820504 • Letter: H
Question
Half life equation for first order reactions t1/2 = 0.693/k where t1/2 is the half life in second and k is the rate constant in inverse seconds
A what is the half life of a first order reaction with a rate constant of 5.10 * 10^-4/s
B what is the rate constant of a first order reaction that takes 6.10 minutes for the reactant concentration to drop to half of its initial value
C a certain first order reaction has a rate constant of 5.10*10^-3/s how long will it take for the reactant concentration to drop to 1/8 of its initial value
Explanation / Answer
Solution
A) Half-life (t1/2) = 0.693 / k
t1/2 = 0.693/5.10 x 10-4 s-1
= 0.13588 x 104 s
Half-life (t1/2) = 1358.8 seconds
B) For first order reaction
ln (N/N0) = -kt
Initial amount (N0) = N0
N= N0/2
t= 6.10 minutes
ln (N0/2 x N0) = -6.10 minutes x k
k = -ln (0.5)/ 6.10 minutes-1 = 0.6931/6.10 minutes-1
k = 0.1136 minutes-1
C)
k = 5.10 x 10-3 s-1
(N/N0) =1/8
ln (N/N0) = -kt
ln (1/8) = - 5.10 x 10-3 s-1 x t
t = ln (0.125)/- 5.10 x 10-3 s
= -2.079/- 5.10 x 10-3 s
= 0.4077 x 103 s
t = 407.7 s
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