Question

Suppose the half-life is 55.0 s for a first order reaction and the reactant concentration is 0.0761 M 40.1 s after the reaction starts. How many seconds after the start of the reaction does it take for the reactant concentration to decrease to 0.0183 M?

Answer 1

half life = 55 s

1st order; implies

Half life = ln(2)/k

C0 = 0.0761 M

t = 40.1 s

time required for.... C = 0.0183

a) easiest way is to model it as 1st order

First = ln(C) vs. t

For first order

dC/dt = k*C^1

dC/dt = k*C

When developed:

dC/C = k*dt

ln(C) = ln(C0) - kt

we need k

so

HL = ln(2)/k

k = ln(2)/HL = ln(2)/(55) = 0.012602 1/s

then

ln(C) = ln(C0) - kt

ln(0.0183 ) = ln(0.0761 ) - 0.012602 *t

solve for t

t = (ln(0.0183 )) -  ln(0.0761 ) )/(-0.012602 )

t = -1.4251/-0.012602 = 113.08 s

now, this is the time from 0.0761 (t = 40.1s) to 0.0183 (t= x+113.08 )

then

from start :

t = 113.08 +40.1 = 153.18 seconds from t = 0; 113.08 s from t = 40.1 s