In: Chemistry
It takes 41.0 min for the concentration of a reactant in a
first-order reaction to drop from 0.430 M
to 0.310 M at 25.0 oC. How long will it take
for the concentration to reach 0.126 Molar?
153 min |
||
60.2 min |
||
87.2 min |
||
0.368 min |
||
71.3 min |
||
50.4 min |
||
5.20 x 102 min |
||
16.9 min |
Consider a given reaction, A Products
For above first order reaction, we have k = [2.303 / t ] log [A] 0 / [A] t
Where, t is a time of reaction, k is a rate constant, [A] 0 is a initial concentration of reactant at time t=0 s, [A] t is a concentration of reactant at time t.
Here t =41.0 min, [A] 0 = 0.430 M and [A] t = 0.310 M .
k = ( 2.303 / 41.0 min ) log 0.430 / 0.310
k = 0.05617 min -1 log 0.430/0.310
k = 0.05617 min -1 ( 0.1421 )
k = 0.00798 min -1
Calculation of time required to reach the concentration to 0.126 Molar
We have, k = 0.00798 min -1, [A] 0 = 0.430 M and [A] t = 0.126 M .
we have k = [2.303 / t ] log [A] 0 / [A] t
0.00798 min -1 = ( 2.303 / t ) log 0.430 / 0.126
0.00798 min -1 = ( 2.303 / t ) 0.533
0.00798 min -1 = 1.228 / t
t = 1.228 / 0.00798 min -1
t = 153.8 min
ANSWER : 153 min