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Sagot :
The half life for this reaction when [A]o = 0.757 M is 4.033 s
integrated rate law for 2nd order reaction:
1/[A]o = 1/[A] - k*t
so, for 2nd order reaction, 1/[A] vs t will be straight line
integrated rate law for 1st order reaction:
ln [A] = –kt + ln [A]o
So, for 1st order reaction, ln[A] vs t will be straight line
integrated rate law for zero order reaction:
[A] = [A]o – k*t
So, for zero order reaction, [A] vs t will be straight line
Here,
1/[A] vs t is straight line
so, order of A is 2
use integrated rate law for 2nd order reaction:
1/[A] = 1/[A]o + k*t
1/(0.107) = 1/(0.757) + k*24.5
9.346 = 1.321 +k*24.5
k*24.5 = 8.025
k = 0.328 M-1.s-1
Given:
rate constant, k = 0.328 M-1.s-1
use relation between rate constant and half life of 2nd order reaction
t1/2 = 1/([A]o*k)
= 1/(0.757*0.328)
= 4.033 s
The half-life of a chemical reaction can be defined as the time it takes for the concentration of a given reactant to reach 50% of its initial concentration (that is, the time it takes for the concentration of a reactant to reach half its initial value). This is represented by the symbol "t1/2" and is usually expressed in seconds.
Learn more about half life of a chemical reaction:
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