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  • A first-order reaction is 40% complete after 8 min. How long will it take for the reaction to be 90% complete? ln 5/3=0.511 ln 10=2.303 &nbs

A first-order reaction is 40% complete after 8 min. How long will it take for the reaction to be 90% complete?
ln 5/3=0.511

ln 10=2.303

 

Option: 1

50.20 min


Option: 2

36.36 min


Option: 3

24.52 min


Option: 4

12.52 min


Answers (1)

best_answer

For a first order reaction we have:
\mathrm{k=\frac{1}{t} \ln \frac{a}{a-x} }
where,
k= rate constant of reaction
t= time
a= initial concentration of the reactant
x= amount of reactant reacted in time t
Since, the reactant has reacted 40 % in 8 min, the remaining reactant is 60 %
\mathrm{\Rightarrow } At t=8 min ,(a-x)=0.6 a
Time when the reaction is 90 % complete i.e. (a-x)=0.1 a

So
Case I: t=8 min; (a-x)=0.6 i a
Case II: t=? , (a-x)=0.1 a
Since the reaction is of the first order k is same in both cases so using the formula (i) for both the cases and equating k,


\mathrm{\frac{1}{8} \times \ln \left(\frac{1}{0.6}\right)=\frac{1}{t} \times \ln \left(\frac{1}{0.1}\right) \\ }
\mathrm{ \Rightarrow t=8 \times \frac{\ln (1 / 0.1)}{\ln (1 / 0.6)}=8 \times \frac{2.302}{0.511}=36.36 }.
 

Posted by

rishi.raj

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