Question

Half life of first order reaction is 3 hrs. what percentage reaction is completed in 9hrs ?​

Answer 1

92.6%

Step-by-step explanation:

Hello there!

In this case, according to the first-order kinetics, we can write:

`(A)/(Ao)=exp(-kt)`

Whereas we need to know the rate constant in order to compute the percentages, given the half-life:

`k=(ln(2))/(9hrs)=0.077hrs^(-1)`

In such a way, we can now compute the fraction in which the reaction has been completed after 3 hrs:

`(A)/(Ao)=exp(-0.077hrs^(-1)*3hrs)\\\\(A)/(Ao)=0.926`

And the percent would be 92.6% after multiplying the fraction by 100 %.

Best regards!