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The half life of a first order reaction can be solved using the expression:
`t(1/2) = (ln(2))/(k)`
since we are given the value of the t1/2, we can solve the value of k.
`k = (ln(2))/(t 1/2)`
`k = (ln(2))/(667seconds)`
`k = 0.0010391 sec^(-1)`
From this point, we can get the ratio of the the initial amount and the final amount using the expression:
`[A] = [A]o e^(-kt)`
`([A])/([A]o) = e^(-kt)`
`([A])/([A]o) = e^(-0.0010391*400)`
`([A])/([A]o) = e^(-0.41564)`
`([A])/([A]o) = 0.6599177`
0.6599177 is the ratio of the initial amount [A]o and the present amount [A]. So to get the percentage of the initial amount that remains unreacted, we just have to multiply it by 100.
%unreacted = 65.99177 % = 66%
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