# A certain first-order rxn A->products has a half life of 677 seconds. What % of an initial amount of the reactant will remain unreacted after 400 sec?

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### 1 Answer

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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