What is the rate constant of a first-order reaction when 20.0% of a reactant remains after 35.5 s?

Answer 1

For a first order reaction we know that t_1/2 = 0.693/k

So, from the information provided, we can find the half life (t_1/2) and then plug it in to that equation.

Fraction remaining = 0.5n where n = # of half lives that have elapsed.

0.20 = 0.5ⁿ

log 0.20 = n log 0.5

-0.699 = -0.301 n

n = 2.32 half lives

2.32 half lives = 35.5 sec

1 half life = 15.3 sec

15.3 sec = 0.693/k

k = 0.0453 sec^-1