How long will it take for the concentration of A to decrease from 0.500 M to 0.200 M in the first-order reaction A → B? (k = 0.800 s⁻¹)

Answer 1

rate=-0.800*[A], the rate law for a first order reaction

d[A]/dt=-0.800*[A]

After integrating this equation, we get

ln(A2/A1)=-0.800*(t-0)

A2 is final concentration or 0.200 M, A1 is initial concentration or 0.500 M

ln(0.200/0.500)=-0.800*t

1.15 s=t

Answer 2

Integrated rate law for first order reaction is [A] = [A]_oe^-kt . This is also ln (A_t/A_o) = -kt

ln (0.2/0.5) = -0.800t

-0.916 = -0.800t

t = 1.15 sec

Or done another way:

t_1/2 = 0.693/k = 0.693/0.800 = 0.866 sec

fraction remaining = 0.5ⁿ where n = number of half lives elapsed

fraction remaining = 0.2/0.5 = 0.4

0.4 = 0.5ⁿ

log 0.4 = n log 0.5

n = 1.32 half lives x 0.866 sec/half life = 1.15 sec