Question

The first-order rate constant for the decomposition of a certain antibiotic in water at 20^o C is 1.65 yr-1 .

(a) If a 6.0 x 10-3 M solution of the antibiotic is stored at 20^o C, what will its concentration be after 3 months?

(b) After 1 year?

(c) How long will it take for the concentration of the solution to drop to 1.0 x 10-3 M?

(d) What is the half-life of the antibiotic solution?

Answer 1

a)

assume:

ln(A) = ln(A0) - kt

k = 1.65 1/y

so

a)

ln(A) = ln(I6*10^-3) - 1.65*t

time = 3 months, but we need it in years, so 3/12 = 0.25

so

ln(A) = ln(6*10^-3) - 1.65*0.25

A = exp(-5.528) = 0.003973 M

b)

by definition, 1 year:

ln(A) = ln(6*10^-3) - 1.65*1

A = exp(ln(6*10^-3) - 1.65*1) = 0.0011522 M

c)

A = 1*10^-3 so

ln(10^-3) = ln(6*10^-3) - 1.65*t

t = (ln(10^-3) -ln(6*10^-3) ) /(1.65) = 1.08591 years

d)

for half life:

t 1/2 = ln(2) / k

so

t 1/2 = ln(2) / (1.65) = 0.4200 years is half life