Reacting with water in an acidic solution at a particular temperature, compound A decomposes into compounds B and C according to the law of uninhibited decay. An initial amount of 0.60 M of compound A decomposes to 0.56 M in 30 minutes. How much of compound A will remain after 3 hours? How long will it take until 0.10 M of compound A remains?

after t = 3 hours compound A remain 0.396 M

t = 12.98 hours

Explanation:

P = P_0*e^ (-kt)

t = 30 minutes = 30/60 hours = 0.5 hours

and P_0 = 0.60M

and P after t = 0.5 hours is 0.56 M

(0.56/0.6) = e^ (-kt)

0.933 = e^ (-kt)

k = 0.138

P = P_0*e^ (-0.138t)

Now after t = 3 hours,

P = 0.6 * e^ ((-0.138) * 3) = 0.396 M

time when P = 0.1 M

0.1 = 0.6*e^ (-0.138t)

t = 12.98 hours