Question

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?

Answer 1

Answer:

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