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 0.30 M of compound A decomposes to 0.25 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

Explanation:

We would apply the formula,

A(t) = Aoe^kt

Where

Ao represents the initial concentration of compound A.

k represents the rate of decomposition.

t represents the decomposition time.

A(t) represents the concentration after t hours.

From the information given

Ao = 0.30 M

t = 30 minutes = 0.5 hour

A(0.5) = 0.25 M

Therefore,

0.25 = 0.3e^0.5k

0.25/0.3 = e^0.5k

0.833 = e^0.5k

Taking ln of both sides, it becomes

ln0.833 = ln e^0.5k = 0.5k

k = - 0.1827/0.5

k = -0.3654

The expression becomes

A(t) = 0.35e^-0.3654t

1) When t = 3 hours,

A(3) = 0.35e^-0.3654 × 3

A(3) = 0.35e^- 1.0962

A(3) = 0.117M

2) When A = 0.1, then

0.1 = 0.35e^-0.3654 × t

0.1/0.35 = e^-0.3654t

0.2857 = e^-0.3654t

Taking ln of both sides, it becomes

Ln 0.2857 = - 0.3654t

- 1.2528 = - 0.3654t

t = - 1.2528/- 0.3654

t = 3.43 hours