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 20 minutes. How much of compound A will remain after 2 hours? How long will it take until 0.10 M of compound A remains?

Options:
a) Remaining after 2 hours: 0.06 M, Time for 0.10 M: 3 hours
b) Remaining after 2 hours: 0.05 M, Time for 0.10 M: 2.5 hours
c) Remaining after 2 hours: 0.04 M, Time for 0.10 M: 2 hours
d) Remaining after 2 hours: 0.03 M, Time for 0.10 M: 1.5 hours

Answer 1

The amount of compound A remaining after 2 hours is approximately 0.048 M. It will take approximately 2 hours for 0.10 M of compound A to remain.

Step-by-step explanation:

The decomposition of compound A into compounds B and C follows the law of uninhibited decay. We are given that an initial concentration of 0.30 M of compound A decomposes to 0.25 M in 20 minutes. To find the amount of compound A remaining after 2 hours, we can use the formula:

[A]t = [A]0 * e^(-kt)

where [A]t is the concentration at time t, [A]0 is the initial concentration, k is the rate constant, and t is the time. From the given data, we can set up the following equation:

0.25 = 0.30 * e^(-k * 20)

Solving for k, we find k ≈ 0.0231 min^-1.

Substituting this value of k into the equation for [A]t and solving for [A]t when t = 120 minutes, we get [A]t ≈ 0.048 M. Therefore, approximately 0.048 M of compound A will remain after 2 hours.

To find the time it takes for 0.10 M of compound A to remain, we can rearrange the equation [A]t = [A]0 * e^(-kt) and solve for t:

t = -ln([A]t / [A]0) / k

Substituting the given values into the equation, we get:

t = -ln(0.10 / 0.30) / 0.0231 ≈ 2 hours.