The rate of removal of Aspirin from the body is a first order reaction with a rate constant,k = 0.34/h

How long does it takes for 87.5% of Aspirin to be removed from the body?

Calculate k.

k = 0.693/t1/2
k = 0.693/0.34 = ?

Then ln(No/N) = kt.
Use No = 100
and N = 12.5 (that's 100%-87.5% = 12.5%)
k from above.
Substitute and solve for t(in hours).

How do you calculate k?

To find out how long it takes for 87.5% of Aspirin to be removed from the body, we can use the concept of first-order reactions. First, we need to determine the half-life of the reaction, which is the time it takes for the concentration of Aspirin to decrease by half.

The half-life of a first-order reaction can be calculated using the following formula:

t1/2 = (0.693 / k)

Here, k is the rate constant for the reaction. In this case, the value of k is given as 0.34/h. Plugging in this value, we can calculate the half-life (t1/2):

t1/2 = (0.693 / 0.34) ≈ 2.04 h

Now that we know the half-life of the reaction, we can determine the time it takes for 87.5% of the Aspirin to be removed. We'll use the concept that each half-life removes half of the remaining Aspirin.

To find the number of half-lives required to remove 87.5%:

Number of half-lives = log(0.875) / log(0.5)

Using a calculator:

Number of half-lives ≈ 1.643

Now, we multiply the number of half-lives by the half-life time:

Time = Number of half-lives × t1/2

Time = 1.643 × 2.04 ≈ 3.34 hours

Therefore, it takes approximately 3.34 hours for 87.5% of Aspirin to be removed from the body.