Question

The reaction A → B follows first-order kinetics with a half-life of 21.7 hours. If the concentration of A is 0.023 M after 48.0 hours, what is the initial concentration of A?

a. 0.046 M
b. 0.069 M
c. 0.012 M
d. 0.023 M

Answer 1

To find the initial concentration of A in a first-order reaction with a given half-life, we use the formula ln([A]_0/[A]) = kt. Substituting the given values, we can solve for the initial concentration of A.

Step-by-step explanation:

To find the initial concentration of A, we can use the formula for first-order kinetics:

ln([A]_0/[A]) = kt

Where:

[A]_0 is the initial concentration of A,

[A] is the concentration of A after a certain time,

k is the rate constant, and

t is the time.

Given that the half-life of the reaction is 21.7 hours, we can use this information to find the rate constant:

t1/2 = (ln2)/k

Substituting the given half-life:

21.7 = (ln2)/k

Solving for k:

k = (ln2)/21.7

Now, we can use the concentration of A after 48 hours to find the initial concentration:

[A] = [A]_0 * e^(-kt)

Substituting the given time and concentration:

0.023 = [A]_0 * e^(-k*48)

Substituting the value of k:

0.023 = [A]_0 * e^(-((ln2)/21.7)*48)

Simplifying the equation:

0.023 = [A]_0 * e^(-2.516)

Dividing both sides by e^(-2.516):

[A]_0 = 0.023 / e^(-2.516)

This gives us an initial concentration of approximately 0.046 M.