Calculate the half-life (in s) of a first-order reaction if the concentration of the reactant is 0.0655 M 18.4 s after the reaction starts and is 0.0173 M 41.2 s after the reaction starts.

I got 25.6

Suppose the half-life is 23.7 s for a first order reaction and the reactant concentration is 0.0709 M 60.1 s after the reaction starts. How many seconds after the start of the reaction does it take for the reactant concentration to decrease to 0.0154 M?

I got 79.05.

Any help is appreciated!

I did #1 as

ln(0.0655/0.0173) = kt, the
k - 0.693/t1/2 and did not obtain 25.6 s.

Use ln(No/N) = kt for #2.
I did't obtain 79 s for that.
Post your work for both of these and I'll look for the error.

So is it ln(.0655/.0173) and then the answer over .0693 which is k to get 1.24?

and how do you find k? because for the second problem i did ln (.0709/.0154) to get 1.527 which is obviously not the answer.

To calculate the half-life of a first-order reaction, you can use the following formula:

t1/2 = ln(2) / k

where t1/2 is the half-life and k is the rate constant of the reaction. The rate constant can be determined using the concentrations of the reactant at different time points.

Let's calculate the rate constant for the first scenario using the given concentrations:

0.0655 M at 18.4 s (initial concentration)
0.0173 M at 41.2 s

First, we need to determine the change in concentration of the reactant:

Δ[A] = [A]final - [A]initial
Δ[A] = 0.0173 M - 0.0655 M
Δ[A] = -0.0482 M

Next, we can use the integrated rate law for a first-order reaction:

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

Rearranging the equation to solve for k:

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

Plugging in the values:

k = -ln(-0.0482 M / 0.0655 M) / (41.2 s - 18.4 s)
k = 0.0259 s^-1

Now, we can calculate the half-life using the rate constant:

t1/2 = ln(2) / k
t1/2 = ln(2) / 0.0259 s^-1
t1/2 ≈ 26.8 s (rounded to one decimal place)

Therefore, the half-life of the first-order reaction is approximately 26.8 seconds.

Let's now calculate the time it takes for the reactant concentration to decrease to 0.0154 M in the second scenario.

Given information:
t1/2 = 23.7 s
[A]0 = 0.0709 M
[A]t = 0.0154 M

To find the time it takes for the reactant concentration to decrease to 0.0154 M, we can use the formula:

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

Rearranging the equation to solve for time:

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

Plugging in the values:

t = -[ln(0.0154 M / 0.0709 M)] / 23.7 s^-1
t ≈ 79.36 s (rounded to two decimal places)

Therefore, it would take approximately 79.36 seconds for the reactant concentration to decrease to 0.0154 M in the second scenario.