The rate constant for a certain reaction is k = 5.00×10−3 s-1 . If the initial reactant concentration was 0.950 M , what will the concentration be after 8.00 minutes?
The units of k are s^-1 which says that the reaction is first order. The integrated rate equation for a first order reaction is
ln([Ao]/[A]) = akt
I presume you assume a = 1 if it isn't given in the problem.
Don't forget to change minutes to seconds.
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To find the concentration of the reactant after a certain time, we can use the first-order rate equation:
ln([A]t/[A]0) = -kt
Where [A]t is the concentration of the reactant at time t, [A]0 is the initial concentration of the reactant, k is the rate constant, and t is the time.
First, we need to convert the time from minutes to seconds:
t = 8.00 minutes = 8.00 x 60 seconds = 480 seconds
Plugging in the values to the rate equation, we have:
ln([A]t/0.950 M) = - (5.00x10^-3 s^-1) x (480 s)
Now, let's solve for [A]t:
ln([A]t/0.950) = -2.4
To isolate [A]t, we exponentiate both sides:
[A]t/0.950 = e^(-2.4)
[A]t = (0.950 M) x e^(-2.4)
Using a calculator, we find:
[A]t = 0.950 M x 0.0912
[A]t ≈ 0.0865 M
Therefore, the concentration of the reactant after 8.00 minutes is approximately 0.0865 M.