What would be the concentration of dinitrogen pentoxide after 6.00 X 10(2) seconds and how long would it take for the concentration of N2O2 to decrease to 10.0% of its initial value in the following equation...
N2O5 to NO2 and O2
initial concentration is 1.65 X 10-2mol/L
rate constant is 4.80 X 10-4/s
time is 825s
According to my book, the answer is 1.24 X 10-2 mol/L and t= 4.80 X 10 (3)s
I can't figure out how to solve t=4.80 X 10 (3)s
Do I set up like this...
ln (0.00124/0.0165)=4.80 X 10-4 X t?
What are you doing? I worked this for you last night (at least part a). I described how to do the second part. Did you read that? You aren't approaching it in the manner I told you it could be worked.
To determine the concentration of dinitrogen pentoxide (N2O5) after 6.00 X 10^2 seconds, you can use the integrated rate equation for a first-order reaction:
[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 time.
In this case, the initial concentration [A]0 is given as 1.65 X 10^-2 mol/L, and the rate constant k is 4.80 X 10^-4/s. The time t is given as 6.00 X 10^2 seconds.
Let's calculate the concentration [A]t:
[A]t = (1.65 X 10^-2 mol/L) * e^(-4.80 X 10^-4/s * 6.00 X 10^2 s)
[A]t ≈ 1.24 X 10^-2 mol/L
So, the concentration of N2O5 after 6.00 X 10^2 seconds is approximately 1.24 X 10^-2 mol/L, as given in your book.
Now, let's move on to the second part of the question.
If we want to find the time it takes for the concentration of N2O5 to decrease to 10.0% of its initial value, we can rearrange the integrated rate equation:
[A]t = [A]0 * e^(-kt)
Dividing both sides by [A]0:
[A]t / [A]0 = e^(-kt)
Taking the natural logarithm (ln) of both sides:
ln([A]t / [A]0) = -kt
Now, we can substitute the values into the equation and solve for t:
ln(0.10) = -4.80 X 10^-4/s * t
t ≈ ln(0.10) / (-4.80 X 10^-4/s)
t ≈ 4.80 X 10^3 seconds
Thus, it would take approximately 4.80 X 10^3 seconds for the concentration of N2O5 to decrease to 10.0% of its initial value, as stated in your book.