Need some help. have the A part of the question but do not know how to set B or C.
Butadiene (C4H6) reacts with itself to form a dimer with the formula C8H12. The reaction is second order in butadiene. If the rate constant at a particular temp is 0.04/(M s) and the initial concentration of butadiene is 0.02M
A. what is its molarity after a reaction time of 1.00 h?
60min x 60 sec= 3600s
1/A = 0.04M/s(3600s)+1/0.02
1/A = 144 + 50
1/A = 194
A = 1/194
=5.2x10^-3M
B. what is the time in hours when butadiene concentration reaches a value of 0.002 M?
C. What is the half life of the reaction?
i figured out part C as well
t1/2 = 1/K(A)
t1/2 = 1/(0.04)x(0.02)
t1/2 = 1/8x10^-4
t1/2 = 1250sec
t1/2 = 1250s/60m
half life = 20.833min
Still do not know how to set up part B. Can anybody help out?
To solve Part B, you need to rearrange the rate equation for a second-order reaction and solve for time (t). The rate equation for a second-order reaction is:
Rate = k * [butadiene]^2
We know the rate constant (k = 0.04/M*s) and the final concentration of butadiene (0.002 M). Let's substitute these values into the rate equation:
0.04/M*s = k * [butadiene]^2
0.04/M*s = 0.04/M*s * (0.002 M)^2
Now, you can solve for t by rearranging and isolating it:
t = [1 / (k * [butadiene]^2)] * [1 / (M*s)]
t = [1 / (0.04/M*s * (0.002 M)^2)] * [1 / (M*s)]
t = 1 / (0.04 * (0.002)^2)
Now, you can calculate the value for t.
To solve Part C, you need to find the half-life of the reaction. The half-life of a second-order reaction can be found using the equation:
t1/2 = 1 / (k * [butadiene]0)
where [butadiene]0 is the initial concentration of butadiene. In this case, [butadiene]0 is given as 0.02 M.
t1/2 = 1 / (0.04/(M*s) * (0.02 M))
t1/2 = 1 / (0.04 * 0.02)
Now, calculate the value for t1/2.
By following these steps, you should be able to find the answers for Part B and Part C of the question.