Asked by Anonymous
If the initial concentration of is 0.250 , and the reaction mixture initially contains no products, what are the concentrations of and after 80 ?
Answers
Answered by
DrBob222
You need to rewrite the question and include all of the details.
Answered by
Anonymous
If the initial concentration of AB is .250 M, and the reaction mixture initially contains no products, what are the concentrations of A and B after 80s, k=5.6*10^-2 ( second order)
AB-A + B
AB-A + B
Answered by
DrBob222
Isn't the second order equation something like this?
[1/(A)]-[1/(A)<sub>o</sub>] = kt
You don't have any units on k in the problem. Make sure the units for k and unit for t agree.
[1/(A)]-[1/(A)<sub>o</sub>] = kt
You don't have any units on k in the problem. Make sure the units for k and unit for t agree.
Answered by
Anonymous
[1/(A)]-[1/(A)o] = kt
k=5.6*10^-2 M^-1 s^-1
k=5.6*10^-2 M^-1 s^-1
Answered by
DrBob222
Substitute and solve for A.
Answered by
Anonymous
My answer is still wrong. Can you explain in more details
Answered by
Anonymous
If I remember correctly. Your intial AB is .250 your k is 5.6 *10^-2 and t is 80s plug this all into the problem to.get the final equilibrium of concentration AB so you'll have 1/AB=whatever answer is then divide 1 by your answer to get AB by itself then you have to subtract that answer from the initial .250 and that will be your a and your b...they will be the same.....think.of the equation like this .... [1/AB]-[1/ABo]=Kt
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