The following reaction was monitored as a function of time: ABA+B A plot of 1/[A
ID: 877477 • Letter: T
Question
The following reaction was monitored as a function of time: ABA+B
A plot of 1/[AB] versus time yields a straight line with slope 5.8×102 (Ms)1 .
B. Write the rate law for the reaction. Rate =k[AB]^2
C.What is the half-life when the initial concentration is 0.54 M -t1/2 = 32 s (The half-life of a second order reaction is calculated by the formula t1/2= 1k [AB])
*** Here's the tricky one>D.If the initial concentration of AB is 0.280 M , and the reaction mixture initially contains no products, what are the concentrations of A and B after 80 s ? The answer was 0.16, 0.16. Please, someone show me how? Here are the instructions mastering gave after using all my tries!
[AB]t = 1/kt + 1/[AB]
The concentration of [AB] that reacted is found using [AB] = [AB]t=0 s [AB]t=80 s
The concentration of each product was the same as the concentration of AB that reacted because of the 1:1 ratio between each product and the reactant.
Recall that the integrated rate law for a second order reaction is 1/[AB]t = kt + 1/[AB]o
Substitute the values for [AB]o and t with the initial concentration and time, respectively. Then, simplify the equation to find [AB]t. Finally, use the 1:1 ratio of reactant to product to find the concentrations of the products.
This is answered wrong around 10X on here and all over Yahoo etc...
Explanation / Answer
The solution is discussed below.
[C] t1/2 = 1/k[AB]o = 1/5.8 x 10^-2 x 0.54 = 31.93 s-1
[D] 1/[A+B]t = 1/[AB] + kt
Feed the values we have,
1/[A+B]t = 1/0.28 + 5.8 x 10^-2 x 80
[A+B]t = 0.12
[A] = [B] = 0.06
Check the answer, I think their might be a mistake, as the answer with the values given is 0.06, 0.06 each A and B.
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.