The following reaction was monitored as a function of time: ABA+B A plot of 1/[A
ID: 877481 • 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...
2nd posting... first one was spammed.
Explanation / Answer
AB---> A+ B
d[AB] /dt = - k[AB]^2
for 2nd oder raction
slope = k = 5.8x10^-2 M.s^-1
the integrated rate equation
1/[AB]t - 1/[AB]o = kt
b- half life for 2nd oder reaction.
[AB]o=0.54M
t1/2= 1/k[AB]o= 1/ 5.8x10^-2(Ms)^-1 x 0.54M = 32 s
c- [AB]o= 0.28M t= 80
1/[AB]t = kt + 1/[AB]o
1/[AB]t = 5.8 x10^-2(Ms)^-1 x 80 + 1/ 0.28
1/[AB]t = 4.64+3.5714=8.211M^-1
[AB]t= 0.121
x= 0.28- 0.121=0.159=0.16
AB-----> A+ B
0.28 0 0
0.28-x x x
0.121 0.16 0.16
amount of A & B = 0.16M
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