Your rate = k(A)(B)^2 is correct(that's #1).
For #2 plug rate for any ONE of the trials, plug (A) and (B) for that trial and solve for k.
#3. You now know k, plug in 0.3 and 0.3 and solve for rate but I would feel better if I saw an equation for the reaction of A and B to produce C.
Experiment 1: A has .20 M, B has .20 M and the initial rate is 2.0*10^-4M/min
Experiment 2: A has .20 M, B has .40 M, and the initial rate is 8.0*10^-4M/min
Experiment 3: A has .40 M, B has .40 M, and the initial rate is 1.6*10^-3M/min
Using the data above, answer the following questions:
Determine the rate law for the reaction
Calculate the value of the specific rate constant
If the initial concentrations of both A and B are .3 M at what initial rate is C formed?
For the first question I got (a)(b)^2, but the other 2 I have no clue on how to find those out.
6 answers
Its A + 2B --> C
So for 2 would I do (a)(b)^2=Rate and plug in the numbers for the correspond letters from whichever trial I choose?
yes.
#2 is like this.
8E-4 = k*(0.2)(0.4)^2 and solve for k or
2E-4 = k*(0.2)(0.2)^2 and solve for k or
1.6E-3 = k*(0.4)(0.4)^2. Note that all of them give the same value for k.
For #3, the rate of disappearance of A = rate of appearance of C since the coefficient of each = 1.
#2 is like this.
8E-4 = k*(0.2)(0.4)^2 and solve for k or
2E-4 = k*(0.2)(0.2)^2 and solve for k or
1.6E-3 = k*(0.4)(0.4)^2. Note that all of them give the same value for k.
For #3, the rate of disappearance of A = rate of appearance of C since the coefficient of each = 1.
So if I used 8E-4 = k*(0.2)(0.4)^2
My answer would be .025 M/min
Then from there I would do .025(.30)(.3)^2 which would be 6.75E-4 right?
My answer would be .025 M/min
Then from there I would do .025(.30)(.3)^2 which would be 6.75E-4 right?
That's what I would do.
2 answers