[kir]

Hi, this is a two part question, I believe I have the first part correct but am stuck on the second.
1. If the rate of formation of C at a certain time in the reaction 2A + B  3C + 2D was found to be 205 mol/L*s. What is the rate of reaction of A and B and the rate of formation of D at this same time?
So this is how I proceeded:
Rate of Formation C= -3/1(Rate of Decomposition B) = -3/2 (Rate of Decomposition A) = 3/2 (Rate of Formation D)
2.5 M/s= -3x
Rate of Decomposition B= -.833 M/s
2.5 M/s= -3/2x
Rate of Decomposition A=-1.67M/s
2.5 M/s= 3/2x
Rate of Formation D = 1.67 M/s
Now on to part 2:
2. Consider the above reaction. If the initial concentration of A is 0.30M and 0.40 M for B. What is the concentration of A B and D, if the concentration of C after 20 minutes is found to be 0.10M?
First, I convert 20 min (60s/1min) = 1200s
I know Rate of Formation C =  :delta:[C]/ :delta:t    :delta:[C]= Concentration of C at time t2 -Concentration of C at t1  :delta: t = t2-t1
so, I set up the following: 2.5 M/s= .10M-x/1200s-0s
3000M=.10M-x
x= -2999.9 M
I know this is not correct because there is no way M could be that big even after 20 minutes!
If you could please point me in the right direction I would greatly appreciate it.
Thanks! 

[juanrga]

Hi, this is a two part question, I believe I have the first part correct but am stuck on the second.
1. If the rate of formation of C at a certain time in the reaction 2A + B  3C + 2D was found to be 205 mol/L*s. What is the rate of reaction of A and B and the rate of formation of D at this same time?
So this is how I proceeded:
Rate of Formation C= -3/1(Rate of Decomposition B) = -3/2 (Rate of Decomposition A) = 3/2 (Rate of Formation D)
2.5 M/s= -3x
Rate of Decomposition B= -.833 M/s
2.5 M/s= -3/2x
Rate of Decomposition A=-1.67M/s
2.5 M/s= 3/2x
Rate of Formation D = 1.67 M/s
Now on to part 2:
2. Consider the above reaction. If the initial concentration of A is 0.30M and 0.40 M for B. What is the concentration of A B and D, if the concentration of C after 20 minutes is found to be 0.10M?
First, I convert 20 min (60s/1min) = 1200s
I know Rate of Formation C =  :delta:[C]/ :delta:t    :delta:[C]= Concentration of C at time t2 -Concentration of C at t1  :delta: t = t2-t1
so, I set up the following: 2.5 M/s= .10M-x/1200s-0s
3000M=.10M-x
x= -2999.9 M
I know this is not correct because there is no way M could be that big even after 20 minutes!
If you could please point me in the right direction I would greatly appreciate it.
Thanks! 

Why do you assume that the average rate of formation for a time span of 20 minutes is the same than the instantaneous rate 205 mol/L*s of C at a certain time?

[kir]

Thanks for the point in the right direction, I realized that the time period was too long to use the change in [ ] over  :delta:t method. But I still had to go see my professor because I was still stuck! So he explained the first part as follows to me.
assume in 1 L of solution, 0.10M C formed
so 0.10 M C * 2 mol A / 3 mol C = 0.067 mol A reacted
A left = 0.30 M - .067= 0.233 M

then:
.10 M C * 1 mol B / 3 Mol C = .033 M
B Left = .40 M - .033 M = .367 M

And finally:
.10 M C * 2 mol D / 3 mol C = 0.067 M = [D]