[methic]

'ello all. New semester, new problems.

I have the actual answer, but am having trouble figuring out how to arrive at it.

2HBr H₂ + Br₂

First part:
In the first 25.0s of this reaction, the concentration of HBr dropped from 0.600M to 0.512M. Calculate the average rate of the reaction in this time interval.

My attempt:
-½(0.512M - 0.600M)/25s =1.8 x 10^-3

Okay, I think that's pretty correct. This second part is what's confusing me.

Second part:
If the volume of the reaction vessel in part b was 1.50L, what amount of Br₂ was formed during the first 15.0s of the reaction?

So, I know there were 0.9 moles of HBr at the beginning (1.50L * 0.600M). And at the end, there should be .45 moles of Br₂. After 15 seconds, the number of moles should be 0.040 moles (answer from back of book); but I have no idea how to set up the problem to arrive at that. Any help would be appreciated.

Regards,

methic.

[methic]

Hi. It's me again. I went for a walk and think I may have figured it out.

Is this the correct way to finish the problem?

rate is 1.8 x 10^-3M/s
time is 15 seconds

15s * (1.8 x 10^-3) = .027M

.027moles/Liter * 1.50L = .0405 moles Br₂

I know the answer is the same. But I'm just wondering if it's a fluke or if I did the problem correctly.

[rabolisk]

That's how you're supposed to solve it. Good job.