Chemical Forums
Chemistry Forums for Students => Undergraduate General Chemistry Forum => Topic started by: methic on January 23, 2011, 06:49:11 PM
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'ello all. New semester, new problems. :(
I have the actual answer, but am having trouble figuring out how to arrive at it.
2HBr :rarrow: 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.
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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.
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That's how you're supposed to solve it. Good job.