Chemical Forums
Chemistry Forums for Students => High School Chemistry Forum => Topic started by: CJP4 on September 28, 2005, 01:25:23 AM
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I believe i have numbers put in wrong position or something. Kind of confused.
This is what is given and what I have done.
A 2.200 g sample of quinone, C6H4O2, is burned in a bomb calorimeter whose total heat capacity is 7.854 kJ/°C. The temperature of the calorimeter increases from 22.52°C to 29.65°C.
What is the heat of combustion per gram of quinone?
So i use H=M C delta T right?
2.200g
7.854kJ/C
7.13C
I did (2.200)(7.854)(7.13) = 123.2kJ/g
Says that the answer is wrong and i'm kinda confused. The next part of the question is What is the heat of combustion per mole of quinone?
Which i'm about to attempt, but will probably get wrong since i got this part incorrect.
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Hrm.. just tried it a buncha diff ways got all wrong answers. But still if you know how to get the correct answer tell me because it will help me figure out all these other problems.
- Ty
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Your units don't add up so it can't be right. Are you sure you don't need the specific heat of water or something?
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I did (2.200)(7.854)(7.13) = 123.2kJ/g
If it is per gram, mass should be in denominator.
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i tried ((7.854)(7.13))/(2.200)=25.45kJ/g
said it was wrong.
Anyone know if i'm supposed to use this equation or a different one?
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Well, it said the answer is -25.5kJ/g.
Which is what it would be in significant figures. But i kept getting it wrong because I didn't put negative.
Why is it negative?? Especially when the temperature increased not decreased. I missed a lot of problems now checking seeing that answers that needed to be positive i had negative, and vise versa.
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Convention. During combustion system looses internal energy, so its change in negative.
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Convention. During combustion system looses internal energy, so its change in negative.
that's convention for chemists ;) physicists use the convention by looking at the system, not looking from within the system. So for physicists, the system gains heat and thus the sign is positive. However, for chemists the sign is negative because the reaction taking place in the system releases energy, so the potential energy of the molecules decrease and thus the sign is negative. That's why the first law of thermodynamics can be looked at in 2 different ways: lokking at the system or looking from within the system.
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That makes sense. I guess i should check if the reaction is exothermic or endothermic to help me decipher the sign. Otherwise i'll just get lost again. TY
:o