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Kimo Khaled

Bomb calorimetry: heat of formation

Brian Arbenz  Follow
So using the heat capacity of the bomb (9.887 kJ/mol) and the change in temperature for one run of benzene (2.6576 °C) I get:

ΔcH=-(9.887 kJ/mol)(2.6576 °C)=-26.28 kJ/mol °C

This is nowhere near the given literature values of -3273 kJ/mol.

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Barbara Davis  Follow
You're still only providing fragments of what you're doing, so it's hard to troubleshoot your work. But it's also very clear you're being sloppy with your units, which probably isn't helping.

Take a look at examples 6 and 7 at this page and see if it doesn't help you understand the proper way to do this kind of calculation.

http://www.science.uwaterloo.ca/~cchieh/cact/c120/calorimetry.html

If it doesn't, please provide a step-by-step account of what you did in your experiment and a full account of your calculations (without omitting steps). This should make it much easier to see what you're doing wrong.

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Dahny Patel  Follow
Just for future reference, it's always helpful to know what exactly you're trying to do in your lab. That said, it seems clear a major part of your problem is this:

Quote
(9886.8899 J/°C)(0.239 cal/1J)(1°C/274.15K)

That's not an appropriate conversion factor for °C to K. These units are pretty much converted 1:1 (a 1° increase in temperature is the same as a 1 K increase in temperature).  9887 J is approximately 2361 calories. So it doesn't look like you're actually that far off.

For the second part, the change in temperature (usually of a water reservoir) is measured because this is what allows you to determine the amount of heat liberated by the reaction being combusted.

http://en.wikipedia.org/wiki/Calorimeter#Bomb_calorimeters

Oh, you probably realize this now but I'll say it anyway: it helps not to wait weeks before you write up your labs. :)

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