Posted by **Liz** on Sunday, July 20, 2014 at 2:01pm.

A calorimeter is filled with 10.0 millimole (mmole) of methane gas and an excess of oxygen. When burned, the ignition wire releases 140.77 J of heat. The heat capacity of the calorimeter is 4.319 KJ°C-1. qwire= 140.77 J

Calculate ΔHcomb. Recall that ideal gas law is PV=nRT where R=0.08206 Latm/molK (since you are working with atm not kPa) and the temperature is in Kelvin

Ccal= 4.319 KJ°C-1

Initial temp= 21.1 degrees celsius

Final temp= 23.4 degrees celsius

v= 271 mL

P= 1.000 atm

I calculated delta H and got 979 KJ. I'm also not sure why ideal gas law formula would be used

- Chemistry -
**DrBob222**, Sunday, July 20, 2014 at 3:46pm
I don't know the context of the problem but your 979 kJ/mol appears to be right if you have the number of significant figures correct. Does the problem say anything about "about 10.0 millimoles" methane? You have P, V, R, and T and you can calculate n. If I can calculate n I get something like 0.0112 and if the q is divided by 0.0112 instead of 0.01 to obtain per mol the number you get is much closer to 890 kJ/mol than if you use 0.01. Perhaps that 10.0 mmols is "close" and you are to use PV = nRT to solve for the real number of mmols. Here is the site I talked about a couple days ago where this experiment is delineated. There is nothing in the material about using PV = nRT BUT the instruction say to add a known amount of methane and I don't know how you would do that without using PV = nRT. I suspect you are to use PV = nRT to solve for the real number of mols CH4.

http://www.chm.davidson.edu/vce/calorimetry/heatofcombustionofmethane.html

- Chemistry -
**Liz**, Sunday, July 20, 2014 at 4:26pm
Ok thanks! I used PV=nRT to calculate the mols but I got a larger number. Here's what I did:

P= 1.000 atm

V= 0.271 L

R= 0.08206 Latm/molK

T= 2.3 K

and I got 1.436 as my answer

- Chemistry -
**DrBob222**, Sunday, July 20, 2014 at 4:57pm
NOPE. T is not 2.3, that's delta T for the experiment. The initial T is 23.1 so that must be the T of the methane when it's added to the bomb or 294.25K and that gives n = 0.0112 or about 11.2 millimoles.

- Chemistry -
**Liz**, Sunday, July 20, 2014 at 5:14pm
ok I got it now!

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