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
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
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
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.
ok I got it now!
To calculate ΔHcomb (the enthalpy change of combustion for methane gas), you need to consider the heat released during the combustion reaction.
The heat released can be determined by finding the difference in temperature between the initial and final states of the calorimeter. Since the calorimeter has a known heat capacity (Ccal), the equation to calculate the heat released is:
qreaction = -(qcal + qwire)
where qreaction is the heat released by the reaction, qcal is the heat absorbed by the calorimeter, and qwire is the heat released by the ignition wire (which is negative since it releases heat).
Given:
qwire = 140.77 J
Ccal = 4.319 kJ/°C = 4.319 × 10^3 J/°C
Initial temp = 21.1°C = 21.1 K (converted to Kelvin)
Final temp = 23.4°C = 23.4 K (converted to Kelvin)
First, calculate the heat absorbed by the calorimeter (qcal) using the equation:
qcal = Ccal × ΔT
where ΔT is the change in temperature (Final temp - Initial temp).
ΔT = 23.4 K - 21.1 K = 2.3 K
qcal = 4.319 × 10^3 J/°C × 2.3 K = 9.9337 × 10^3 J
Next, substitute the given values into the equation:
qreaction = -(qcal + qwire)
qreaction = -(9.9337 × 10^3 J + 140.77 J)
qreaction = -10.07447 × 10^3 J
Finally, convert the heat released to kilojoules by dividing by 1000:
qreaction = -10.07447 × 10^3 J / 1000 = -10.07447 kJ
Thus, ΔHcomb is approximately -10.07447 kJ. Note that the negative sign indicates that the reaction is exothermic.
The ideal gas law formula, PV = nRT, is not used directly in this calculation. It is used to relate the pressure, volume, temperature, and number of moles of an ideal gas, and is typically used for different purposes in gas law problems. In this case, the molar heat capacity and the volume of the methane gas are not relevant to calculating ΔHcomb.