posted by Jess on .
A few classmates and I were arguing about a problem we had to do for homework.
The problem states:
Limestone, CaCO3, when subjected to a temperature of 900 degrees C in a kiln, decomposes to calcium oxide and carbon dioxide . How much heat is evolved/absorbed when one gram of limestone decomposes?
Some of the students thought that 900 degrees C was not important to the calculation of the heat, so they did their work like this:
CaCO3 -> CaO + CO2
635.1 kj/mol + -393.5 kj/mol - (-1206.9 kj/mol)= 178.3 = delta H
1.00 g x 1mol/100.05 g = .009995 mol CaCO3
178.3 x .009995 mol = 1.78 kJ
Another student believed that the 900 degrees C, and explained his approach to the problem like this:
CaCO3 -> CaO + CO2
635.1 kj/mol + -393.5 kj/mol - (-1206.9 kj/mol)= 178.3 = delta H at 25 degrees C and 1 atm
1=1 atm x V/.08206 x 298.2
n= 1 atm x 24.47 / .08206 x 1173.2
178.3 x .254 = 45.28 kj/mol x .009995 mol CaCO3 = .45 [although he claimed to get 7. as an answer, so maybe I missed a step.]
For the most part, which idea is correct? And what changes [if any] need to be made to the set-up of the problem?
I don't have the delta Hf values for these materials except for CO2. Assuming those values are correct, then the first method is the correct one to use. We COULD calculate the volume at 25 C if we wanted to do so and that is about 0.244 L or at 900 C it would be about 0.962 L; however, there is nothing about this being a closed system and we don't need to know the volume. But the real clinker here is that the second method then takes the volume and uses that to calculate the mols as about 0.254. But you KNOW the mols of CO2 will be 0.009995. The 1.0 g CaCO3 (0.009995 mol) taken initally sets the CO2 that is evolved and no other calculation can change that. There will be 0.009995 mol CO2 and 0.009995 mol CaO.