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
n=PV/RT
1=1 atm x V/.08206 x 298.2
v= 24.47
n= 1 atm x 24.47 / .08206 x 1173.2
n= .254
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.
The first method is the correct one to use. The calculation correctly determines the heat evolved when one gram of limestone decomposes to calcium oxide and carbon dioxide. It uses the stoichiometry of the reaction and the enthalpy values for the formation of calcium oxide and carbon dioxide.
However, there are a couple of things to note regarding the second method. Firstly, the assumption that the temperature of 900 degrees C is not important is incorrect. Temperature does affect the enthalpy of a reaction, so it should be taken into account. Secondly, the calculation of the number of moles of reactant and products is incorrect. The molar amount of CO2 evolved during the decomposition of 1 gram of limestone should be 0.009995 moles, not 0.254 moles.
Therefore, the changes that need to be made to the setup of the problem are as follows:
- Use the proper enthalpy values for the reaction at the given temperature of 900 degrees C.
- Calculate the correct number of moles for the reactants and products, which should be 0.009995 moles of CO2 and CaO.
By making these changes, you will arrive at the correct answer for the heat evolved or absorbed during the decomposition of one gram of limestone.