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January 30, 2015

January 30, 2015

Posted by **Jess** on Thursday, February 1, 2007 at 10:41pm.

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 H

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