How much heat is evolved or absorbed in this problem?
Question Details
Limestone, CaCo3, when subjected to a temperature of 900 degrees Celsius in a kiln, decomposes to calcium oxide and carbon dioxide. How much heat is evolved or absorbed when one gram of limestone decomposes?
The standard enthalpies of formation at 25 degrees Celsius(kj/mol) of these compounds at 1 atm, Aqueous ions at 1 M is:
CaCO3 -1206.9
CaO -635.1
CO2 -393.5
See your post above. n = 1 in this case.
To calculate the heat evolved or absorbed when one gram of limestone decomposes, we need to use the enthalpy of formation values given.
First, we convert the given data from kJ/mol to kJ/g by using the molar mass of each compound. The molar mass of CaCO3 is 100.09 g/mol, CaO is 56.08 g/mol, and CO2 is 44.01 g/mol.
So, we can calculate the enthalpy change for the decomposition of one gram of limestone as follows:
1. Calculate the moles of CaCO3 in one gram:
moles of CaCO3 = 1 g / 100.09 g/mol = 0.00999 mol (approximately 0.01 mol)
2. Calculate the enthalpy change for the decomposition of CaCO3:
enthalpy change = (enthalpy of formation of CaO + enthalpy of formation of CO2) - enthalpy of formation of CaCO3
enthalpy change = [(-635.1 kJ/mol) + (-393.5 kJ/mol)] - (-1206.9 kJ/mol)
3. Convert the enthalpy change into kJ/g:
enthalpy change per gram = enthalpy change / moles of CaCO3
enthalpy change per gram = (enthalpy change in kJ) / 0.00999 mol
Now, you can evaluate the expression to find the answer.