10. The Romans used calcium oxide, CaO, to produce a strong mortar to build stone structures. The CaO was mixed with water to give Ca(OH)2, which reacted slowly with CO2 in the air to give CaCO3. Using the values from Appendix 4 of your textbook, calculate the standard enthalpy change for this reaction Ca(ОН)2 (s) + СO2 (g) СаCОз (s) + Н20 (g) -69.2
Thermochemistry
Thermochemistry can be considered as a branch of thermodynamics that deals with the connections between warmth, work, and various types of energy, formed because of different synthetic and actual cycles. Thermochemistry describes the energy changes that occur as a result of reactions or chemical changes in a substance.
Exergonic Reaction
The term exergonic is derived from the Greek word in which ‘ergon’ means work and exergonic means ‘work outside’. Exergonic reactions releases work energy. Exergonic reactions are different from exothermic reactions, the one that releases only heat energy during the course of the reaction. So, exothermic reaction is one type of exergonic reaction. Exergonic reaction releases work energy in different forms like heat, light or sound. For example, a glow stick releases light making that an exergonic reaction and not an exothermic reaction since no heat is released. Even endothermic reactions at very high temperature are exergonic.
This is a practice sheet for an exam tomorrow, this is concerning enthalpy of formation and using that to find the change of enthalpy for a reaction.
The question states:
The Romans used calcium oxide, CaO, to produce strong mortar to build stone structures. The CaO was mixed with water to give Ca(OH)2 which reacted slowly with CO2 in the air to give CaCO3. Using the values in Appendix 4 of your textbook, calculate the standard of enthalpy change for this reaction.
Ca(OH)2(s) + CO2(g) -> CaCO3(s) + H2O(g)
Which is a balanced equation, and the enthalpys of formation for each of these substances according to my appendix is:
Ca(OH)2(s) = -985.2 kJ/mol
CO2(g) = -393.5 kJ/mol
CaCO3(s) = -1206.9 kJ/mol
H2O(g) = -241.8 kJ/mol
and the change in enthalpy for the reaction is suppose to be the sum of the enthalpy of formation for the products minus the sum of the enthalpy of formation from the reactant side.
So Hf(products) = -1206.9 kJ/mol + (-241.8 kJ/mol) = -1448.7 kJ/mol
and Hf(reactants) = -985.2 kJ/mol + (-393.5 kJ/mol) = -1378.7 kJ/mol
so Hrxn = Hf(products) - Hf(reactants) = -1448.7 kJ/mol - (-1378.7 kJ/mol) = -70 kJ/mol (exactly)
however my answer key says that it is suppose to be -69.2 kJ/mol but I cannot see how to get there from the given information. Perhaps my profressor used a different book with more precise enthalpys of formation but is this not the correct way to do this problem? If not please let me know what I did wrong. I have a test tomorrow afternoon and would greatly appreciate an explaination!
Thank You!
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