Question

Part A

Determine the enthalpy for this reaction:

Ca(OH)2(s)+CO2(g)→CaCO3(s)+H2O(l)

Express your answer in kilojoules per mole to one decimal place.

ΔHrxn∘=		kJ/mol

Part B

Consider the reaction

Ca(OH)2(s)→CaO(s)+H2O(l)

with enthalpy of reaction

ΔHrxn∘=65.2kJ/mol

What is the enthalpy of formation of CaO(s)?

Express your answer in kilojoules per mole to one decimal place.

± Enthalpy

Enthalpy H is a measure of the energy content of a system at constant pressure. Chemical reactions involve changes in enthalpy, ΔH, which can be measured and calculated:

ΔHrxn∘=∑productsmΔHf∘−∑reactantsnΔHf∘

where the subscript "rxn" is for "enthalpy of reaction" and "f" is for "enthalpy of formation" and m and nrepresent the appropriate stoichiometric coefficients for each substance.

The following table lists some enthalpy of formation values for selected substances.

Substance	ΔHf∘ (kJ/mol)
CO2(g)	−393.5
Ca(OH)2(s)	−986.1
H2O(l)	−285.8
CaCO3(s)	−1207.0
H2O(g)	−241.8

Answer 1

Enthalpy is a thermodynamic quantity, it is a measure of the energy content of a system at constant pressure. It is denoted by H The standard enthalpy change for a reaction (AH ° is calculated by using the following rxn formula AH° = ДН? - Х. (1) m reactants AH, products n is the appropriate Here, mis the appropriate stoichiometric coefficients of the products, stoichiometric coefficients of the reactants, AH is the standard enthalpy of formation (лH;) The definition of the standard enthalpy of formation is as follows: The change in the enthalpy when one mole of a substance is formed from its pure elements under standard conditions is known as standard enthalpy of formation. Part A: Solid calcium hydroxide (Ca(OH),) solid calcium carbonate (CaCO3) and liquid water (H,0) is reacted with carbon dioxide gas (CO,) to produce

The reaction between solid calcium hydroxide and carbon dioxide gas is as follows: Ca (OH), (s) CO2 (g) Н.О) CaCO, (s) Here, Ca(ОH), , со, are the reactants, CaCO3, and H,O are the products 1, 1, , and 1 are the stoichiometric coefficients of Ca(OH),, CO,, CaCO,, and H20 respectively

Calculate the standard enthalpy change for reaction (2) as shown below: дн дн; дн; = > m products reactants rxn .-[(maco,c Aco,)(00AH (O)AH )+(co,A.co) AH mCaCoa(s) тн,00) Пco,(g) Ca(OH).(s) ,CaCO3 (s) f,Ca(OH(s) rn (Do'н' H kJ + 1x| -285.8 mol kJ -986.1- mol kJ 1x -1207.0 mol kJ 1x| -393.5 mol AH® rxn kJ 1207.0 mol kJ -285.8 mol kJ 986.1 mol kJ -393.5 mol ДН. rxn kJ - 285.8 mol kJ kJ -393.5 mol kJ ДН -1207.0 986.1 mol mol kJ -1492.8 mol kJ -1379.6 mol ДН rxn kJ +1379.6 mol kJ =-1492.8 ДН rxn mol kJ =-113.2 mol (round off to one decimal place) ДНО rxn kJ is113.2 mo Therefore, the standard enthalpy change for reaction (2), AH

Part B: Solid calcium hydroxide (Ca (OH),) on decomposition produces solid calcium oxide (CaO) (н.о). and liquid water The reaction as a result for the decomposition of solid calcium hydroxide is as follows Ca(Он), (6) CaO (s) Н,О() (3) Here, Ca(ОH), is the reactant, CaO , and H,O are the products 1, 1, and 1 are the stoichiometric coefficients of Ca(OH),, CaO, and H,O respectively The standard enthalpy change for reaction (3) is as follows: kJ AH 65.2 mol (4) rxn

Calculate the standard enthalpy of formation of CaO (s) as shown below products AH m reactants ran A(m0 )+(m0 AR)H(arOc mcao(s) nCa(OH), (s) f,H20() f,CaO(s) 1,0(1) f,Ca(OH), (s) kJ 65.2 mol kJ -285.8 mol kJ 1x-986.1 mol (1x AHCao(s) Ix kl 65.2 mol kJ 285.8 mol kJ -986.1 mol AH f,CaO(s) H kJ 65.2 mol kJ kJ 986.1 mol f,CaO(s)285.8 mol AH k.J 65.2 mol kJ +986.1 mol kJ f,CaO(s) mol kJ kJ f,CaO(s)700 3 mol AHO 65.2 mol kJ -700.3 mol kJ AH. f,CaO(s) 65.2 mol kJ =-635.1 mol (round off to one decimal place AHO f,CaO (s) kJ is-635.1 mol Therefore, the standard enthalpy of formation of CaO (s), AH

Answer 2

Ca(OH)2(s)+CO2(g)→CaCO3(s)+H2O(l)

Check if bbalanced:

it is balanced

H = Hprod - Hreact = (−1207.0 + −285.8)-(−986.1+-393.5) = -113.2 kJ/mol

B)

when

Ca(OH)2(s)→CaO(s)+H2O(l) ΔHrxn∘=65.2kJ/mol

then

Hrxn = Hprod - Hreact

65.2 = HCaO + HH2O - HCa(OH)2

65.2 = HCaO + −285.8 - −986.1 = 700.3

65.2 = HCaO + 700.3

HCaO = 65.2-700 = -634.8

Answer 3

[1(−1207.0kJ/mol) + 1(−285.8kJ/mol)] − [1(−393.5kJ/mol) + 1(−986.1kJ/mol)] = −113.2kJ/mol