Question

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

Substance	ΔfH∘ΔfH∘ (kJ mol−1)(kJ mol−1)
CO2(g)CO2(g)	−393.5−393.5
Ca(OH)2(s)Ca(OH)2(s)	−986.1−986.1
H2O(l)H2O(l)	−285.8−285.8
CaCO3(s)CaCO3(s)	−1207−1207
H2O(g)H2O(g)	−241.8−241.8

Part A: Determine the enthalpy for this reaction: Ca(OH)2(s)+CO2(g)→CaCO3(s)+H2O(l) C a ( O H ) 2 ( s ) + C O 2 ( g ) → C a C O 3 ( s ) + H 2 O ( l ) Express your answer in kJ mol−1 k J m o l − 1 to four significant figures.

Part B: Consider the reaction Ca(OH)2(s)→CaO(s)+H2O(l) C a ( O H ) 2 ( s ) → C a O ( s ) + H 2 O ( l ) with enthalpy of reaction ΔrH∘=65.20kJ mol−1 Δ r H ∘ = 65.20 k J m o l − 1 What is the enthalpy of formation of CaO(s) C a O ( s ) ? Express your answer in kJ mol−1 k J m o l − 1 to four significant figures.

Answer 1

Part A : enthalpy = -113.2 kJ.mol-1

Part B : enthalpy of formation = -635.1 kJ.mol-1

Explanation

Part A

Balanced reaction : Ca(OH)2 (s) + CO2 (g) $\rightarrow$ CaCO3 (s) + H2O (l)

enthalpy of reaction, $\Delta$ Horxn = $\Delta$ Hof products - $\Delta$ Hof reactants

$\Delta$Horxn = [$\Delta$Hof CaCO3 (s) + $\Delta$ Hof H2O (l)] - [$\Delta$Hof Ca(OH)2 (s) + $\Delta$ Hof CO2 (g)]

$\Delta$Horxn = [(-1207 kJ/mol) + (-285.8 kJ/mol)] - [(-986.1 kJ/mol) + (-393.5 kJ/mol)]

$\Delta$Horxn = -1207 kJ/mol - 285.8 kJ/mol + 986.1 kJ/mol + 393.5 kJ/mol

$\Delta$Horxn = -113.2 kJ/mol