Answer:
When the enthalpy of this overall chemical equation is calculated, the enthalpy of the second intermediate equation is halved and the sign on it reversed, to obtain the required enthalpy of overall reaction.
ΔH(overall) = -178.3 kJ
Explanation:
First intermediate equation
Ca (s) + CO₂ (g) + ½O₂ (g) → CaCO₃ (s) ΔH₁ = -812.8 kJ
Second intermediate equation
2Ca (s) + O₂ (g) → 2CaO (s) ΔH₂ = -1269 kJ
Overall Equation
CaO (s) + CO₂ (g) → CaCO₃ (s) ΔH = ?
When the enthalpy for the overall equation is to be calculated, the intermediate reactions are summed up (this is according to Hess' Law and Bon Haber cycle),
But the sum of the intermediate reactions for this case would not give the overall reaction.
But
(Intermediate reaction 1) + ½(- Intermediate reaction 2) = (Overall reaction)
Intermediate reaction 1
Ca (s) + CO₂ (g) + ½O₂ (g) → CaCO₃ (s) ΔH₁ = -812.8 kJ
½(Intermediate reaction 2)
½[2Ca (s) + O₂ (g) → 2CaO (s) ΔH₂ = -1269 kJ]
= Ca (s) + ½O₂ (g) → CaO (s) ΔH₂' = ½ × -1269 kJ = -634.5 kJ
-½(Intermediate reaction 2)
CaO → Ca (s) + ½O₂ (g) ΔH₂" = 634.5 kJ
Ca (s) + CO₂ (g) + ½O₂ (g) → CaCO₃ (s) ΔH₁ = -812.8 kJ
CaO → Ca (s) + ½O₂ (g) ΔH₂" = 634.5 kJ
Overall Equation
CaO (s) + CO₂ (g) → CaCO₃ (s) ΔH = ?
ΔH = ΔH₁ - ½ΔH₂ = ΔH₁ + ΔH₂"
ΔH = -812.8 + 634.5 = -178.3 kJ
When the enthalpy of this overall chemical equation is calculated, the enthalpy of the second intermediate equation is halved and the sign on it reversed, to obtain the required enthalpy of overall reaction.
Hope this Helps!!!
