Answer:
The Enthalpy for the reaction will be -113.2 kJ/mol
Explanation:
[tex]Ca(OH)_{2} (s) +CO_{2} (g)[/tex] → [tex]CaCO_{3}(s) +H_{2} O( l)[/tex]
We know that, Ideal Enthalpy of formation = ( Net Enthalpy of the products) -( Net Enthalpy of reactants) at 298K temperature.
So, Δ[tex]H_{f} ^{0}[/tex] = ( Δ[tex]H_{f} ^{0} _{CaCO_{3} }[/tex] +Δ[tex]H_{f} ^{0} _{H_{2}O }[/tex]) - (Δ[tex]H_{f}^{0} _{Ca(OH)_{2} }[/tex] + Δ[tex]H_{f}^{0}CO_{2}[/tex])
= (- 1207.0 - 285.8) - ( -986.1 - 393.5)
= -1492.8 +1379.6
= -113.2 kJ/mol
The enthalpy of the reaction Ca(OH)2(s)+CO2(g)→CaCO3(s)+H2O(l)
is -113.2 kJ/mol.
The enthalpy of reaction can be obtained as the difference between the enthalpy of formation of products and enthalpy of formation of reactants. Hence;
ΔHrxn = ∑ΔHproducts - ∑ΔHreactants
The equation of the reaction is;
Ca(OH)2(s)+CO2(g)→CaCO3(s)+H2O(l)
The enthalpies of formation of each specie is shown below in kJ/mol;
ΔHf∘CO2(g) = −393.5
ΔHf∘Ca(OH)2(s) = −986.1
ΔHf∘H2O(l) = −285.8
ΔHf∘CaCO3(s) = −1207.0
Hence;
ΔHrxn = [(−1207.0) + (−285.8)] - [(−986.1) + (−393.5)]
ΔHrxn = -113.2 kJ/mol
Learn more: https://brainly.com/question/11897796
