At what temperature is the following reaction feasible: CaCO3 -> CaO + CO2? Enthalpy data:
CaCO3: -1207 kJ/mol
CaO: -635 kJ/mol
CO2: -394 kJ/mol
Entropy data:
CaCO3: +93 J/K mol
CaO: +40 J/K mol
CO2:+214 J/K mol

1105.6 K

Explanation:

The following data were obtained from the question:

CaCO3 —> CaO + CO2

Enthalpy (H) data:

CaCO3 = -1207 kJ/mol

CaO = -635 kJ/mol

CO2 = -394 kJ/mol

Entropy (S) data:

CaCO3 = +93 J/K mol

CaO = +40 J/K mol

CO2 = +214 J/K mol

Next, we shall determine the enthalphy change (ΔH). This can be obtained as follow:

CaCO3 —> CaO + CO2

CaCO3 = -1207 kJ/mol

CaO = -635 kJ/mol

CO2 = -394 kJ/mol

Heat of product (Hp) = -635 + (-394) = - 1029 KJ/mol

Heat of reactant (Hr) = -1207 kJ/mol

Enthalphy change (ΔH) = Hp – Hr

Enthalphy change (ΔH) = -1029 – (-1207)

Enthalphy change (ΔH) = 178 KJ/mol

Next, we shall determine the entropy change (ΔS). This can be obtained as follow:

CaCO3 —> CaO + CO2

CaCO3 = +93 J/K mol

CaO = +40 J/K mol

CO2 = +214 J/K mol

Entropy of product (Sp) = 40 + 214 = +254 J/Kmol

Entropy of reactant (Sr) = +93 J/Kmol

Entropy change (ΔS) = Sp – Sr

Entropy change (ΔS) = 254 – 93

Entropy change (ΔS) = + 161 J/Kmol

Finally, we can obtain the temperature at which the reaction is feasible as follow:

Enthalphy change (ΔH) = 178 KJ/mol = 178000 J/mol

Entropy change (ΔS) = + 161 J/Kmol

Temperature (T) =..?

Entropy change (ΔS) = Enthalphy change (ΔH) / Temperature (T)

ΔS = ΔH/T

161 = 178000/T

Cross multiply

161 x T = 178000

Divide both side by 161

T = 178000/161

T = 1105.6 K

Therefore, the temperature at which the reaction is feasible is 1105.6 K

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yes

explanation:

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