Answer:
1105.6 K
Step-by-step 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