Answer:
Option C. 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
Temperature (T) =..?
Next, we shall determine the enthalphy change (ΔH).
This is illustrated:
CaCO3 —> CaO + CO2
CaCO3 = -1207 kJ/mol
CaO = -635 kJ/mol
CO2 = -394 kJ/mol
Heat of product (Hp) = -635 + -394 =
- 1029 KJ
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 change in entropy (ΔS).
This is illustrated below:
CaCO3 —> CaO + CO2
Entropy (S) data:
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
Change in entropy (ΔS) = Sp – Sr
Change in entropy (ΔS) = 254 – 93
Change in entropy (ΔS) = 161 J/Kmol
Finally, we shall determine the temperature (T) at which the reaction was feasible.
This is illustrated below:
Enthalphy change (ΔH) = 178 KJ/mol = 178000 J/mol
Change in entropy (ΔS) = 161 J/Kmol
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 was feasible is 1105.6 K