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
A. 1780 K
B.17 K
C. 1105.6 K
OD. 1.1K
Reset Selection

Option C. 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

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

catya that's not even a b or c