Determine the enthalpy change for the decomposition of calcium carbonate. CaCO3 (s) --> CaO (s) + CO2 (g) given the thermochemical equations below:

Ca(OH)2 (s) --> CaO (s) + H2O (l) enthalpy reaction = 65.2 kJ/mol

Ca(OH)2 (s) + CO2(g) --> CaCO3 (s) + H2O (l) enthalpy reaction = -113.8 kJ/mol

C(s) + O2 (g) --> CO2 (g) enthalpy of reation = -393.5 kJ/mol

2Ca(s) + O2(g) --> 2 CaO (s) enthalpy of reaction = -1270.2 kJ/mol

a. 1711.7 kJ/mol rxn

b. 441 kJ/mol rxn

c. 179 kJ/mol rxn

d. 48 kJ/mol rxn

e. 345.5 kJ. mol rxn

Answer : The enthalpy change for the decomposition of calcium carbonate is, 178.1 kJ/mol

Explanation :

According to Hess’s law of constant heat summation, the heat absorbed or evolved in a given chemical equation is the same whether the process occurs in one step or several steps.

According to this law, the chemical equation can be treated as ordinary algebraic expression and can be added or subtracted to yield the required equation. That means the enthalpy change of the overall reaction is the sum of the enthalpy changes of the intermediate reactions.

The given main reaction is,

The intermediate balanced chemical reaction will be,

(1)    

(2)    

(3)    

(4)    

Now we are reversing reaction 1 and then adding reaction 1 and 2, we get :

(1)    

(2)    

The expression for enthalpy of change will be,

Thus, the enthalpy change for the decomposition of calcium carbonate is, 178.1 kJ/mol