Question

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 1

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,

$CaCO_3(s)\rightarrow CaO(s)+CO_2(g)$    $\Delta H=?$

The intermediate balanced chemical reaction will be,

(1) $Ca(OH)_2(s)\rightarrow CaO(s)+H_2O(l)$     $\Delta H_1=65.02kJ/mol$

(2) $Ca(OH)_2(s)+CO_2(g)\rightarrow CaCO_3(s)+H_2O(l)$    $\Delta H_2=-113.8kJ/mol$

(3) $C(s)+O_2(g)\rightarrow CO_2(g)$    $\Delta H_3=-393.5kJ/mol$

(4) $2Ca(s)+O_2(g)\rightarrow 2CaO(s)$    $\Delta H_4=-1270.2kJ/mol$

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

(1) $Ca(OH)_2(s)\rightarrow CaO(s)+H_2O(l)$     $\Delta H_1=65.02kJ/mol$

(2) $CaCO_3(s)+H_2O(l)\rightarrow Ca(OH)_2(s)+CO_2(g)$    $\Delta H_2=113.8kJ/mol$

The expression for enthalpy of change will be,

$\Delta H=\Delta H_1+\Delta H_2$

$\Delta H=(65.02)+(113.08)$

$\Delta H=178.1kJ/mol$

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