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) --> 2CaO(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: c. 179 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 Hess’s law, the chemical equation can be treated as algebraic expressions 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.

Given:

$Ca(OH)_2(s)\rightarrow CaO(s)+H_2O (l)$   $\Delta H_1= 65.2 kJ/mol$     (1)

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

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

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

On subtracting eq (1) from eq (2) we have:

$Ca(OH)_2(s)\rightarrow CaO(s)+H_2O(l)$ - $Ca(OH)_2(s)+CO_2(g)\rightarrow CaCO_3(s)+H_2O(l)$

$CaCO_3(s)\rightarrow CaO(s)+CO_2(g)$

$\Delta H=-113.8-(65.2)kJ/mol=-179kJ/mol$

Hence the enthalpy change for the raection is 179.0 kJ/mol.