Question

Given enthalpy of formation of CO2(g) and CaO(s) are -94.0KJ and -152KJ respectively and the enthalpy of the reaction : CaCO3(s) → CaO(s)+CO2(g)
is 42 KJ. The enthalpy of formation of CaCO3 (s) is ?

Answer 1

The enthalpy of reaction is the difference in enthalpy of product to enthalpy of reactant as follows:

$dH_rxn = dH_ product– dH_ reactant$

where, $dH_rxn$is the enthalpy of reaction

$dH_ product$ is enthalpy of product

$dH_ reactant$  is enthalpy of reactant

$dH_C_a_O = -152 kJ$

$dH_C_O_2 = -94 kJ$

$= (dH_C_a_O + dH_C_O_2) – (dH_C_a_C_O_3)$

$42 = (-152 - 94) - (dH_C_a_C_O_3)$

$(dH_C_a_C_O_3) = - 204 kJ /mol$

Thus, enthalpy of formation of CaCO3 is – 204 kJ /mol