Question

Determine the heat of reaction (ΔHrxn) for the reaction of calcium carbonate (CaCO3) with HCl to produce CO2 by using heat of formation data:

CaCO3 (s) + 2 HCl (g) → CaCl2 (s) + CO2 (g) + H2O (g)

Answer 1

Answer: The heat of reaction (ΔHrxn) for the reaction is -164.9kJ

Explanation:

The given balanced chemical reaction is,

$CaCO_3(s)+2HCl(g)\rightarrow CaCl_2(s)+CO_2(g)+H_2O(g)$

To calculate the enthalpy of reaction $(\Delta H^o)$.

$\Delta H^o=H_f_{product}-H_f_{reactant}$

$\Delta H^o=[n_{CaCl_2}\times \Delta H_f^0_{(CaCl_2)}+n_{CO_2}\times \Delta H_f^0_{(CO_2)}+n_{H_2O}\times \Delta H_f^0_{(H_2O)}]-[n_{CaCO_3}\times \Delta H_f^0_{(CaCO_3)+n_{HCl}\times \Delta H_f^0_{(HCl)}]$

where,

$\Delta H^o_f_{(CaCO_3(s))}=-1206.9kJ/mol\\\Delta H^o_f_{(HCl(g))}=-92.30kJ/mol\\\Delta H^o_f_{(CaCl_2(s))}=-877.1kJ/mol\\\Delta H^o_f_{(H_2O(g))}=-285.8kJ/mol\\\Delta H^o_f_{(CO_2(g))}=-393.51kJ/mol$

Putting values in above equation, we get:

$\Delta H^o_{rxn}=[(1\times -877.1)+(1\times -393.51)+(1\times -285.8)]-[(1\times -1206.9)+(2\times -92.30)]=-164.9kJ$

Therefore the heat of reaction (ΔHrxn) for the reaction is -164.9kJ