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

Step-by-step 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)* \Delta H_f^0_((CaCl_2))+n_(CO_2)* \Delta H_f^0_((CO_2))+n_(H_2O)* \Delta H_f^0_((H_2O))]-[n_(CaCO_3)* \Delta H_f^0_{(CaCO_3)+n_(HCl)* \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* -877.1)+(1* -393.51)+(1* -285.8)]-[(1* -1206.9)+(2* -92.30)]=-164.9kJ`

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