Question

Calculate the standard enthalpy change for the following reaction at 25°C given the enthalpies of formation (in kJ/mol) of the compounds involved:CaCO3(s) + 2HCl(aq) → CaCl2(aq) + CO2(g) + H2O(l)Hf(CaCO3) = -1206.9 kJ/molHf(CaCl2) = -795.8 kJ/molHf(CO2) = -393.5 kJ/molHf(H2O) = -285.8 kJ/molHf(HCl) = -92.31 kJ/mol

Answer 1

To calculate the standard enthalpy change for the reaction, we can use the following formula:H =  [Hf products ] -  [Hf reactants ]For the products:Hf CaCl2  = -795.8 kJ/molHf CO2  = -393.5 kJ/molHf H2O  = -285.8 kJ/molFor the reactants:Hf CaCO3  = -1206.9 kJ/molHf HCl  = -92.31 kJ/mol  there are 2 moles of HCl, so multiply by 2 Now, plug the values into the formula:H = [ -795.8  +  -393.5  +  -285.8 ] - [ -1206.9  + 2   -92.31 ]H =  -1475.1  -  -1391.52 H = 83.58 kJ/molThe standard enthalpy change for the reaction is 83.58 kJ/mol.