Question

Calculate the standard enthalpy change (∆H°) for the reaction: 2HCl(aq) + Na2CO3(aq) → CO2(g) + H2O(l) + 2NaCl(aq) given that the standard enthalpy of formation (∆H°f) for NaCl(aq), H2O(l) and CO2(g) are -407.3 kJ/mol, -285.8 kJ/mol and -393.5 kJ/mol respectively.

Answer 1

To calculate the standard enthalpy change  H  for the reaction, we can use the following equation:H =  Hf products  -  Hf reactants For the given reaction:2HCl aq  + Na2CO3 aq   CO2 g  + H2O l  + 2NaCl aq We have the standard enthalpy of formation  Hf  for NaCl aq , H2O l , and CO2 g  as -407.3 kJ/mol, -285.8 kJ/mol, and -393.5 kJ/mol, respectively. First, we need to find the Hf for the reactants and products:Reactants:2HCl aq : 2 *  -167.2 kJ/mol  = -334.4 kJ/mol  Note: Hf for HCl aq  is -167.2 kJ/mol Na2CO3 aq : -1130.7 kJ/mol  Note: Hf for Na2CO3 aq  is -1130.7 kJ/mol Total Hf for reactants = -334.4 kJ/mol +  -1130.7 kJ/mol  = -1465.1 kJ/molProducts:CO2 g : -393.5 kJ/molH2O l : -285.8 kJ/mol2NaCl aq : 2 *  -407.3 kJ/mol  = -814.6 kJ/molTotal Hf for products = -393.5 kJ/mol +  -285.8 kJ/mol  +  -814.6 kJ/mol  = -1493.9 kJ/molNow, we can calculate the standard enthalpy change  H  for the reaction:H =  Hf products  -  Hf reactants H =  -1493.9 kJ/mol  -  -1465.1 kJ/mol  = -28.8 kJ/molTherefore, the standard enthalpy change  H  for the given reaction is -28.8 kJ/mol.