Question

Calculate the enthalpy change (in kJ/mol) for the chemical reaction: C2H5OH(l) + 3O2(g) → 2CO2(g) + 3H2O(l) Given the standard enthalpies of formation (ΔHf°) for C2H5OH(l) = -277.6 kJ/mol, CO2(g) = -393.5 kJ/mol, and H2O(l) = -285.8 kJ/mol.

Answer 1

To calculate the enthalpy change  H  for the given reaction, we can use the formula:H = nHf products  - nHf reactants where n is the stoichiometric coefficient of each species in the reaction, and Hf is the standard enthalpy of formation of each species.For the given reaction:C2H5OH l  + 3O2 g   2CO2 g  + 3H2O l The enthalpy change  H  can be calculated as follows:H = [2   -393.5 kJ/mol  + 3   -285.8 kJ/mol ] - [ -277.6 kJ/mol  + 3  0 kJ/mol]Note that the standard enthalpy of formation for O2 g  is 0 kJ/mol because it is in its standard state.H = [-787.0 kJ +  -857.4 kJ ] - [-277.6 kJ]H = -1644.4 kJ + 277.6 kJH = -1366.8 kJ/molSo, the enthalpy change for the given reaction is -1366.8 kJ/mol.