Question

Calculate the standard enthalpy change for the reaction of the combustion of 1 mole of liquid ethanol (C2H5OH) at 298K and 1 atm. The balanced chemical equation for the reaction is:  C2H5OH(l) + 3O2(g) → 2CO2(g) + 3H2O(l)Given:ΔHf°(C2H5OH(l)) = -277.7 kJ/molΔHf°(CO2(g)) = -393.5 kJ/molΔHf°(H2O(l)) = -285.8 kJ/molΔHf°(O2(g)) = 0 kJ/mol

Answer 1

To calculate the standard enthalpy change for the reaction, we can use the following formula:H reaction  =  Hf products  -  Hf reactants For the products, we have 2 moles of CO2 g  and 3 moles of H2O l : Hf products  =  2  -393.5 kJ/mol  +  3  -285.8 kJ/mol  = -787.0 kJ/mol + -857.4 kJ/mol = -1644.4 kJ/molFor the reactants, we have 1 mole of C2H5OH l  and 3 moles of O2 g : Hf reactants  =  -277.7 kJ/mol  +  3  0 kJ/mol  = -277.7 kJ/molNow, we can calculate the standard enthalpy change for the reaction:H reaction  = -1644.4 kJ/mol -  -277.7 kJ/mol  = -1366.7 kJ/molThe standard enthalpy change for the combustion of 1 mole of liquid ethanol at 298K and 1 atm is -1366.7 kJ/mol.