Question

Calculate the standard enthalpy change for the reaction: C2H5OH(l) + 3 O2(g) → 2 CO2(g) + 3 H2O(l), given the following standard enthalpies of formation: ΔHf°[C2H5OH(l)] = -277.69 kJ/mol, ΔHf°[CO2(g)] = -393.51 kJ/mol, ΔHf°[H2O(l)] = -285.83 kJ/mol.

Answer 1

To calculate the standard enthalpy change for the reaction, we can use the formula:H reaction  =  Hf products  -  Hf reactants For the given reaction:C2H5OH l  + 3 O2 g   2 CO2 g  + 3 H2O l The standard enthalpies of formation for the products are:Hf[CO2 g ] = -393.51 kJ/mol  2 moles Hf[H2O l ] = -285.83 kJ/mol  3 moles The standard enthalpies of formation for the reactants are:Hf[C2H5OH l ] = -277.69 kJ/mol  1 mole Hf[O2 g ] = 0 kJ/mol  3 moles   since O2 is in its standard state Now, we can plug these values into the formula:H reaction  = [ 2  -393.51  +  3  -285.83 ] - [ -277.69  +  3  0 ]H reaction  = [ -787.02  +  -857.49 ] - [ -277.69 ]H reaction  = -1644.51 + 277.69H reaction  = -1366.82 kJ/molThe standard enthalpy change for the reaction is -1366.82 kJ/mol.