Question

Calculate the standard enthalpy change for the following reaction involving liquids at 298 K: C2H5OH(l) + 3O2(g) → 2CO2(g) + 3H2O(l)Given the standard enthalpies of formation of C2H5OH(l), CO2(g) and H2O(l) are −277.6 kJ/mol, −393.5 kJ/mol and −285.8 kJ/mol respectively.

Answer 1

To calculate the standard enthalpy change for the reaction, we can use the following formula:H =  Hf products  -  Hf reactants where H is the standard enthalpy change, and Hf is the standard enthalpy of formation.For the given reaction:C2H5OH l  + 3O2 g   2CO2 g  + 3H2O l The standard enthalpies of formation are:C2H5OH l : -277.6 kJ/molCO2 g : -393.5 kJ/molH2O l : -285.8 kJ/molNow, we can plug these values into the formula: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 since 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 kJTherefore, the standard enthalpy change for the reaction is -1366.8 kJ.