Question

What is the standard enthalpy change (ΔH°) for the reaction of 1 mole of liquid ethanol (C2H5OH) and 1 mole of liquid oxygen (O2) to produce 2 moles of gaseous carbon dioxide (CO2) and 3 moles of liquid water (H2O)?

Answer 1

To calculate the standard enthalpy change  H  for the reaction, we need to use the following formula:H =  Hf products  -  Hf reactants where Hf represents the standard enthalpy of formation of each substance.First, let's write the balanced chemical equation for the reaction:C2H5OH  l  + 3 O2  g   2 CO2  g  + 3 H2O  l Now, we need the standard enthalpies of formation  Hf  for each substance:Hf  C2H5OH, l  = -277.7 kJ/molHf  O2, g  = 0 kJ/mol  since O2 is in its standard state Hf  CO2, g  = -393.5 kJ/molHf  H2O, l  = -285.8 kJ/molNow, we can plug these values into the formula:H = [2   -393.5  + 3   -285.8 ] - [-277.7 + 3  0]H = [-787 +  -857.4 ] -  -277.7 H = -1644.4 + 277.7H = -1366.7 kJ/molSo, the standard enthalpy change  H  for the reaction is -1366.7 kJ/mol.