Question

Calculate the standard enthalpy change for the reaction of 2 moles of liquid ethanol with 3 moles of gaseous oxygen to form 2 moles of liquid acetic acid and 3 moles of gaseous water at 1 atm and 25°C, given the standard enthalpy of formation of ethanol, acetic acid, and water are -277.6 kJ/mol, -487.5 kJ/mol, and -285.8 kJ/mol respectively.

Answer 1

To calculate the standard enthalpy change for the reaction, we can use the 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:2 C2H5OH  l  + 3 O2  g   2 CH3COOH  l  + 3 H2O  g The standard enthalpy of formation for the products is: 2 moles CH3COOH  -487.5 kJ/mol  +  3 moles H2O  -285.8 kJ/mol The standard enthalpy of formation for the reactants is: 2 moles C2H5OH  -277.6 kJ/mol  +  3 moles O2  0 kJ/mol   since the standard enthalpy of formation for an element in its standard state is zero Now, we can calculate the standard enthalpy change:H = [  2  -487.5  +  3  -285.8  ] - [  2  -277.6  +  3  0  ]H = [  -975  +  -857.4  ] - [  -555.2  ]H = -1832.4 + 555.2H = -1277.2 kJThe standard enthalpy change for the reaction is -1277.2 kJ.