Question

Calculate the standard enthalpy change for the reaction of 2 moles of liquid ethanol (C2H5OH) reacting with 1 mole of gaseous oxygen (O2) to form 2 moles of gaseous carbon dioxide (CO2) and 3 moles of liquid water (H2O) at a temperature of 298K. Assume the reactants and products are at standard state conditions.

Answer 1

To calculate the standard enthalpy change for the reaction, we can use the following equation:H =  Hf products  -  Hf reactants where H is the standard enthalpy change, and Hf is the standard enthalpy of formation for each substance.First, we need to find the standard enthalpy of formation for each substance involved in the reaction:C2H5OH  l : -277.69 kJ/molO2  g : 0 kJ/mol  since it is in its standard state CO2  g : -393.51 kJ/molH2O  l : -285.83 kJ/molNow, we can plug these values into the equation:H = [2 -393.51  + 3 -285.83 ] - [2 -277.69  + 1 0 ]H = [-787.02 - 857.49] - [-555.38 + 0]H = -1644.51 + 555.38H = -1089.13 kJThe standard enthalpy change for the reaction is -1089.13 kJ.