Question

Calculate the standard enthalpy change (∆H°) for the following reaction involving liquids:C6H12O6 (s) + 6O2 (g) → 6CO2 (g) + 6H2O (l)

Answer 1

To calculate the standard enthalpy change  H  for the given reaction, we need to use the following formula:H =  Hf products  -  Hf reactants Here, Hf refers to the standard enthalpy of formation of each substance involved in the reaction. The standard enthalpy of formation values for the substances involved in the reaction are:C6H12O6  s : -1273 kJ/molO2  g : 0 kJ/mol  since it is an element in its standard state CO2  g : -393.5 kJ/molH2O  l : -285.8 kJ/molNow, we can plug these values into the formula:H = [6 -393.5  + 6 -285.8 ] - [-1273 + 6 0 ]H = [-2361 - 1714.8] - [-1273]H = -4075.8 + 1273H = -2802.8 kJ/molSo, the standard enthalpy change for the given reaction is -2802.8 kJ/mol.