Graphite example from lecture

[Olivia D 3B]

During lecture, Dr. Lavelle used the example of Graphite (most stable form of C) and gaseous H2 and 02 to form Ethanol. The standard reaction enthalpy was found to be -555.38kJ. This was then used to find the standard enthalpy of formation to be -277.69kJ.

Could someone walk me through the reasoning behind why the -555.38kJ was divided by 2 to obtain the -277.69kJ?

[Anna Wilborn-1L]

All the other molecules in the equation are in their standard state so ethanol is the only molecule contributing to the enthaply of the reaction. So enthaply of the reaction is equal to the enthalpy of ethanol in the reaction but as seen in the balanced equation, there was two moles of ethanol in the reaction so the enthalpy of ethanol is the enthaply of the reaction divided by the amount of moles of ethanol or -555.38 KJ divided by 2 moles which is -277.69 KJ per mole.