When 9.41 g of phenol, C6H5OH, are burned in a bomb calorimeter at 25 oC, 305.1 kJ of heat are given off. How to calculate the standard enthalpy of combustion, in kJ/mol, of phenol?
The standard enthalpy of combustion of phenol is
You can solve this problem without the balanced chemical equation and without standard enthalpies of formation, all you have to do is use the heat absorbed by the water.
So, you're burning phenol,
The actual heat given off by the combustion of phenol will be equal to this value but have a negative sign - heat released, not absorbed
This is how much heat is given off in the combustion of
You need to determine exactly how many moles of phenol you've burned
This means that you give off -305.1 kJ when 0.100 moles are burned; therefore, when 1 mole is burned you'll give off 10 times as much heat
SInce this value represents the heat given off per mole, you can write it as
SIDE NOTE Rounded to 3 sig figs, the answer wil be -3050 kJ/mol.
I assume the pressure to be at 1 atm, since standard state implies a pressure of 1 atm and a temperature of
Here is a video which discusses how to calculate the enthalpy change when 0.13g of butane is burned.
Video from: Noel Pauller