i have been asked to calculate the standard enthalpies of formatino of sulphur dioxide, on burning in excess oxygen under standard conditions 1g of sulphur evolves 9.28KJ.
Number of mole of SO2 = 2nS2 = 2*1/64 = 1/32.
So dH = -9.28/(1/32) = -9.82*32 kJ/mol (because it's exothermic reaction)
(1/8)S8 + O2 --> SO2
while the enthalpy of combusion of sulphur concerns the reaction
S8 + 8O2 --> 8SO2
The two reactions are in a 1:8 ratio, so will their enthalpies. So if you find how many moles of S8 you have, you then know the combustion enthalpy for that many moles. From there, you can find the combustion enthalpy for one mole of S8. Then convert that by the correct ratio for the enthalpy of formation of SO2, and you've got your enthalpy of formation
If Sulphur is in gas state, it should be volume of gas.
It'd better if we solve like that
number of moles of SO2 = number of moles of S = 1/32.
So the results are the same, that makes me confused ....
Ah, right ... if S2 is used instead of S8, energy released maybe smaller (different condition)