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    Hi
    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.

    Please help
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    1/2S2 + O2 -> SO2
    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)
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    The enthalpy of formation of SO2 concerns the 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
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    Yeah, sorry... if sulphur is in solid state, it's S8, and mass is 1g.
    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)
 
 
 

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