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Calculating unknown bond enthalpies

Please could somebody help me with an explanation of how to work out this question. Been trying for ages and I can't get the same answer as the text book...

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The only data relevant in table 1 is the O=O avg. bond enthalpy being 498kj per mol.

Thanks
Original post by nathanjeffs
Please could somebody help me with an explanation of how to work out this question. Been trying for ages and I can't get the same answer as the text book...

Attachment not found


The only data relevant in table 1 is the O=O avg. bond enthalpy being 498kj per mol.

Thanks


Bond breaking in the reactants is positive (endothermic) and bond formation in the products is negative (exothermic).


You have all of the bond enthalpies, do the sums.
Reply 2
Original post by charco
Bond breaking in the reactants is positive (endothermic) and bond formation in the products is negative (exothermic).


You have all of the bond enthalpies, do the sums.


Thanks for your help. Like I said I have attempted the sums, this being as far as I got...

Bonds broken: 4 x S=O bonds, 1 x O=O (enthalpy 498kj/mol)
Bonds formed: 6 x S=O bonds

Enthalpy change = energy to break bonds - energy to form bonds = -192kj/mol

so: (4 x S=O + 498) - 6 x S=O = -192kj/mol

-2 S=O = -690kj/mol giving me an average bond enthalpy of 345kj/mol where the actual answer is 469kj/mol. So if anybody could please give me the correct method to solve this it would be much appreciated.
Original post by nathanjeffs
Thanks for your help. Like I said I have attempted the sums, this being as far as I got...

Bonds broken: 4 x S=O bonds, 1 x O=O (enthalpy 498kj/mol)
Bonds formed: 6 x S=O bonds

Enthalpy change = energy to break bonds - energy to form bonds = -192kj/mol

so: (4 x S=O + 498) - 6 x S=O = -192kj/mol

-2 S=O = -690kj/mol giving me an average bond enthalpy of 345kj/mol where the actual answer is 469kj/mol. So if anybody could please give me the correct method to solve this it would be much appreciated.


According to the data you are given your answer is correct.

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