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1. This is a question that I am finding challenging and I am unaware if this is the correct working or not.

Use the Table of bond enthalpies and the following data
2SO2 (g) + O2 (g) → 2SO3 (g)
ΔH = -192kJ mol-1

Assume that SO2 and SO3 contain only S=O bonds
Calculate the average bond enthalpy of the S=O bond.

Taking S=O as x and using the table data, O=O as 498kJ mol-1

4x + 498 - 6x = -192
498 - 2x = -192
690 = 2x
345 = x

So I thought that S=O was 345kJ mol-1
However the answers and tables I have searched provided different values.

2. (Original post by nwmyname)
This is a question that I am finding challenging and I am unaware if this is the correct working or not.

Use the Table of bond enthalpies and the following data
2SO2 (g) + O2 (g) → 2SO3 (g)
ΔH = -192kJ mol-1

Assume that SO2 and SO3 contain only S=O bonds
Calculate the average bond enthalpy of the S=O bond.

Taking S=O as x and using the table data, O=O as 498kJ mol-1

4x + 498 - 6x = -192
498 - 2x = -192
690 = 2x
345 = x

So I thought that S=O was 345kJ mol-1
However the answers and tables I have searched provided different values.

I agree with you.
I think the discrepancy in the value is because of this statement:
Calculate the average bond enthalpy of the S=O bond.

Bond enthalpies are averaged over a range of compounds. The S=O bond in SO2 will probably have a lower bond enthalpy (or higher, unsure) than the S=O bond in SO3.
3. You must have been given ∆rH per mole of SO2, and in the equation given you’re reacting 2 moles of SO2. So your equation should be:

498 - 2x = -2*192

I reckon it's a bit of a slippery question really if you weren’t told that explicitly as you could (and did!) interpret it as ‘per mole of O2 reacting’ but that would explain why you’re getting the wrong number.

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