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I have answered this question and got a value of $\mathrm {1409 \space kJ\ mol^{-1}}$ however, the answer sheet says $\mathrm{-1009}$.

Using the equation: $$\ce{C2H4 + 3O2 -> 2CO2 + 2H2O}$$

I ended up with: $$\text{Enthalpy of combustion of ethene}+52 = (2\times-285.5) + (2\times-393)$$

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    $\begingroup$ Can just explain the question a little more. $\endgroup$ – sedflix Apr 2 '15 at 12:14
  • $\begingroup$ Hi Matt, your question title leads me to think you would include the values of enthalpy of formation in the question. Would you be able to do an edit and include this? $\endgroup$ – John Snow Apr 3 '15 at 0:29
  • $\begingroup$ The answer sheet is incorrect. Your method is sound and it agrees with the published values I could find within a few percent. Just to be clear, this assumes that the reaction is happening at standard conditions. $\endgroup$ – Jason Patterson Jun 2 '15 at 3:39
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Apply Hess's law: enter image description here

Then make an equation by looking at above working: enter image description here

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