Calculate the standard enthalpy change for the following reaction.

[From: ] [author: ] [Date: 12-03-17] [Hit: ]

Nowyou use Hess Law.Heat of reaction =Sum of heats of formation of products - sum of heats of formation of reactants.Heat of reaction = [2 x -395.7] - [2 x -296.Heat of reaction = -791.4 kJ - (-593.......

2SO(2)g +O(2)g ---> 2SO(3)g

No clue where to start. I have about 10 questions just like this one, so the most information you can give on how you solve this would be greatly appreciated; so I can try to reproduce it on the others.

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You need to use your table of heats of formation. The heat of formation for SO3 is -395.7 kJ/mol and the heat of formation for SO2 is -296.8 kJ/mol

Nowyou use Hess' Law. Heat of reaction = Sum of heats of formation of products - sum of heats of formation of reactants. (Do not include elements)

Heat of reaction = [2 x -395.7] - [2 x -296.8]

Heat of reaction = -791.4 kJ - (-593.6)

Heat of reaction = -791.4 kJ + 593.6 kJ

Heat of reaction = -197.8 kJ

Now that is for two moles of SO3 formed. So for one mole of SO3 it would be 1/2 that. OK?

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