Answer:
ΔH = -196 kJ
Explanation:
To do this, we need to apply the Hess's law, which states the following:
<em>"The heat of any reaction
ΔHf° for a specific reaction is equal to the sum of the heats of reaction for any set of reactions which in sum are equivalent to the overall reaction."</em>
This means that if we have the reaction:
2SO2 + O2 ------> 2SO3
We need to write and rearrange the given reactions, in order, to give this final reaction.
For the first reaction we have:
2S(s) + 3O2(g) -----> 2 SO3(g) ΔH = -790 kJ
As we can see, we have the SO3 as product, (Like the overall reaction), so by logic, we should rearrange the second reaction:
S(s) + O2(g) -----> SO2(g)
To rearrange, we only reverse the reaction and then, double the coefficients. Doing this, we have the following:
2SO2(g) ----> 2 S(s)+ 2 O2(g) 2(ΔH = 297kJ)
As the reaction is reversed, the sign of ΔH is reversed too.
Now, putting both reactions:
2S(s) + 3O2(g) -----> 2SO3(g)
2SO2(g) ----> 2S(s)+ 2O2(g)
Simplyfing we have the overall reaction (Sulfur cancels, and O2 is substrated, leaving one molecule of O2).
2SO2(g) + O2(g) --> 2SO3(g)
So all we have to do is sum the enthalpy of reaction of both equations, and the result would be the enthalpy of reaction of the sulfur trioxide:
ΔHrxn = -790 kJ + 2(297 kJ) = -196 kJ
, ΔH = -297 kJ The enthalpy of the reaction in which sulfur dioxide is oxidized to sulfur trioxide 
is ________ kJ.