Given the following reactions , δh = -790 kj , δh = -297 kj the enthalpy of the reaction in which sulfur dioxide is oxidized to sulfur trioxide is kj.

ΔH = -196 kJ

Explanation:

To do this, we need to apply the Hess's law, which states the following:

"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."

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

you gotta add the following molecules so we can give u an answer

explanation: