Chemistry
posted by Eddie .
Given the following data:
S(s) + 3/2 O2(g)>SO3(g) ∆H = 395.2 kJ
2SO2(g) + O2(g) >2SO3(g)∆H = 198.2 kJ
Calculate ∆H for the reaction
S(s) + O2(g) > SO2(g)
Could someone please tell me how you would go about doing this type of problem?

Multiply eqn 1 by 2. Add the reverse of eqn 2. This give you twice the equation you want so divide everything by 2.
For dH. dH 1 + (dH 2) = 2*dH total and divide by 2.