Given the following data:

Question

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?

Answer 1

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.