Calculate ΔHºrxn for the reaction

Question

3 O2(g) ==> 2 O3(g)

This is the information I am given:
2H2S(g) + SO2g) ==> 3S(s) + 2H2O(l)

Answer 1

ugh my bad the information I'm given is:
O2 ==> 2 O(g)ΔHºrxn=498kJ/mol
O(g) + O2(g) ==> O3ΔHºrxn = /106kJ/mol

I think what I'm supposed to do is change the two reactions I'm given by either reversing them or multiplying by a coefficient, then doing the same thing to the enthalpy, but I'm confused about how i go about doing that

Answer 2

Multiply equation 1 by 1 and equation 2 by 2, add them together and add the delta H for each. (When you add the two multiplied equations, check to make sure the items you don't want cancel and leaves the equation desired.)

Answer 3

1 *(O2 ==> 2 O(g)ΔHºrxn=498kJ/mol)
2 * (O(g) + O2(g) ==> O3ΔHºrxn = -106kJ/mol)
O2 + 2O + 2O2 ==> 2O + 2O3
3O2 ==> 2O3
498-212=286ΔHºrxn
is that right? and if so, how did you know which numbers to multiply each equation by?

Answer 4

1 *(O2 ==> 2 O(g)ΔHºrxn=498kJ/mol)
2 * (O(g) + O2(g) ==> O3ΔHºrxn = -106kJ/mol)
O2 + 2O + 2O2 ==> 2O + 2O3
3O2 ==> 2O3
498-212=286ΔHºrxn
is that right? and if so, how did you know which numbers to multiply each equation by?

Yes, that number is correct.
How to know what to multiply by.
Look at the final equation. We MUST have 2O3 on the right side and the ONLY way to get that is to multiply the second equation by 2. So I did that, added the two equations to see what I had; lo and behold, the final equation added up to what I wanted. (To emphasize the point, I should point out that wasn't my first try. FIRST, I said I needed 3 O2 and 2O3 so I FIRST multiplied equation 1 by 3 and equation 2 by 2 and added. Of course, I had too many O2 (because some comes from both 1 and 2) AND I didn't cancel all the O atoms. So I regrouped and tried the next approach. It worked. :)]