Answer:
Explanation:
We are going to manipulate the equations:
SO2(g) → S(s) + O2(g) ΔH°rxn = +296.8
2 SO2(g) + O2(g) → 2 SO3(g) ΔH°rxn = -197.8 kJ
So that after we add them together we will get the desired equation:
4 SO3(g) → 4 S(s) + 6 O2(g) ΔH°rxn = ?
The ΔH°rxn will then be added according to Hess´s law to get our answer.
Notice that S has a coefficient of 4, and it only appears in the first as 1 mole. Therefore we will multiply it by 4. Also, notice in the final equation we want the sulfur as product, so we need to multiply by 4 the first equation:
4 SO2(g) → 4 S(s) + 4 O2(g) ΔH°rxn =( +296.8 kJ ) x 4 = 1,187.20 kJ
Working with the second equation, we can see that if we multiply this equation by 2 and reverse it, when we add it to the previously manipulated equation, we can cancel the SO2s which are not present in our desired reaction:
4 SO3(g) → 4 SO2(g) + 4 O2(g) ΔH°rxn = ( + 197.8 kJ) x 2 = 395.6 kK
adding our worked equations:
4 SO2(g) → 4 S(s) + 4 O2(g) ΔH°rxn = 1187.2 kJ
4 SO3(g) → 4 SO2(g) + 2 O2(g) ΔH°rxn = 395.6 kJ
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4 SO3(g) → 4 S(s) + 6 O2(g) ΔH°rxn = 1582.8 kJ
