To calculate the ΔH° for the reaction S(s) + O2(g) → SO2(g), we can use the given enthalpy information for the other two reactions and manipulate them to match the desired reaction.
First, let's reverse the first reaction and change the sign of its enthalpy value:
SO2(g) → S(s) + O2(g)
ΔH° = +395 kJ
Since we want the equation S(s) + O2(g) → SO2(g), we need to add these two reactions together. To do that, we can multiply the first reaction by 2 so that the moles of SO2 cancel out:
2(SO2(g) → S(s) + O2(g)) (1st reaction * 2)
+ (2SO2(g) + O2(g) → 2SO3(g)) (3rd reaction)
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2S(s) + 2O2(g) → 2SO2(g) + 2SO3(g)
Now, we can sum up the enthalpy values of the two reactions:
(2 times the enthalpy of the 1st reaction) + (enthalpy of the 3rd reaction) = ΔH°
2 × (+395 kJ) + (-198.2 kJ) = ΔH°
790 kJ - 198.2 kJ = ΔH°
591.8 kJ = ΔH°
Therefore, the ΔH° for the reaction S(s) + O2(g) → SO2(g) is 591.8 kJ.