calculate ΔH° for the reaction: 2SO2(g) + O2(g) → 2SO3(g) Table: ΔH° (kJmol) S(s) + O2(g) → SO2(g) = -297 2S(s) + 3O2(g) → 2SO3(g) = -792?

S(s) + O2(g) → SO2(g) = -297
2S(s) + 3O2(g) → 2SO3(g) = -792
Supposing most of the lacking products is “kJ”:
You’ll need “2SO2(g)” on remaining. Really the only offered equation that mentions SO2(g) could be the very first one. Therefore increase initial offered equation by 2, after that compose it backwards:
2 SO2(g) → 2 S(s) + 2 O2(g), ΔH= +594 kJ
Additionally you require “2SO3(g)” on right. Really the only offered equation that mentions SO3(g) could be the 2nd one. Therefore copy the next given equation:
2 S(s) + 3 O2(g) → 2 SO3(g), ΔH= -792 kJ
Include the final two equations right here:
2 SO2(g) + 2 S(s) + 3 O2(g) → 2 S(s) + 2 O2(g) + 2 SO3(g), ΔH= +594 kJ -792 kJ
Cancel like quantities on contrary edges associated with the arrow, and perform some arithmetic for ΔH:
2 SO2(g) + O2(g) → 2 SO3(g), ΔH= -198 kJ

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