Given the following reactions and subsequent ∆H values what is the ∆H for the reaction below?
Zn(s) + 1/8S8(s) + 2O2(g) → ZnSO4(s)
Given:
Zn(s) + 1/8S8(s) → ZnS(s)
∆H = -183.92kJ
2ZnS(s) + 3O2(g) → 2ZnO(s) + 2SO2(g)
∆H = -927.54kJ
2SO2(g) + O2(g) → 2SO3(g)
∆H = -196.04kJ
ZnO(s) + SO3(g) → ZnSO4(s)
∆H = -230.32kJ
Answers options:
A) 244.01 kJ
B) 488.02 kJ
C) -488.02 kJ
D) -976.03 kJ
E) 976.03 kJ
