Given these values of ∆H°?

Question

CS2 (l) + 3O2(g) -> CO2 (g) + 2SO2 (g) ∆H° = -1077 KJ

H2(g) + O2 (g) -> H2O2 (l) ∆H° = -188 kj

H2(g) + 1/2 O2 (g) -> ∆H° = -286 kj

What is the value of ∆H° for this reaction?

CS2 (l) + 6H2O2 (L) -> CO2 (g) + 6H2O(l) + 2SO2 (g)

Answer 1

The steps of reactions :

CS2 + 3O2 --> CO2 + 2SO2

6H2O2 --> 6H2 + 6O2

6H2 + 3O2 --> 6H2O

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CS2 + 6H2O2 --> CO2 + 6H2O + 2SO2

Therefore, DH = -1077 kJ + 6(188 kJ) + 6(-286 kJ)

= -1077 kJ + 1128 kJ - 1716 kJ

= -1665 kJ