Question

3. Given the following equations, calculate the ΔH of reaction in which sulfur dioxide (SO2) is reduced by hydrogen disulfide (H2S) to give elemental sulfur and water (H2O).

Answer 1

We are given the enthalpy of formation of 3 compounds; sulfur dioxide (SO₂), water (H₂O) and hydrogen sulfide (H₂S): (we need to flip the second reaction)
H₂ (g) + S (s) → H₂S (g) ΔH = -20.6 kJ
S (s) + O₂ (g) → SO₂ (g) ΔH = -296.8 kJ
H₂ (g) + O₂ (g) → H₂O (g) ΔH = -285.8 kJ

We are asked to find the enthalpy for the following reaction:
SO₂ (g) + 2H₂S (g) → 3S (s) + 2H₂O (g)

Recall that ΔH of a reaction is (n and m are the coefficients of the species):
m(Σ ΔH formation of products) - n(Σ ΔH formation of reactants)

In our case:
ΔH = [3*ΔH of S(s) + 2*ΔH of H₂O(g)] - [2*ΔH of H₂S + ΔH of SO₂] =
[3*0 kJ + 2*(-285.8 kJ)] - [2*(-20.6 kJ) + (-296.8 kJ)]

Source: http://au.answers.yahoo.com/question/index?qid=20130410152319AArYL5Y

Answer 2

did this answer your question?

Answer 3

just test