Answer:
ΔH = -20kJ
Explanation:
The enthalpy of formation of a compound is defined as the change of enthalpy during the formation of 1 mole of the substance from its constituent elements. For H₂S(g) the reaction that describes this process is:
H₂(g) + S(g) → H₂S(g)
Using Hess's law, it is possible to sum the enthalpies of several reactions to obtain the change in enthalpy of a particular reaction thus:
(1) H₂S(g) + ³/₂O₂(g) → SO₂(g) + H₂O(g) ΔH = -519 kJ
(2) H₂(g) + ¹/₂O₂(g) → H₂O(g) ΔH = -242 kJ
(3) S(g) + O₂(g) → SO₂(g) ΔH = -297 kJ
The sum of -(1) + (2) + (3) gives:
-(1) SO₂(g) + H₂O(g) → H₂S(g) + ³/₂O₂(g) ΔH = +519 kJ
(2) H₂(g) + ¹/₂O₂(g) → H₂O(g) ΔH = -242 kJ
(3) S(g) + O₂(g) → SO₂(g) ΔH = -297 kJ
-(1) + (2) + (3): H₂(g) + S(g) → H₂S(g)
ΔH = +519kJ - 242kJ - 297kJ = -20 kJ
I hope it helps!