Hydrogen sulfide, h2s, is a very toxic gas with a smell of rotten eggs. using the following: h2s+3/2 o2=so2+h2o h2+1/2o2=h2o s+o2=so2 determine the enthalpy of formation for h2s g

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!

b is the correct