Question

Calculate the standard entropy change for the reaction:

2H₂S(g) +SO₂(g)→3S_rh(s) +2H₂O(g), where:

1. S⁰[S_rh]=32 J/k mol

2. S⁰[H₂O(g)]=189 J/k mol

3. S⁰[H₂S(g)]=206 J/k mol

4. S⁰[SO₂]=248 J/k mol

a. +1134 J/K

b. -1134 J/K

c. -186 J/K

d +186 J/K

Answer 1

we have:

Sof(H2S (g)) = 206.0 J/mol.K

Sof(SO2 (g)) = 248.0 J/mol.K

Sof(SrH (s)) = 32.0 J/mol.K

Sof(H2O (g)) = 189.0 J/mol.K

we have the Balanced chemical equation as:

2 H2S (g) + SO2 (g) ---> 3 SrH (s) + 2 H2O (g)

deltaSo rxn = 3*Sof(SrH (s)) + 2*Sof(H2O (g)) - 2*Sof( H2S (g)) - 1*Sof(SO2 (g))

deltaSo rxn = 3*(32.0) + 2*(189.0) - 2*(206.0) - 1*(248.0)

deltaSo rxn = -186 J/K

Answer: c