Calculate the standard entropy change for the reaction:
2H2S(g) +SO2(g)→3Srh(s)
+2H2O(g), where:
1. S0[Srh]=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
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
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