Question

For the following process:
H2O(g) + Cl2O(g) à 2HOCl(g)
Ho(H2O(g))= 242KJ/mol Ho(Cl2O(g))=80KJ/mol Ho(HOCl(g))= -80KJ/mol So(H2O(g))= 0.189 KJ/K So(Cl2O(g))=0.266 KJ/K So(HOCl(g))= 0.237 KJ/K
A) Calculate ΔHo, ΔSo and ΔGo
B)Calculate ΔG at 100K and P(H2O)=P(Cl2O)= P(HOCL)=10atm

Answer 1

(A)

ΔH^o_rxn = 2 ΔH^o_f(HOCl) - ΔH^o_f (H₂O) - ΔHo (Cl₂O) = 2(-80) - (-242) - (80) = +2 kJ

ΔS_o^rxn = 2 ΔS^o_f(HOCl) - ΔS^o_f (H₂O) - ΔS^o (Cl₂O) = 2 (0.237) - 0.189 - 0.266 = + 0.019 kJ/K

ΔG^o = ΔH^o_rxn - TΔS^o_rxn = + 2 - 298 (0.019) = - 3.662 kJ

(B)

ΔG = ΔG^o + R T lnK

= ΔG^o + R T lnP_HOCl²/P_H2O*P_Cl2O

ΔG = - 3.662 + (0.008314 * 373.15* ln(10)²/(10*10))

ΔG = - 3.662 kJ