Using the given data,determine ΔrG° at 500.0 K for the reaction below. Ba(s) + H₂O(g) → BaO(s) + H₂(g)
Substance
Δ_fH°(kJ/mol-rxn) at 298 K
S° (J/K·mol-rxn) at 298 K
Ba(s)
0
62) 8
H₂O(g)
-241.8
188) 7
BaO(s)
-592
70) 4
H₂(g)
0
130) 6

Question

Quizplus · Accepted Answer

ΔrG° = ΔrG°(298 K) + (ΔrS°) (T - 298 K)ΔrG°(298 K) = ΔrG°f(products) - ΔrG°f(reactants)ΔrG°f(products) = ΔG°f(BaO) + ΔG°f(H2) = -592 kJ/mol + 0 kJ/mol = -592 kJ/molΔrG°f(reactants) = ΔG°f(Ba) + ΔG°f(H2O) = 0 kJ/mol + (-241.8 kJ/mol) = -241.8 kJ/molΔrG°(298 K) = -592 kJ/mol - (-241.8 kJ/mol) = -350.2 kJ/molΔrS° = S°(products) - S°(reactants)S°(products) = S°(BaO) + S°(H2) = 70.4 J/K·mol + 130.6 J/K·mol = 201 J/K·molS°(reactants) = S°(Ba) + S°(H2O) = 62.8 J/K·mol + 188.7 J/K·mol = 251.5 J/K·molΔrS° = 201 J/K·mol - 251.5 J/K·mol = -50.5 J/K·molΔrG° = -350.2 kJ/mol + (-50.5 J/K·mol) (500.0 K - 298 K) = -325.0 kJ/mol

Using the Given Data,determine ΔRG° at 500