Final answer:
The reactions are as follows:
(a) MnO₂ (s) → Mn(s) + O₂ (g) is nonspontaneous.
(b) H₂ (g) + Br₂ (l) → 2HBr(g) is spontaneous.
(c) Cu(s) + S(g) → CuS(s) is spontaneous.
Step-by-step explanation:
To determine the free energy change for each of the given reactions, we can use the standard free energy of formation data in Appendix G. The standard free energy change (ΔG°) is given by the equation:
ΔG° = Σ nΔGf°(products) - Σ mΔGf°(reactants)
where ΔGf° is the standard free energy of formation for each compound and n and m are the stoichiometric coefficients of the products and reactants respectively. If ΔG° is negative, the reaction is spontaneous; if it is positive, the reaction is nonspontaneous at the given conditions.
Let's calculate the free energy change for each reaction:
(a) MnO₂ (s) → Mn(s) + O₂ (g)
Using the standard free energy of formation data:
ΔG° = [ΔGf°(Mn) + ΔGf°(O₂)] - ΔGf°(MnO₂)
ΔG° = [0 + 0] - (-520.32 kJ/mol) = 520.32 kJ/mol
Since ΔG° is positive, the reaction is nonspontaneous at these conditions.
(b) H₂ (g) + Br₂ (l) → 2HBr(g)
ΔG° = [ΔGf°(2HBr)] - [ΔGf°(H₂) + ΔGf°(Br₂)]
ΔG° = (-53.44 kJ/mol) - (0 + 0) = -53.44 kJ/mol
Since ΔG° is negative, the reaction is spontaneous at these conditions.
(c) Cu(s) + S(g) → CuS(s)
ΔG° = [ΔGf°(CuS)] - [ΔGf°(Cu) + ΔGf°(S)]
ΔG° = (-89.80 kJ/mol) - (0 + 0) = -89.80 kJ/mol
Since ΔG° is negative, the reaction is spontaneous at these conditions.