Which statement is correct for the reaction below under standard conditions? 2Cl2 (g) + 2NO (g) → N2 (g) + 2Cl2O (g); ΔH° = -22.0 kJ/mol

Which statement is correct for the reaction below under standard conditions? 2Cl2 (g) + 2NO (g) → N2 (g) + 2Cl2O (g); ΔH° = -22.0 kJ/mol

• A. The reaction is spontaneous at all temperatures.
• B. The reaction is spontaneous only at low temperatures.
• C. The reaction is spontaneous only at high temperatures.
• D. The reaction is not spontaneous at any temperature.

Answer: (A)

Gibbs free energy governs spontaneity through ΔG° = ΔH° − TΔS°. Here the reaction is exothermic (ΔH° = −22.0 kJ/mol), which by itself favors spontaneity. However, the entropy change is likely negative because the number of gas molecules decreases from reactants (two Cl2 and two NO, four moles of gas) to products (one N2 and two Cl2O, about three moles of gas, assuming Cl2O is gaseous). A negative ΔS° means the term −TΔS° adds a positive amount to ΔG°, growing with temperature. Therefore, as temperature rises, the positive contribution from TΔS° can overcome the negative ΔH°, making ΔG° less negative and potentially positive. At low temperatures, the negative enthalpy dominates and the reaction is spontaneous; at sufficiently high temperatures, it can become nonspontaneous. The takeaway is that this reaction is spontaneous at low temperatures, but not at all temperatures. The given statement claiming spontaneity at every temperature would require a positive ΔS°, which isn’t indicated here.

Gibbs free energy governs spontaneity through ΔG° = ΔH° − TΔS°. Here the reaction is exothermic (ΔH° = −22.0 kJ/mol), which by itself favors spontaneity. However, the entropy change is likely negative because the number of gas molecules decreases from reactants (two Cl2 and two NO, four moles of gas) to products (one N2 and two Cl2O, about three moles of gas, assuming Cl2O is gaseous). A negative ΔS° means the term −TΔS° adds a positive amount to ΔG°, growing with temperature.

Therefore, as temperature rises, the positive contribution from TΔS° can overcome the negative ΔH°, making ΔG° less negative and potentially positive. At low temperatures, the negative enthalpy dominates and the reaction is spontaneous; at sufficiently high temperatures, it can become nonspontaneous.

The takeaway is that this reaction is spontaneous at low temperatures, but not at all temperatures. The given statement claiming spontaneity at every temperature would require a positive ΔS°, which isn’t indicated here.