Consider the reaction 2A(g)⇌ A 2 (g).The following pictures represent two possible initial states and the equilibrium state of the system.
-What are the signs (+ or -)of ΔH,ΔS,and ΔG when the system spontaneously goes from initial state 1 to the equilibrium state?

Question

Consider the reaction 2A(g)⇌ A 2 (g).The following pictures represent two possible initial states and the equilibrium state of the system. 
-What are the signs (+ or -)of ΔH,ΔS,and ΔG when the system spontaneously goes from initial state 1 to the equilibrium state?

Quizplus · Accepted Answer

Since the reaction is at equilibrium, ΔG = 0. 
At equilibrium, ΔS should also be 0, which means the entropy of the system does not increase or decrease. The only way that equilibrium can be achieved is if the forward and reverse reactions have the same enthalpy (ΔH) so that the energy absorbed by the endothermic reaction can be compensated by the energy released by the exothermic reaction. Therefore, ΔH = 0. Since the reaction is endothermic in the forward direction, ΔH must be positive, and since the reaction increases the number of gas molecules from 2 to 3, ΔS must also be positive. Thus, the sign of ΔH is +, the sign of ΔS is +, and the sign of ΔG is - (since ΔG = ΔH - TΔS and T is positive).

Option D is the best choice as it satisfies all the above conditions.

Consider the Reaction 2A(g)⇌ A2(g)

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