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Professor Lavelle did an example on Wednesday where he asked at what temperature is spontaneous at 1 atm given and . He said that we were looking to find the temperature when and said that neither the forward or reverse reaction is favored when equals zero. If this is the case, I'm confused as to why the reaction would go from a liquid to gas spontaneously if neither the forward or reverse reaction is favored, and was hoping for some clarification. Thanks!
The reason why we set G° = 0 is to find out the temperature that the Br2(l) is in equilibrium with the Br2(g). The answer we got for T was 333K. So yes, at 333K, the temperature is at equilibrium and neither the forward nor reverse equation is favored. However, when T > 333K, the forward process is favored, making this reaction spontaneous. Conversely, when T < 333K, the reverse process is favored and is spontaneous.
Making (delta)G=0 was an assumption we had made to find out the temperature for phase changes. When we set (delta)G=0, it knocks out a potential variable and lets us see the temperature for where the in between is. Past this temperature (greater), the forward reaction is favored; below this temperature (less), the reverse reaction is favored. So you're right to question why we assumed (delta)G=0 when a phase change that needs energy is the reaction. But remember that when (delta)G=0, the reaction is at equilibrium and the liquid and gas phases are coexisting and neither one is favored.
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