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In the lecture, Dr Lavelle went over the phase change problem from Br(l)--> Br(g). There, he talked about how deltaH + is a endothermic reaction that does not favor forward processes. I understand that deltaH + refers to reactions being spontaneous at high temperature but what did Dr Lavelle mean by "DeltaH + endothermic does not favor forward process"?
I believe that this just means that when delta H is positive, the reaction is endothermic. This doesn't favor the forward process because you have to add heat to the reactants; it isn't favorable or spontaneous in this fashion. However, it can become favorable when the temperature is high enough.
When delta H is positive, it takes a very high temperature for entropy to dominate and cause delta G to become negative. So delta H + is only spontaneous at high temperatures.
It would probably help to put this in context of the equation, ∆G=∆H-T∆S. In a process with a positive ∆H and a positive ∆S, having a high T(for specific calculations, T>(∆H/∆S)) would make ∆G negative, and thus the reaction would be spontaneous and favorable.
When delta H is positive, the reaction is endothermic and is only spontaneous at high temperatures (unless delta S is negative, in which case the reaction is not spontaneous at any temperature, but the reverse reaction is spontaneous). Because this reaction requires heat, it is not favorable, and the forward process can only be spontaneous at high temperatures.
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