Example from Wednesday's Lecture

$\Delta G^{\circ}= \Delta H^{\circ} - T \Delta S^{\circ}$

$\Delta G^{\circ}= -RT\ln K$

$\Delta G^{\circ}= \sum \Delta G_{f}^{\circ}(products) - \sum \Delta G_{f}^{\circ}(reactants)$

Andres Reynoso 1J
Posts: 30
Joined: Fri Sep 29, 2017 7:06 am

Example from Wednesday's Lecture

Professor Lavelle did an example on Wednesday where he asked at what temperature is $Br(l)\rightarrow Br(g)$ spontaneous at 1 atm given $\Delta H^{o}= 31.0 kJ/mol$ and $\Delta S^{o}= 93.0 J/Kmol$. He said that we were looking to find the temperature when $\Delta G^{o}= 0$ and said that neither the forward or reverse reaction is favored when $\Delta G^{o}$ 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!

Varsha Sivaganesh 1A
Posts: 51
Joined: Thu Jul 13, 2017 3:00 am

Re: Example from Wednesday's Lecture

The reason why we set $\Delta$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.

Michelle Lee 2E
Posts: 64
Joined: Thu Jul 27, 2017 3:01 am

Re: Example from Wednesday's Lecture

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.

Kevin Tabibian 1A
Posts: 42
Joined: Fri Sep 29, 2017 7:06 am

Re: Example from Wednesday's Lecture

I believe he was just demonstrating that that specific temperature is a threshold. A student asked and professor said that you could also set up an inequality; the final answer would be the same .

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