## Spontaneous delta G

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

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

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

Alex Tchekanov Dis 2k
Posts: 118
Joined: Sat Aug 24, 2019 12:16 am

### Spontaneous delta G

Can someone please explain why in the practice problem today in class we set delta G equal to zero to find out at what temperature the reaction will be spontaneous?

sarahsalama2E
Posts: 164
Joined: Fri Aug 30, 2019 12:16 am

### Re: Spontaneous delta G

when we set delta G equal to 0, we find the boiling point at which the liquid turns into a gas state. Any temperature greater than what we find once we set delta G=0 using the formula delta G= delta H -TdeltaS, causes the delta G value to be less than one (aka negative), therefore spontaneous. Remember that Delta G always equals 0 at phase changes.

Eva Zhao 4I
Posts: 101
Joined: Sun Sep 29, 2019 12:16 am

### Re: Spontaneous delta G

For this problem, we're using the equation: ∆G˚ = ∆H˚ - T∆S˚. The key point to keep in mind here is that the reaction is spontaneous when ∆G˚ is negative. Thus, we're trying to find the temperature, T, at which T∆S˚ > ∆H˚ such that ∆G˚ is negative. To do so, we can set ∆G˚ = 0 to find the minimum temperature in which ∆H˚ = T∆S˚ (where Br2(l) is in equilibrium with Br2(g)). The values are given for both ∆H˚ and ∆S˚, so we can rearrange as T = ∆H˚/∆S˚ and plug in the values to solve for T. We get a value of T = 333K, essentially the boiling point. The temperature at which ∆G˚ is negative is then T > 333K; when T > 333K, T∆S˚ > ∆H˚, the conditions for a negative ∆G˚.

105335337
Posts: 103
Joined: Sat Aug 24, 2019 12:15 am

### Re: Spontaneous delta G

WE set delta G to zero in the practice problem because by setting it to 0, it gave us the minimum temperature needed to make the reaction spontaneous, therefore answering the question.

Samuel Tzeng 1B
Posts: 103
Joined: Sat Aug 24, 2019 12:15 am

### Re: Spontaneous delta G

finding the temperature at delta G=0 gives the minimum temperature for the reaction to be spontaneous

Rhea Shah 2F
Posts: 97
Joined: Thu Jul 25, 2019 12:17 am

### Re: Spontaneous delta G

a reaction is spontaneous if delta G is less than zero. thus, we set the equation for gibbs free energy equal to 0 to find the temperature at which delta g changes sign and becomes negative. the temperature at which delta g is 0 is the temperature at which the two phases coexist, so the temperature at which the forward process is spontaneous is greater than the temperature at which delta g is zero.

Kallista McCarty 1C
Posts: 212
Joined: Wed Sep 18, 2019 12:18 am

### Re: Spontaneous delta G

A reaction is spontaneous when delta G is negative, so by setting it to zero we were able to find the value and then determine the answer as "above 333K"

805394719
Posts: 104
Joined: Wed Sep 11, 2019 12:16 am

### Re: Spontaneous delta G

When delta is smaller than zero, the reaction is spontaneous, meaning that it will occur without energy input. We knew that delta G must have been spontaneous, but we did not know the exact temperature at which it would turn spontaneous. The critical point here is delta G = 0 since it is the point after which delta G can be positive or negative. By finding the temperature at delta G we could conclude that at temperatures above 333K, the reaction would be spontaneous. The reason why we decided that it is above that value and not below is that the entropy value was positive and enthalpy was negative. Since delta G = enthalpy - entropy x temperature, if the temperature was higher, the second term would be greater, causing delta G to be negative.