## 9.51 [ENDORSED]

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

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

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

nelms6678
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### 9.51

Why are so many exothermic reactions spontaneous?

Using the Gibbs formula is it the negative delta G and delta H values that cause spontaneity?

Chem_Mod
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### Re: 9.51  [ENDORSED]

Exothermic reactions transfer heat from the system to the surroundings. One of the common formulations of the 2nd Law of Thermodynamics is that the entropy of the universe always increases. This means that exothermic reactions are more likely to be spontaneous.

If you consider the relationship ΔG = ΔH - TΔS, where a negative value of ΔG means the process is spontaneous, then having a large negative value for ΔH (i.e., an exothermic reaction) means that only a very large negative ΔS for the system could possibly result in the process becoming nonspontaneous.

Angel Ni 2K
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Joined: Sat Jul 22, 2017 3:01 am

### Re: 9.51

deltaG = deltaH - TdeltaS. If deltaG is negative, then the reaction is spontaneous. Since deltaH is negative, the reaction will be spontaneous for all positive values of deltaS and small negative values of deltaS.

JamesAntonios 1E
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Joined: Fri Sep 29, 2017 7:04 am

### Re: 9.51

Chem_Mod wrote:Exothermic reactions transfer heat from the system to the surroundings. One of the common formulations of the 2nd Law of Thermodynamics is that the entropy of the universe always increases. This means that exothermic reactions are more likely to be spontaneous.

If you consider the relationship ΔG = ΔH - TΔS, where a negative value of ΔG means the process is spontaneous, then having a large negative value for ΔH (i.e., an exothermic reaction) means that only a very large negative ΔS for the system could possibly result in the process becoming nonspontaneous.

If a reaction is exothermic, according to ΔS=ΔH/T, does that not make the ΔS of the surroundings positive as they are accepting heat, and the ΔH of the system negative, since it is losing heat? And so, the ΔS (total) in the Gibbs Free Energy would have to take that into account, correct?