## Increasing temperature to make G negative

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

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

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

Mary Becerra 2D
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Joined: Fri Sep 29, 2017 7:06 am

### Increasing temperature to make G negative

The textbook (example 9.16) presents an example of an endothermic reaction becoming exothermic due to an increase in temperature. Is this the case for all endothermic processes? Or are there some cases where an increase in temperature will never make an endothermic process exothermic?

Kevin Ru 1D
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Joined: Thu Jul 13, 2017 3:00 am

### Re: Increasing temperature to make G negative

Generally this should be true for all endothermic processes as raising the temperature of the reaction favors the products. This means that the reverse reaction is favored which is exothermic. Hope that helps!

Lena Nguyen 2H
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### Re: Increasing temperature to make G negative

The example 9.16 refers to increasing the temperature to make the reaction spontaneous, not exothermic. However, raising the temperature will not always make an endothermic reaction spontaneous. Since $\Delta G^{o} = \Delta H^{o} - T\Delta S^{o}$, if the endothermic reaction has a positive change in entropy, raising the temperature high enough will make the reaction spontaneous.

Andrea ORiordan 1L
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### Re: Increasing temperature to make G negative

You cannot change whether or not a reaction is endothermic or exothermic (at least not in this course). If a reaction is exothermic, its reverse reaction will be endothermic. However, you can change temperature to make a reaction more energetically favorable. Given that ∆G=∆H-T∆S and reactions are spontaneous when ∆G is negative:
If ∆H is negative (exothermic) and ∆S is positive, the reaction will be spontaneous only at high temperatures, when the -T∆S value can overcome the positive ∆H value.
If ∆H is positive and ∆S is negative, the reaction will be spontaneous only at low temperatures, when the value of -T∆S is not sufficiently large to overcome the ∆H value.

Hope this helps!