## 9.52

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

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

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

Hailey Johnson
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Joined: Wed Nov 16, 2016 3:04 am

### 9.52

This problem from the book asks " Explain how an endothermic reaction can be spontaneous." Since this is an even numbered problem we don't have the answer to this question. Can someone explain this concept for me?

Curtis Wong 2D
Posts: 62
Joined: Sat Jul 22, 2017 3:00 am

### Re: 9.52

An endothermic process is usually not spontaneous due to the fact that by putting energy into the system, there is an external influence now acting on the system. However, the only way that an endothermic process can be spontaneous is if the increase in entropy will be greater than the increase in enthalpy that occurs.

One example found online is table salt dissolving in water. Salt dissolves on its own in water, but that requires energy to do so (endothermic reaction), however the increase in entropy from dissolving that salt in water is greater than the enthalpy change from the change in energy.

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If deltaS is positive and deltaH is positive, that would be an endothermic reaction, but it would become spontaneous at high temperatures. This is because G = H - (delta T x S), so we would want out delta T x S to be very big because that would be the only way G could be negative because when G is negative, the reaction is spontaneous.
An example would be ice melting. You need to add heat, thus making it endothermic, and since it goes from a solid to a liquid, entropy increases.

Aijun Zhang 1D
Posts: 53
Joined: Tue Oct 10, 2017 7:13 am

### Re: 9.52

Endothermic reaction means that $\Delta H > 0$ since it absorbs energy. The Gibb's free energy equation is $\Delta G = \Delta H - T\Delta S$ and we know that this equation determines the spontaneity of a reaction. We want $\Delta G$to be negative. Now it comes to algebra.
There is actually a chart to organize all the possibilities with $\Delta G$ with different signs of $\Delta H$ and $\Delta S$.
If $\Delta H > 0$, we want $T\Delta S$ to be very large, larger than $\Delta H$. So $\Delta S$ must be positive, so the reaction can be spontaneous at high temperature. If $\Delta S$< 0, the the reaction will not be spontaneous at any temperature.