# Question #34319

##### 1 Answer
Nov 6, 2016

$\Delta H > 0$, $\Delta S > 0$

#### Explanation:

As you know, the only criterion that determines the spontaneity of a reaction is the Gibbs free energy change, $\Delta G$, which is defined as

$\textcolor{b l u e}{\overline{\underline{| \textcolor{b l a c k}{\Delta G = \Delta H - T \cdot \Delta S} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

Here

$\Delta H$ - the enthalpy change of reaction
$T$ - the absolute temperature at which the reaction takes place
$\Delta S$ - the entropy change of reaction

Now, in order for a reaction to be spontaneous at a given temperature, it must have

$\Delta G < 0$

This, of course, implies that a non-spontaneous reaction will have

$\Delta G > 0$

A positive Gibbs free energy change corresponds to

$\Delta H - T \cdot \Delta S > 0$

This means that at low temperatures, you have

$\Delta H > T \cdot \Delta S$

Now, this can be true for $\Delta H < 0$ and $\Delta S < 0$. However, you are told that at high temperatures the reaction becomes spontaneous.

This means that you need

$\Delta H - T \cdot \Delta S < 0$

or

$\Delta H < T \cdot \Delta S$

As you can see, this cannot be true if $\Delta H < 0$ and $\Delta S < 0$ because increasing the value of $T$ would simply make

${\overbrace{\Delta H}}^{\textcolor{b l u e}{\text{negative")) > overbrace(T * DeltaS)^(color(blue)("even more negative}}} \to$ non-spontaneous reaction

However, if $\Delta H > 0$ and $\Delta S > 0$, increasing the value of $T$ would make

${\overbrace{\Delta H}}^{\textcolor{\mathrm{da} r k g r e e n}{\text{positive")) < overbrace(T * DeltaS)^(color(darkgreen)("even more positive}}} \to$ spontaneous reaction

Remember, $T$ is always positive because it expresses absolute temperature.

In general terms, you can have four possible scenarios when dealing with the Gibbs free energy change

• $\Delta H < 0$, $\Delta S > 0 \to$ spontaneous at any temperature
• $\Delta H > 0$, $\Delta S < 0 \to$ non-spontaneous regardless of temperature
• $\Delta H > 0$, $\Delta S > 0 \to$ spontaneous at a certain temperature range
• $\Delta H < 0$, $\Delta S < 0 \to$ spontaneous at a certain temperature range As you can see, reactions that have $\Delta H > 0$ and $\Delta S > 0$ are only spontaneous at high temperatures.

In this particular case, the reaction is endothermic, since $\Delta H > 0$, but the entropy change of the system overcomes the energy requirement at high temperatures.

A classic example would be the melting of ice, for which

• $\Delta H > 0 \to$ you need to add heat to melt ice
• $\Delta S > 0 \to$ the entropy of the system is increasing because you're going from solid to liquid

However, the melting of ice is only spontaneous when $T > \text{273.15 K}$, i.e. at temperatures above ${0}^{\circ} \text{C}$. When the temperature falls below ${0}^{\circ} \text{C}$, the melting of ice is a non-spontaneous process.