Animated Solution for Chemistry - Chemical Thermodynamics: A process has $\Delta H=200{\text{ J mol}}^{-1}$ and $\Delta S=40{\text{ J K}}^{-1}{\text{ mol}}^{-1}$. Out of the values given below, choose the minimum temperature above which the process will be spontaneous

Enthalpy vs. Entropy

Imagine a chemical reaction as a grand tug-of-war. On one side, we have Enthalpy ($\Delta H$), which represents the heat energy absorbed or released. On the other side, we have Entropy ($\Delta S$), which measures the randomness or disorder of the system.

The ultimate referee in this match is the Gibbs Free Energy ($\Delta G$). The master equation that governs whether a reaction will happen on its own (spontaneously) is:

For any process to be spontaneous, the change in Gibbs free energy, $\Delta G$, must be strictly negative ($\Delta G<0$).

Before we can determine the temperature range for spontaneity, we need to find the exact tipping point. This is the equilibrium temperature ($T_{eq}$) where the system is perfectly balanced, and $\Delta G$ is exactly zero. At this point, the enthalpy change perfectly balances the entropy term:

Now, let's substitute the values given in our problem. We have $\Delta H=200{\text{ J mol}}^{-1}$ and $\Delta S=40{\text{ J K}}^{-1}{\text{ mol}}^{-1}$.

This tells us that at exactly $5\text{ K}$, the system is at equilibrium.

So, what happens when we change the temperature? Since both $\Delta H$ and $\Delta S$ are positive, the temperature $T$ acts as an amplifier for the entropy term.

We want $\Delta G<0$, which means:

The $T\Delta S$ term becomes larger than $\Delta H$ only at temperatures higher than $5\text{ K}$. Therefore, the minimum temperature above which the process becomes spontaneous is $5\text{ K}$. Always remember to check the signs of your thermodynamic parameters, as they completely dictate the slope of the $\Delta G$ curve!