# Gibbs free energy and spontaneity

How the second law of thermodynamics helps us determine whether a process will be spontaneous, and using changes in Gibbs free energy to predict whether a reaction will be spontaneous in the forward or reverse direction (or whether it is at equilibrium!).

## Key points

• The second law of thermodynamics says that the entropy of the universe always increases for a spontaneous process: $\Delta \text {S}_{\text{universe}}=\Delta \text {S}_{\text{system}} + \Delta \text {S}_{\text{surroundings}} > 0$
• At constant temperature and pressure, the change in Gibbs free energy is defined as $\Delta \text G = \Delta \text H - \text{T}\Delta \text S$.
• When $\Delta \text G$ is negative, a process will proceed spontaneously and is referred to as exergonic.
• The spontaneity of a process can depend on the temperature.

## Spontaneous processes

In chemistry, a spontaneous processes is one that occurs without the addition of external energy. A spontaneous process may take place quickly or slowly, because spontaneity is not related to kinetics or reaction rate. A classic example is the process of carbon in the form of a diamond turning into graphite, which can be written as the following reaction:
$\text C(s, \text {diamond}) \rightarrow \text C(s, \text{graphite})$
On left, multiple shiny cut diamonds. On right, chunk of black graphitic carbon.
Ever heard the saying, "graphite is forever"? If we waited long enough, we would observe a diamond spontaneously turn into the more stable form of carbon, graphite. Picture from Wikipedia, CC BY-SA 3.0
This reaction takes so long that it is not detectable on the timescale of (ordinary) humans, hence the saying, "diamonds are forever." If we could wait long enough, we should be able to see carbon in the diamond form turn into the more stable but less shiny, graphite form.
Another thing to remember is that spontaneous processes can be exothermic or endothermic. That is another way of saying that spontaneity is not necessarily related to the enthalpy change of a process, $\Delta \text H$.
Dissolving sodium chloride, $\text{NaCl}$, in water at $25~^\circ \text C$ is an endothermic reaction! The heat of solution, $\Delta\text H_{soln}$, for sodium chloride at $25~^\circ \text C$ is $+3.87\,\dfrac{\text{kJ}}{\text{mol}}$. If you aren't sure whether this is actually a spontaneous process, you should test it yourself!
How do we know if a process will occur spontaneously? The short but slightly complicated answer is that we can use the second law of thermodynamics. According to the second law of thermodynamics, any spontaneous process must increase the entropy in the universe. This can be expressed mathematically as follows:
$\Delta \text {S}_{\text{universe}}=\Delta \text {S}_{\text{system}} + \Delta \text {S}_{\text{surroundings}} > 0~~~~~~~~\text{For a spontaneous process}$
Great! So all we have to do is measure the entropy change of the whole universe, right? Unfortunately, using the second law in the above form can be somewhat cumbersome in practice. After all, most of the time chemists are primarily interested in changes within our system, which might be a chemical reaction in a beaker. Do we really have to investigate the whole universe, too? (Not that chemists are lazy or anything, but how would we even do that?)
Luckily, chemists can get around having to determine the entropy change of the universe by defining and using a new thermodynamic quantity called Gibbs free energy.

## Gibbs free energy and spontaneity

When a process occurs at constant temperature $\text T$ and pressure $\text P$, we can rearrange the second law of thermodynamics and define a new quantity known as Gibbs free energy:
$\text{Gibbs free energy}=\text G =\text H - \text{TS}$
where $\text H$ is enthalpy, $\text T$ is temperature (in kelvin, $\text K$), and $\text S$ is the entropy. Gibbs free energy is represented using the symbol $\text G$ and typically has units of $\dfrac{\text {kJ}}{\text{mol-rxn}}$.
If you are curious about where this equation came from, see this video that uses pressure-volume (PV) diagrams to derive the Gibbs free energy equation.
For an article that explains how this equation might be used from a biological context, see this article on free energy in biology.
A "mole of reaction," which is abbreviated as $\text{mol-rxn}$ or $\text{mol-reaction}$, is defined as occurring when the number of moles given by the coefficients in your balanced equation react. That definition can sound rather confusing, but the idea is hopefully more clear in the context of an example. If we had the following balanced reaction:
$2 \text{Al}(s)+ 3\text{Cl}_2(g) \rightarrow 2 \text{AlCl}_3(s)$
we would say that $1$ mole of reaction is when $2$ moles of $\text{Al}$ react with $3$ moles of $\text{Cl}_2$ to produce $2$ moles of $\text{AlCl}_3$. We can write these relationships mathematically as well:
$1\,\text{mol-rxn}=2\,\text{mol Al}=3\,\text{mol Cl}_2=2\,\text{mol AlCl}_3$
When using Gibbs free energy to determine the spontaneity of a process, we are only concerned with changes in $\text G$, rather than its absolute value. The change in Gibbs free energy for a process is thus written as $\Delta \text G$, which is the difference between $\text G_{\text{final}}$, the Gibbs free energy of the products, and $\text{G}_{\text{initial}}$, the Gibbs free energy of the reactants.
$\Delta \text G =\text G_{\text{final}} - \text{G}_{\text{initial}}$
For a process at constant $\text T$ and constant $\text P$, we can rewrite the equation for Gibbs free energy in terms of changes in the enthalpy ($\Delta \text H_{\text{system}}$) and entropy ($\Delta \text S_{\text{system}}$) for our system:
$\Delta \text G_{\text{system}} =\Delta \text H_{\text{system}} - \text{T}\Delta \text S_{\text{system}}$
You might also see this reaction written without the subscripts specifying that the thermodynamic values are for the system (not the surroundings or the universe), but it is still understood that the values for $\Delta \text H$ and $\Delta \text S$ are for the system of interest. This equation is exciting because it allows us to determine the change in Gibbs free energy using the enthalpy change, $\Delta \text H$, and the entropy change , $\Delta \text S$, of the system. We can use the sign of $\Delta \text G$ to figure out whether a reaction is spontaneous in the forward direction, backward direction, or if the reaction is at equilibrium.
• When $\Delta \text G<0$, the process is exergonic and will proceed spontaneously in the forward direction to form more products.
• When $\Delta \text G>0$, the process is endergonic and not spontaneous in the forward direction. Instead, it will proceed spontaneously in the reverse direction to make more starting materials.
• When $\Delta \text G=0$, the system is in equilibrium and the concentrations of the products and reactants will remain constant.
When a reaction is at equilibrium, the forward reaction and reverse reaction are occurring at the same rate. That means the concentrations of the reactants and products will remain constant at equilibrium.
For more details, you can watch this video on reactions at equilibrium.

## Calculating change in Gibbs free energy

Although $\Delta \text G$ is temperature dependent, it's generally okay to assume that the $\Delta \text H$ and $\Delta \text S$ values are independent of temperature as long as the reaction does not involve a phase change. That means that if we know $\Delta \text H$ and $\Delta \text S$, we can use those values to calculate $\Delta \text G$ at any temperature. We won't be talking in detail about how to calculate $\Delta \text H$ and $\Delta \text S$ in this article, but there are many methods to calculate those values including:
When the process occurs under standard conditions (all gases at $1\,\text {bar}$ pressure, all concentrations are $1\,\text M$, and $\text T=25\,^\circ\text C$), we can also calculate $\Delta \text G$ using the standard free energy of formation, $\Delta_{f} \text G^\circ$.
Problem-solving tip: It is important to pay extra close attention to units when calculating $\Delta \text G$ from $\Delta \text H$ and $\Delta \text S$! Although $\Delta \text H$ is usually given in $\dfrac{\text{kJ}}{\text{mol-reaction}}$, $\Delta \text S$ is most often reported in $\dfrac{\text{J}}{\text{mol-reaction}\cdot \text K}$. The difference is a factor of $1000$!!

## When is $\Delta \text G$ negative?

If we look at our equation in greater detail, we see that $\Delta \text G_\text{system}$ depends on $3$ values:

$\Delta \text G_{\text{system}} =\Delta \text H_{\text{system}} - \text{T}\Delta \text S_{\text{system}}$
• the change in enthalpy $\Delta \text H_{\text{system}}$
• the temperature $\text T$
• the change in entropy $\Delta \text S_{\text{system}}$
Temperature in this equation always positive (or zero) because it has units of $\text K$. Therefore, the second term in our equation, $\text T \Delta \text S_\text{system}$, will always have the same sign as $\Delta \text S_\text{system}$. We can make the following conclusions about when processes will have a negative $\Delta \text G_\text{system}$:

• When the process is exothermic ($\Delta \text H_{\text{system}}<0$), and the entropy of the system increases ($\Delta \text S_{\text{system}}>0$), the sign of $\Delta \text G_{\text{system}}$ is negative at all temperatures. Thus, the process is always spontaneous.
• When the process is endothermic, $\Delta \text H_{\text{system}}>0$, and the entropy of the system decreases, $\Delta \text S_{\text{system}}<0$, the sign of $\Delta \text G$ is positive at all temperatures. Thus, the process is never spontaneous.
For other combinations of $\Delta \text H_\text{system}$ and $\Delta \text S_\text{system}$, the spontaneity of a process depends on the temperature.
• Exothermic reactions ($\Delta \text H_\text{system}<0$) that decrease the entropy of the system ($\Delta \text S_\text{system}<0$) are spontaneous at low temperatures.
• Endothermic reactions ($\Delta \text H_\text{system}>0$) that increase the entropy of the system ($\Delta \text S_\text{system}>0$) are spontaneous at high temperatures.
Can you think of any reactions in your day-to-day life that are spontaneous at certain temperatures but not at others?

## Example $1$: Calculating $\Delta \text G$ for melting ice

Three melting ice cubes in a puddle of water on a mirrored surface.
At what temperatures (if any) is the melting of ice a spontaneous process? Photo of ice cubes from flickr, CC BY 2.0.
Let's consider an example that looks at the effect of temperature on the spontaneity of a process. The enthalpy of fusion and entropy of fusion for water have the following values:
$\Delta_\text{fus} \text H=6.01 \dfrac{\text{kJ}}{\text{mol-rxn}}$
$\Delta_\text{fus} \text S=22.0 \dfrac{\text{J}}{\text{mol-rxn}\cdot \text K}$
What is $\Delta \text G$ for the melting of ice at $20\,^\circ \text C$?
The process we are considering is water changing phase from solid to liquid:
$\text H_2 \text O(s) \rightarrow \text H_2 \text O(l)$
For this problem, we can use the following equation to calculate $\Delta \text G_\text{rxn}$:
$\Delta \text G =\Delta \text H - \text{T}\Delta \text S$
Luckily, we already know $\Delta \text H$ and $\Delta \text S$ for this process! We just need to check our units, which means making sure that entropy and enthalpy have the same energy units, and converting the temperature to Kelvin:
$\text T=20\,^\circ \text C+273 =293\,\text K$
If we plug the values for $\Delta \text H$, $\text T$, and $\Delta \text S$ into our equation, we get:
\begin{aligned} \Delta \text G &= \Delta \text H - \text{T}\Delta \text S \\ \\ &= 6.01 \dfrac{\text{kJ}}{\text{mol-rxn}}-(293\,\cancel{\text K})(0.022\,\dfrac{\text{kJ}}{\text{mol-rxn}\cdot \cancel{\text K})} \\ \\ &= 6.01\, \dfrac{\text{kJ}}{\text{mol-rxn}}-6.45\, \dfrac{\text{kJ}}{\text{mol-rxn}}\\ \\ &= -0.44 \, \dfrac{\text{kJ}}{\text{mol-rxn}}\end{aligned}
Since $\Delta \text G$ is negative, we would predict that ice spontaneously melts at $20\,^\circ \text C$. If you aren't convinced that result makes sense, you should go test it out!
Concept check: What is $\Delta \text G$ for the melting of ice at $-10\,^\circ \text C$?
First we need to calculate our temperature to kelvins:
$\text T=-10\,^\circ \text C+273 =263\,\text K$:
If we plug the values into our equation to calculate $\Delta \text G$, we get:
\begin{aligned} \Delta \text G &= \Delta \text H - \text{T}\Delta \text S \\ \\ &= 6.01 \dfrac{\text{kJ}}{\text{mol-rxn}}-(263\,\cancel{\text K})(0.022\,\dfrac{\text{kJ}}{\text{mol-rxn}\cdot \cancel{\text K})} \\ \\ &= 6.01\, \dfrac{\text{kJ}}{\text{mol-rxn}}-5.79\, \dfrac{\text{kJ}}{\text{mol-rxn}}\\ \\ &= 0.22 \, \dfrac{\text{kJ}}{\text{mol-rxn}}\end{aligned}
Thus, we see that at $-10\,^\circ \text C$ the Gibbs free energy change $\Delta \text G$ is positive for the melting of water. Therefore, we would predict that the reaction is not spontaneous at $-10\,^\circ \text C$. In fact, we would predict that the reverse reaction should be spontaneous: we would expect to see a puddle of water turn into ice.
We could test these results by observing what happens to an ice cube outside on a chilly winter day (or perhaps in your freezer, though it is harder to see what is happening when the door is closed). If the temperature outside is $-10\,^\circ \text C$ (or $14\,^\circ \text F$), we would expect the ice cube to stay solid.

## Other applications for $\Delta \text G$: A sneak preview

Being able to calculate $\Delta \text G$ can be enormously useful when we are trying to design experiments in lab! We will often want to know which direction a reaction will proceed at a particular temperature, especially if we are trying to make a particular product. Chances are we would strongly prefer the reaction to proceed in a particular direction (the direction that makes our product!), but it's hard to argue with a positive $\Delta \text G$!
Thermodynamics is also connected to concepts in other areas of chemistry. For example:

## Summary

• The second law of thermodynamics says that the entropy of the universe always increases for a spontaneous process: $\Delta \text {S}_{\text{universe}}=\Delta \text {S}_{\text{system}} + \Delta \text {S}_{\text{surroundings}} > 0$
• At constant temperature and pressure, the change in Gibbs free energy is defined as $\Delta \text G = \Delta \text H - \text{T}\Delta \text S$.
• When $\Delta \text G$ is negative, a process will proceed spontaneously and is referred to as exergonic.
• Depending on the signs of $\Delta \text H$ and $\Delta \text S$, the spontaneity of a process can change at different temperatures.

## Try it!

For the following reaction, $\Delta \text H_\text{rxn}=-120 \dfrac{\text{kJ}}{\text{mol-rxn}}$ and $\Delta \text S_\text{rxn}=-150 \dfrac{\text{J}}{\text{mol-rxn}\cdot \text K}$:
$2\text{NO}(g)+\text O_2(g) \rightarrow 2\text{NO}_2(g)$
At what temperatures will this reaction be spontaneous?
Note: Remember that we can assume that the $\Delta \text H$ and $\Delta \text S$ values are approximately independent of temperature.
Since $\Delta \text H$ is negative and $\Delta \text S$ is negative, the reaction can't be spontaneous at all temperatures because there will be at least some values for $\text T$ that will make $\Delta \text G$ positive.
To determine when the reaction will be spontaneous, we need to know when $\Delta \text G$ is negative. We can begin by finding the temperature where $\Delta \text G=0$.
$\Delta \text G =\Delta \text H - \text{T}\Delta \text S=0$
Rearranging our equation to solve for $\text T$, we get:
\begin{aligned}\Delta \text H - \text{T}\Delta \text S &= 0 ~~~~~~~~~~~~\text{Move }\text T\Delta \text {S term to other side} \\ \\ \Delta \text H&= \text{T}\Delta \text S~~~~~~\text{Divide both sides by } \Delta \text S\\ \\ \text T&= \dfrac{\Delta \text H}{\Delta \text S}=\dfrac{-120 \dfrac{\text{kJ}}{\text{mol-rxn}}}{-0.15 \dfrac{\text{kJ}}{\text{mol-rxn}\cdot \text K}}=800\,\text K\end{aligned}
Thus, at $\text T=800\,\text K$, we know that $\Delta\text G=0$ and the reaction is at equilibrium.
When $\Delta \text S$ is negative, any increase in temperature from the equilibrium point will make the value for $\Delta \text G$ more positive which makes it less likely to be spontaneous.
Therefore, we can conclude that the reaction will be spontaneous when $\text T<800\,\text K$.