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: ΔSuniverse=ΔSsystem+ΔSsurroundings>0\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 ΔG=ΔHTΔS\Delta \text G = \Delta \text H - \text{T}\Delta \text S.
  • When ΔG\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:
C(s,diamond)C(s,graphite)\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, ΔH\Delta \text H.
Dissolving sodium chloride, NaCl\text{NaCl}, in water at 25 C25~^\circ \text C is an endothermic reaction! The heat of solution, ΔHsoln\Delta\text H_{soln}, for sodium chloride at 25 C25~^\circ \text C is +3.87kJmol+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:
ΔSuniverse=ΔSsystem+ΔSsurroundings>0        For a spontaneous process\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 T\text T and pressure P\text P, we can rearrange the second law of thermodynamics and define a new quantity known as Gibbs free energy:
Gibbs free energy=G=HTS\text{Gibbs free energy}=\text G =\text H - \text{TS}
where H\text H is enthalpy, T\text T is temperature (in kelvin, K\text K), and S\text S is the entropy. Gibbs free energy is represented using the symbol G\text G and typically has units of kJmol-rxn\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 mol-rxn\text{mol-rxn} or mol-reaction\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:
2Al(s)+3Cl2(g)2AlCl3(s)2 \text{Al}(s)+ 3\text{Cl}_2(g) \rightarrow 2 \text{AlCl}_3(s)
we would say that 11 mole of reaction is when 22 moles of Al\text{Al} react with 33 moles of Cl2\text{Cl}_2 to produce 22 moles of AlCl3\text{AlCl}_3. We can write these relationships mathematically as well:
1mol-rxn=2mol Al=3mol Cl2=2mol AlCl31\,\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 G\text G, rather than its absolute value. The change in Gibbs free energy for a process is thus written as ΔG\Delta \text G, which is the difference between Gfinal\text G_{\text{final}}, the Gibbs free energy of the products, and Ginitial\text{G}_{\text{initial}}, the Gibbs free energy of the reactants.
ΔG=GfinalGinitial\Delta \text G =\text G_{\text{final}} - \text{G}_{\text{initial}}
For a process at constant T\text T and constant P\text P, we can rewrite the equation for Gibbs free energy in terms of changes in the enthalpy (ΔHsystem\Delta \text H_{\text{system}}) and entropy (ΔSsystem\Delta \text S_{\text{system}}) for our system:
ΔGsystem=ΔHsystemTΔSsystem\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 ΔH\Delta \text H and ΔS\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, ΔH\Delta \text H, and the entropy change , ΔS\Delta \text S, of the system. We can use the sign of ΔG\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 ΔG<0\Delta \text G<0, the process is exergonic and will proceed spontaneously in the forward direction to form more products.
  • When ΔG>0\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 ΔG=0\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 ΔG\Delta \text G is temperature dependent, it's generally okay to assume that the ΔH\Delta \text H and ΔS\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 ΔH\Delta \text H and ΔS\Delta \text S, we can use those values to calculate ΔG\Delta \text G at any temperature. We won't be talking in detail about how to calculate ΔH\Delta \text H and ΔS\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 1bar1\,\text {bar} pressure, all concentrations are 1M1\,\text M, and T=25C\text T=25\,^\circ\text C), we can also calculate ΔG\Delta \text G using the standard free energy of formation, ΔfG\Delta_{f} \text G^\circ.
Problem-solving tip: It is important to pay extra close attention to units when calculating ΔG\Delta \text G from ΔH\Delta \text H and ΔS\Delta \text S! Although ΔH\Delta \text H is usually given in kJmol-reaction\dfrac{\text{kJ}}{\text{mol-reaction}}, ΔS\Delta \text S is most often reported in Jmol-reactionK\dfrac{\text{J}}{\text{mol-reaction}\cdot \text K}. The difference is a factor of 10001000!!

When is ΔG\Delta \text G negative?

If we look at our equation in greater detail, we see that ΔGsystem\Delta \text G_\text{system} depends on 33 values:

ΔGsystem=ΔHsystemTΔSsystem\Delta \text G_{\text{system}} =\Delta \text H_{\text{system}} - \text{T}\Delta \text S_{\text{system}}
  • the change in enthalpy ΔHsystem\Delta \text H_{\text{system}}
  • the temperature T\text T
  • the change in entropy ΔSsystem\Delta \text S_{\text{system}}
Temperature in this equation always positive (or zero) because it has units of K\text K. Therefore, the second term in our equation, TΔSsystem\text T \Delta \text S_\text{system}, will always have the same sign as ΔSsystem\Delta \text S_\text{system}. We can make the following conclusions about when processes will have a negative ΔGsystem\Delta \text G_\text{system}:
  • When the process is exothermic (ΔHsystem<0\Delta \text H_{\text{system}}<0), and the entropy of the system increases (ΔSsystem>0\Delta \text S_{\text{system}}>0), the sign of ΔGsystem\Delta \text G_{\text{system}} is negative at all temperatures. Thus, the process is always spontaneous.
  • When the process is endothermic, ΔHsystem>0\Delta \text H_{\text{system}}>0, and the entropy of the system decreases, ΔSsystem<0\Delta \text S_{\text{system}}<0, the sign of ΔG\Delta \text G is positive at all temperatures. Thus, the process is never spontaneous.
For other combinations of ΔHsystem\Delta \text H_\text{system} and ΔSsystem\Delta \text S_\text{system}, the spontaneity of a process depends on the temperature.
  • Exothermic reactions (ΔHsystem<0\Delta \text H_\text{system}<0) that decrease the entropy of the system (ΔSsystem<0\Delta \text S_\text{system}<0) are spontaneous at low temperatures.
  • Endothermic reactions (ΔHsystem>0\Delta \text H_\text{system}>0) that increase the entropy of the system (ΔSsystem>0\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 11: Calculating ΔG\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:
ΔfusH=6.01kJmol-rxn\Delta_\text{fus} \text H=6.01 \dfrac{\text{kJ}}{\text{mol-rxn}}
ΔfusS=22.0Jmol-rxnK\Delta_\text{fus} \text S=22.0 \dfrac{\text{J}}{\text{mol-rxn}\cdot \text K}
What is ΔG\Delta \text G for the melting of ice at 20C20\,^\circ \text C?
The process we are considering is water changing phase from solid to liquid:
H2O(s)H2O(l)\text H_2 \text O(s) \rightarrow \text H_2 \text O(l)
For this problem, we can use the following equation to calculate ΔGrxn\Delta \text G_\text{rxn}:
ΔG=ΔHTΔS\Delta \text G =\Delta \text H - \text{T}\Delta \text S
Luckily, we already know ΔH\Delta \text H and ΔS\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:
T=20C+273=293K\text T=20\,^\circ \text C+273 =293\,\text K
If we plug the values for ΔH\Delta \text H, T\text T, and ΔS\Delta \text S into our equation, we get:
ΔG=ΔHTΔS=6.01kJmol-rxn(293K)(0.022kJmol-rxnK)=6.01kJmol-rxn6.45kJmol-rxn=0.44kJmol-rxn\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 ΔG\Delta \text G is negative, we would predict that ice spontaneously melts at 20C20\,^\circ \text C. If you aren't convinced that result makes sense, you should go test it out!
Concept check: What is ΔG\Delta \text G for the melting of ice at 10C-10\,^\circ \text C?
First we need to calculate our temperature to kelvins:
T=10C+273=263K\text T=-10\,^\circ \text C+273 =263\,\text K:
If we plug the values into our equation to calculate ΔG\Delta \text G, we get:
ΔG=ΔHTΔS=6.01kJmol-rxn(263K)(0.022kJmol-rxnK)=6.01kJmol-rxn5.79kJmol-rxn=0.22kJmol-rxn\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 10C-10\,^\circ \text C the Gibbs free energy change ΔG\Delta \text G is positive for the melting of water. Therefore, we would predict that the reaction is not spontaneous at 10C-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 10C-10\,^\circ \text C (or 14F14\,^\circ \text F), we would expect the ice cube to stay solid.

Other applications for ΔG\Delta \text G: A sneak preview

Being able to calculate ΔG\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 ΔG\Delta \text G!
Thermodynamics is also connected to concepts in other areas of chemistry. For example:


  • The second law of thermodynamics says that the entropy of the universe always increases for a spontaneous process: ΔSuniverse=ΔSsystem+ΔSsurroundings>0\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 ΔG=ΔHTΔS\Delta \text G = \Delta \text H - \text{T}\Delta \text S.
  • When ΔG\Delta \text G is negative, a process will proceed spontaneously and is referred to as exergonic.
  • Depending on the signs of ΔH\Delta \text H and ΔS\Delta \text S, the spontaneity of a process can change at different temperatures.


This article is adapted from the following articles:
The modified article is licensed under a CC-BY-NC-SA 4.0 license.

Try it!

For the following reaction, ΔHrxn=120kJmol-rxn\Delta \text H_\text{rxn}=-120 \dfrac{\text{kJ}}{\text{mol-rxn}} and ΔSrxn=150Jmol-rxnK\Delta \text S_\text{rxn}=-150 \dfrac{\text{J}}{\text{mol-rxn}\cdot \text K}:
2NO(g)+O2(g)2NO2(g)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 ΔH\Delta \text H and ΔS\Delta \text S values are approximately independent of temperature.
Choose 1 answer:
Choose 1 answer:
Since ΔH\Delta \text H is negative and ΔS\Delta \text S is negative, the reaction can't be spontaneous at all temperatures because there will be at least some values for T\text T that will make ΔG\Delta \text G positive.
To determine when the reaction will be spontaneous, we need to know when ΔG\Delta \text G is negative. We can begin by finding the temperature where ΔG=0\Delta \text G=0.
ΔG=ΔHTΔS=0\Delta \text G =\Delta \text H - \text{T}\Delta \text S=0
Rearranging our equation to solve for T\text T, we get:
ΔHTΔS=0            Move TΔS term to other sideΔH=TΔS      Divide both sides by ΔST=ΔHΔS=120kJmol-rxn0.15kJmol-rxnK=800K\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 T=800K\text T=800\,\text K, we know that ΔG=0\Delta\text G=0 and the reaction is at equilibrium.
When ΔS\Delta \text S is negative, any increase in temperature from the equilibrium point will make the value for ΔG\Delta \text G more positive which makes it less likely to be spontaneous.
Therefore, we can conclude that the reaction will be spontaneous when T<800K\text T<800\,\text K.