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### Course: Physical Chemistry (Essentials) - Class 11>Unit 6

Lesson 13: Gibb's free energy

# 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: $\mathrm{\Delta }{\text{S}}_{\text{universe}}=\mathrm{\Delta }{\text{S}}_{\text{system}}+\mathrm{\Delta }{\text{S}}_{\text{surroundings}}>0$
• At constant temperature and pressure, the change in Gibbs free energy is defined as $\mathrm{\Delta }\text{G}=\mathrm{\Delta }\text{H}-\text{T}\mathrm{\Delta }\text{S}$.
• When $\mathrm{\Delta }\text{G}$ 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:
$\text{C}\left(s,\text{diamond}\right)\to \text{C}\left(s,\text{graphite}\right)$
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, $\mathrm{\Delta }\text{H}$.
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:
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 $\frac{\text{kJ}}{\text{mol-rxn}}$.
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 $\mathrm{\Delta }\text{G}$, 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.
$\mathrm{\Delta }\text{G}={\text{G}}_{\text{final}}-{\text{G}}_{\text{initial}}$
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 ($\mathrm{\Delta }{\text{H}}_{\text{system}}$) and entropy ($\mathrm{\Delta }{\text{S}}_{\text{system}}$) for our system:
$\mathrm{\Delta }{\text{G}}_{\text{system}}=\mathrm{\Delta }{\text{H}}_{\text{system}}-\text{T}\mathrm{\Delta }{\text{S}}_{\text{system}}$
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 $\mathrm{\Delta }\text{H}$ and $\mathrm{\Delta }\text{S}$ 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, $\mathrm{\Delta }\text{H}$, and the entropy change , $\mathrm{\Delta }\text{S}$, of the system. We can use the sign of $\mathrm{\Delta }\text{G}$ to figure out whether a reaction is spontaneous in the forward direction, backward direction, or if the reaction is at equilibrium.
• When $\mathrm{\Delta }\text{G}<0$, the process is exergonic and will proceed spontaneously in the forward direction to form more products.
• When $\mathrm{\Delta }\text{G}>0$, 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 $\mathrm{\Delta }\text{G}=0$, the system is in equilibrium and the concentrations of the products and reactants will remain constant.

## Calculating change in Gibbs free energy

Although $\mathrm{\Delta }\text{G}$ is temperature dependent, it's generally okay to assume that the $\mathrm{\Delta }\text{H}$ and $\mathrm{\Delta }\text{S}$ values are independent of temperature as long as the reaction does not involve a phase change. That means that if we know $\mathrm{\Delta }\text{H}$ and $\mathrm{\Delta }\text{S}$, we can use those values to calculate $\mathrm{\Delta }\text{G}$ at any temperature. We won't be talking in detail about how to calculate $\mathrm{\Delta }\text{H}$ and $\mathrm{\Delta }\text{S}$ in this article, but there are many methods to calculate those values including:
When the process occurs under standard conditions (all gases at $1\phantom{\rule{0.167em}{0ex}}\text{bar}$ pressure, all concentrations are $1\phantom{\rule{0.167em}{0ex}}\text{M}$, and $\text{T}=25{\phantom{\rule{0.167em}{0ex}}}^{\circ }\text{C}$), we can also calculate $\mathrm{\Delta }\text{G}$ using the standard free energy of formation, ${\mathrm{\Delta }}_{f}{\text{G}}^{\circ }$.
Problem-solving tip: It is important to pay extra close attention to units when calculating $\mathrm{\Delta }\text{G}$ from $\mathrm{\Delta }\text{H}$ and $\mathrm{\Delta }\text{S}$! Although $\mathrm{\Delta }\text{H}$ is usually given in $\frac{\text{kJ}}{\text{mol-reaction}}$, $\mathrm{\Delta }\text{S}$ is most often reported in $\frac{\text{J}}{\text{mol-reaction}\cdot \text{K}}$. The difference is a factor of $1000$!!

## When is $\mathrm{\Delta }\text{G}$‍  negative?

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

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

• When the process is exothermic ($\mathrm{\Delta }{\text{H}}_{\text{system}}<0$), and the entropy of the system increases ($\mathrm{\Delta }{\text{S}}_{\text{system}}>0$), the sign of $\mathrm{\Delta }{\text{G}}_{\text{system}}$ is negative at all temperatures. Thus, the process is always spontaneous.
• When the process is endothermic, $\mathrm{\Delta }{\text{H}}_{\text{system}}>0$, and the entropy of the system decreases, $\mathrm{\Delta }{\text{S}}_{\text{system}}<0$, the sign of $\mathrm{\Delta }\text{G}$ is positive at all temperatures. Thus, the process is never spontaneous.
For other combinations of $\mathrm{\Delta }{\text{H}}_{\text{system}}$ and $\mathrm{\Delta }{\text{S}}_{\text{system}}$, the spontaneity of a process depends on the temperature.
• Exothermic reactions ($\mathrm{\Delta }{\text{H}}_{\text{system}}<0$) that decrease the entropy of the system ($\mathrm{\Delta }{\text{S}}_{\text{system}}<0$) are spontaneous at low temperatures.
• Endothermic reactions ($\mathrm{\Delta }{\text{H}}_{\text{system}}>0$) that increase the entropy of the system ($\mathrm{\Delta }{\text{S}}_{\text{system}}>0$) 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 $\mathrm{\Delta }\text{G}$‍  for melting ice

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:
${\mathrm{\Delta }}_{\text{fus}}\text{H}=6.01\frac{\text{kJ}}{\text{mol-rxn}}$
${\mathrm{\Delta }}_{\text{fus}}\text{S}=22.0\frac{\text{J}}{\text{mol-rxn}\cdot \text{K}}$
What is $\mathrm{\Delta }\text{G}$ for the melting of ice at $20{\phantom{\rule{0.167em}{0ex}}}^{\circ }\text{C}$?
The process we are considering is water changing phase from solid to liquid:
${\text{H}}_{2}\text{O}\left(s\right)\to {\text{H}}_{2}\text{O}\left(l\right)$
For this problem, we can use the following equation to calculate $\mathrm{\Delta }{\text{G}}_{\text{rxn}}$:
$\mathrm{\Delta }\text{G}=\mathrm{\Delta }\text{H}-\text{T}\mathrm{\Delta }\text{S}$
Luckily, we already know $\mathrm{\Delta }\text{H}$ and $\mathrm{\Delta }\text{S}$ 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:
$\text{T}=20{\phantom{\rule{0.167em}{0ex}}}^{\circ }\text{C}+273=293\phantom{\rule{0.167em}{0ex}}\text{K}$
If we plug the values for $\mathrm{\Delta }\text{H}$, $\text{T}$, and $\mathrm{\Delta }\text{S}$ into our equation, we get:
$\begin{array}{rl}\mathrm{\Delta }\text{G}& =\mathrm{\Delta }\text{H}-\text{T}\mathrm{\Delta }\text{S}\\ \\ & =6.01\frac{\text{kJ}}{\text{mol-rxn}}-\left(293\phantom{\rule{0.167em}{0ex}}\overline{)\text{K}}\right)\left(0.022\phantom{\rule{0.167em}{0ex}}\frac{\text{kJ}}{\text{mol-rxn}\cdot \overline{)\text{K}}\right)}\\ \\ & =6.01\phantom{\rule{0.167em}{0ex}}\frac{\text{kJ}}{\text{mol-rxn}}-6.45\phantom{\rule{0.167em}{0ex}}\frac{\text{kJ}}{\text{mol-rxn}}\\ \\ & =-0.44\phantom{\rule{0.167em}{0ex}}\frac{\text{kJ}}{\text{mol-rxn}}\end{array}$
Since $\mathrm{\Delta }\text{G}$ is negative, we would predict that ice spontaneously melts at $20{\phantom{\rule{0.167em}{0ex}}}^{\circ }\text{C}$. If you aren't convinced that result makes sense, you should go test it out!
Concept check: What is $\mathrm{\Delta }\text{G}$ for the melting of ice at $-10{\phantom{\rule{0.167em}{0ex}}}^{\circ }\text{C}$?

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

Being able to calculate $\mathrm{\Delta }\text{G}$ 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 $\mathrm{\Delta }\text{G}$!
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: $\mathrm{\Delta }{\text{S}}_{\text{universe}}=\mathrm{\Delta }{\text{S}}_{\text{system}}+\mathrm{\Delta }{\text{S}}_{\text{surroundings}}>0$
• At constant temperature and pressure, the change in Gibbs free energy is defined as $\mathrm{\Delta }\text{G}=\mathrm{\Delta }\text{H}-\text{T}\mathrm{\Delta }\text{S}$.
• When $\mathrm{\Delta }\text{G}$ is negative, a process will proceed spontaneously and is referred to as exergonic.
• Depending on the signs of $\mathrm{\Delta }\text{H}$ and $\mathrm{\Delta }\text{S}$, the spontaneity of a process can change at different temperatures.

## Try it!

For the following reaction, $\mathrm{\Delta }{\text{H}}_{\text{rxn}}=-120\frac{\text{kJ}}{\text{mol-rxn}}$ and $\mathrm{\Delta }{\text{S}}_{\text{rxn}}=-150\frac{\text{J}}{\text{mol-rxn}\cdot \text{K}}$:
$2\text{NO}\left(g\right)+{\text{O}}_{2}\left(g\right)\to 2{\text{NO}}_{2}\left(g\right)$
At what temperatures will this reaction be spontaneous?
Note: Remember that we can assume that the $\mathrm{\Delta }\text{H}$ and $\mathrm{\Delta }\text{S}$ values are approximately independent of temperature.

## Want to join the conversation?

• Well I got what the formula for gibbs free energy is. but what's the nature of this energy and why is it called 'free'? It does free work is what textbooks say but didn't get the intuitive feel.
• The word "free" is not a very good one! In fact, IUPAC recommend calling it Gibbs energy or the Gibbs function, although most chemists still refer to it as Gibbs free energy.

Gibbs originally called it available energy and that is a good term because it is the energy associated with a chemical reaction that is available (or you could say free) to do work, assuming constant T and P.
• Is there a difference between the notation ΔG and the notation ΔG˚, and if so, what is it?
• STP is not standard conditions. Standard conditions are 1.0 M solutions and gases at 1.0 atm. Standard conditions does not actually specify a temperature but almost all thermodynamic data is given at 25C (298K) so many people assume this temperature.
• Hey I´m stuck: The ∆G in a reaction is negative but the ∆H was positive and it is assumed that a change temperature doesn´ t significantly affect entropy and entalpy. What does this do to 1) spontanity 2) spontanity at high temp 3) value or sign of ∆S

i read it 3 times now but i´m still insecure - :(
• This looks like a homework question, so I'll give you some hints to get you on the riht path rather than answering directly.

3) We know that ∆G = ∆H - T∆S. Solving for ∆S, we have:

∆S=(∆H-∆G)/T.

We know (from the question) that ∆G is negative and that ∆H is positive. Temperature is always positive (in Kelvin). From these values, we can know for certain whether ∆S is positive or negative (hint: remember that we are subtracting ∆G!).

1) Knowing the sign of ∆G is enough to say whether the reaction is spontaneous or not under these conditions. If ∆G is negative (from the question), is the reaction spontaneous or non-spontaneous?

2) Let's use ∆G = ∆H - T∆S again. Since ∆H and ∆S don't change significantly with temperature (given in the question), we can assume that they keep the same signs and values: i.e. ∆H is still positive and ∆S is still whatever sign you figured out above. As T increases, the T∆S component gets bigger. T is always positive, so if ∆S is positive then a bigger T∆S will make ∆G more negative (since we subtract T∆S). If ∆S is negative, then the negative signs (from the subtraction and the sign of ∆S) will cancel out, and so as T∆S gets bigger, ∆G will get more positive.
• In the subject heading, 'When is ΔG is negative?', is it a typo that it says
'When the process is endothermic, ΔHsystem > 0, and the entropy of the system decreases, ΔSsystem>0, the sign of ΔG is positive at all temperatures. Thus, the process is never spontaneous' shouldn't the entropy be < 0? if there is a decrease in entropy?
• I think you are correct. Change in entropy must be smaller than zero, for the entropy to decrease. It is a typo. You can cross-check from the figure.
• The Entropy change is given by Enthalpy change divided by the Temperature. Then how can the entropy change for a reaction be positive if the enthalpy change is negative?
• Great question! Figuring out the answer has helped me learn this material.

One way to define entropy is Q/T (where Q is the heat associated with a reversible process {also note that ∆H is only equal to Q when P is constant}).

For spontaneous and thus irreversible reactions, the ∆S is the same as for a reversible reaction (because S is a state variable it doesn't depend on how we got from one condition to another). In contrast, ∆H is not equal to Q (because this is not a reversible reaction).

A later video helps explain this:
• Hi all, Sal sir said we would 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! ( located before summary at other applications of del G) .can anybody please explain?
• If ΔG is positive, then the only possible option is to vary the temperature but whether that would work depends on whether the reaction is exo- or endothermic and what the entropy change is.
• how do i see the sign of entropy when both reactant and product have the same phase
• We have to look up the ΔS for the whole reaction in a table (or test the reaction ourselves... I'd rather look it up!). The value will be either positive or negative. If the reaction can result in a phase change then we might be lucky enough to find a list that has the reaction with reactant and products in the phases we need.

Otherwise we could calculate the change in energy and the use the specific heat equations to see if the phase would change. The example above with melting ice looks a little different because the reaction was a phase change (ice to water) instead of the usual combining or splitting of molecules.