# Gibbs Free Energy

Thermodynamics | Free Energy.

Tip: This isn't the place to ask a question because the teacher can't reply.

1 of 2 videos by Dr. Hayek

## Key Questions

The Gibbs free energy is classically the energy associated with a chemical reaction that can be used to do work. It includes an enthalpy term, $\Delta H$, and an entropy term, $\Delta S$. It remains the unequivocal criterion for the spontaneity of chemical change.

#### Explanation:

By definition, the Gibbs free energy is defined by the relationship:

$\Delta G = \Delta H - T \Delta S$

We can introduce specified standard state conditions, but if $\Delta G$ is negative, then the reaction as written is spontaneous. If $\Delta G$ is positive, then the reaction as written is non-spontaneous. It may be related to the equilibrium constant, ${K}_{e q}$, by the relationship:

$\Delta G = - R T \ln {K}_{e q}$.

Given this relationship, only negative $\Delta G$ values will give rise to an equilibrium constant that is $> 1$, and hence spontaneous.

Often you see values of $\Delta G$ tabulated for standard state conditions, $298$ $K$, and near atmospheric pressure, in which case $\Delta {G}^{\circ}$ is specified.

I recommend this video that explains in details Gibbs free energy.

Why? Because Gibbs free energy is the single, unequivocal criterion for the spontaneity of chemical change.

#### Explanation:

Gibbs free energy is no longer included on the UK A level syllabus. It includes both an enthalpy term ($\Delta H$), and entropy term ($\Delta S$). Its sign predicts spontaneity for both physical and chemical reactions. It is still widely used.

Gibbs himself was an accomplished polymath, and made prodigious contributions to chemistry, physics, engineering, and mathematics.

• It's not measured. Actually, other MEASURABLE variables (volume and pressure) are measured in a certain way so that we can calculate $\Delta G$.

$\textcolor{b l u e}{\Delta G = {\int}_{{P}_{1}}^{{P}_{2}} V \mathrm{dP}}$

For the same-volume sample, change the pressure while keeping the temperature constant, and you can calculate $\Delta G$ from measurements. I derive this below.

DISCLAIMER: This can be a difficult topic, so ask questions if you are confused.

If we take a look at this "thermodynamic square", we see the following terms:

• $S$ for entropy, a natural variable, placed on a corner.
• $H$ for enthalpy, a state function, placed on a side.
• $P$ for pressure, a natural variable, placed on a corner.
• $U$ for internal energy, a state function, placed on a side.
• $G$ for Gibbs' Free Energy, a state function, placed on a side.
• $V$ for volume, a natural variable, placed on a corner.
• $A$ for Helmholtz Free Energy, a state function, placed on a side.
• $T$ for temperature, a natural variable, placed on a corner.

From these we can derive the Maxwell Thermodynamics Relations.

Since I couldn't find this square anywhere, I actually had to draw it myself. So, let's derive a familiar relationship to show that this square works.

ENTHALPY VS. TEMPERATURE, ENTROPY, VOLUME, AND PRESSURE

Let's use this square to find a Maxwell relation for enthalpy.

Because one arrow points from the side of $H$ towards the temperature, which is a natural variable, we write positive $T$. Likewise, the volume is a natural variable, and the other arrow also points from the side of $H$ towards the volume, so we write positive $V$.

The remaining terms placed right next to $H$ are changing. For those we write $\mathrm{dS}$ and $\mathrm{dP}$ because $S$ and $P$ are right next to $H$ on the square. They pair up with the term on the other end of their arrow.

So, we write:

$\textcolor{g r e e n}{\mathrm{dH} = T \mathrm{dS} + V \mathrm{dP}}$

GIBBS FREE ENERGY VS. ENTROPY, TEMPERATURE, VOLUME, AND PRESSURE

Next, let's get a Maxwell relation for the Gibbs' Free Energy.

Since the arrow is pointing towards temperature AND coming from the side opposite to $G$, $S$ is negative. Entropy IS a natural variable, which is why it said to not be changing in this case. Likewise, volume is a natural variable, and it is being pointed towards, making it positive.

Since $T$ is on the side of $G$, it changes, so we write $\mathrm{dT}$. It is being pointed towards AND it is on the same side as $G$. Similarly, since $P$ is on the side of $G$, it changes, so we use $\mathrm{dP}$. They pair up with the term on the other end of their arrow.

Therefore, we write:

$\textcolor{g r e e n}{\mathrm{dG} = - S \mathrm{dT} + V \mathrm{dP}}$

I promise there's a point to this.

THE FAMILIAR THERMODYNAMICS EQUATION

So, let's compare $\mathrm{dH}$ and $\mathrm{dG}$. Notice how they both contain $V \mathrm{dP}$. Watch this:

$\mathrm{dG} = - S \mathrm{dT} + \stackrel{\mathrm{dH}}{\overbrace{V \mathrm{dP} + T \mathrm{dS}}} - T \mathrm{dS}$

$= - \left(S \mathrm{dT} + T \mathrm{dS}\right) + \mathrm{dH}$

Hm... Familiar. The first term is a result of a product rule. $T \mathrm{dS} + S \mathrm{dT} = d \left(T S\right)$. Therefore, we can rewrite this as:

$\mathrm{dG} = \mathrm{dH} - d \left(T S\right)$

Or, as a more familiar notation:

$\textcolor{g r e e n}{\Delta G = \Delta H - \Delta \left(T S\right)}$

Or, at a constant temperature, like we normally use this equation in this form:

$\Delta G = \Delta H \cancel{- S \Delta T} + T \Delta S$

$\setminus m a t h b f \left(\Delta G = \Delta H - T \Delta S\right) \leftarrow$ FAMILIAR!

And you should know this equation! That means both equations we derived are correct.

HOW TO GET GIBBS' FREE ENERGY FROM MEASUREMENTS OF MEASURABLE VARIABLES

Now that we've established that, let's go back to the Gibbs' free energy Maxwell relation.

$\mathrm{dG} = - S \mathrm{dT} + V \mathrm{dP}$

Let's say we wanted to focus on the $V \mathrm{dP}$ term. We can take the partial derivative with respect to pressure at a constant temperature to do so. Thus, we can say $\mathrm{dT} = 0$ and cross out $- S \mathrm{dT}$.

$\textcolor{g r e e n}{{\left(\frac{\partial G}{\partial P}\right)}_{T} = V}$

Next, at a constant temperature, let us find an expression to use volume and pressure (two MEASURABLE variables) to calculate the Gibbs' Free Energy:

$\mathrm{dG} = V \mathrm{dP}$

$\textcolor{b l u e}{\Delta G = {\int}_{{P}_{1}}^{{P}_{2}} V \mathrm{dP}}$

Therefore, if we know the volume of the sample, and we change the pressure while keeping the experiment temperature constant, we can calculate $\Delta G$ using measurable variables.

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