Gibbs Energy

### Predicting spontaneity

In the pages on enthalpy and entropy, we saw that whether a chemical process is endothermic or exothermic, or leads to a local* increase or decrease in entropy, does not determine whether that process is spontaneous. That is, whether the process will occur all on its own (except perhaps for a little "push" at the beginning, such as igniting a combustion reaction).

It's not that enthalpy and entropy are of no value in predicting spontaneity; they are necessary.

* I say "local" here because we realize that the entropy of the universe always increases.

But what we have to do is come up with another state function that combines the two in order to predict whether a process is spontaneous.

There are several such functions, each serving a role in a particular set of conditions (pressure, temperature, &c.). The most useful of these is the Gibbs energy, or the "Gibbs free energy," or sometimes just the "free energy."

Mr. Gibbs is Josiah Willard Gibbs, an American scientist who made many important contributions to chemistry and physics, particularly in thermodynamics.

### The Gibbs energy defined

The Gibbs energy (G) is defined like this:

$$G = H - TS$$

where H is the enthalpy, T is the temperature in Kelvins (K), and S is the entropy. We're almost always more interested in the change in Gibbs energy, ΔG,

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

ΔG is defined under two very specific conditions.

1. Constant pressure because enthalpy is defined as the heat generated or absorbed by a process occurring at constant pressure, and

2. Constant temperature because the entropy is temperature-dependent. Notice that in this case, Δ(TS) = TΔS.

#### Conditions for spontaneity

I'll cut to the chase here and explain more later: When ΔG < 0 for a process, that process is spontaneous. When ΔG > 0 it's not.

Now we can define some conditions for spontaneity using the Gibbs energy. Recall that when ΔH < 0, the reaction is exothermic and when ΔH > 0 the reaction is endothermic. Those were agreed-upon a long time ago, so we have to use them.

When ΔS > 0, the entropy of the system increases and when ΔS < 0, the entropy decreases. If we apply those conditions to the Gibbs energy we get four possibilities:

#### 1. ΔH < 0, ΔS < 0

When this is true, the sign of ΔG can't be determined because the size of the product entropy change and temperature could swamp the negative enthalpy. In this case we just need to know ΔH and T·ΔS in order to determine spontaneity.

#### 2. ΔH < 0, ΔS > 0

In this case, ΔG is always negative, and the process is spontaneous at any temperature.

#### 3. ΔH > 0, ΔS < 0

In this case, ΔG is always positive, and the process won't be spontaneous at any temperature. These are reactions that just won't go.

#### 4. ΔH > 0, ΔS > 0

This is like case 1. The size of ΔS and the temperature will determine whether the process is spontaneous.

That's a lot of information to absorb. Let's summarize it.

#### The Gibbs energy

The Gibbs energy or Gibbs free energy (G) is a state function that depends on the state functions enthalpy (H) and entropy.

$$G = H - TS$$

The change in Gibbs energy is

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

The Gibbs energy is defined for processes that occur at constant pressure and constant temperature.

### Conditions for spontaneity

ΔH ΔS ΔG Spontaneity
negative negative ? Depends on the T and relative sizes of ΔH, ΔS
negative positive < 0 Spontaneous at any temperature
positive negative > 0 Not spontaneous at any temperature
positive positive ? Depends on the T and relative sizes of ΔH, ΔS

#### Another way to visualize it

Here is one more visual way to decide spontaneity, if it can be done quickly. There is some wiggle room for deciding in the indeterminate cases.

For example, when ΔH < 0 and ΔS < 0, there is a good chance that the free energy change will be negative at low temperatures, where the ΔS term in the Gibbs energy doesn't mean as much.

Likewise, when ΔH > 0 and ΔS > 0 (lower right square), there is some chance that the Gibbs energy change will be negative at high temperature, where a negative TΔS term could swamp a smaller ΔH term.

### The Gibbs energy and thermochemistry

In the sections on enthalpy and entropy, we've already learned how to apply the techniques of thermochemistry to chemical reactions. Recall that the enthalpy of a reaction is

$$\Delta H = \sum H_{f \; products}^o - \sum H_{f \; reactants}^o$$

where Hfo is the standard enthalpy of formation, and long lists of those have been measured, refined and tabulated. Wikipedia is a great place to look them up. The Greek letter (capital) sigma, Σ, means sum.

Likewise, the overall entropy change of a process can be calculated using tabulated standard entropies.

$$\Delta S = \sum S_{f \; products}^o - \sum S_{f \; reactants}^o$$

Remember that the unit of enthalpy is the Joule per mole (J·mol-1) and the unit of entropy is J·mol-1·K-1.

The thermochemistry of the Gibbs energy is very similar. Standard Gibbs energies of compounds have been measured, refined and tabulated, and you can look them up and use them to determine the spontaneity and other properties of chemical reactions and processes.

$$\Delta G = \sum G_{f \; products}^o - \sum G_{f \; reactants}^o$$

### Example 1

Calculate the Gibbs energy change for the reaction   $4 \, KClO_{3 \, (s)} \; \longrightarrow \; 3 \, KClO_{3 \, (s)} + KCl_{(s)} \; \text{ at T = 298 K}$.

Solution: In this example, we'll calculate the Gibbs energy change using tabulated enthalpies of formation and standard entropies.

First the ΔH:

The sum of the enthalpies (don't forget to multiply by the number of moles) of the products minus the the enthalpy of our single reactant is

\begin{align} \Delta H &= [3(-433) - 437] - [4(-398)] \\[5pt] &= -144 \; KJ \end{align}

Now the ΔS:

The entropy change is

\begin{align} \Delta S &= [3(151) + 82.6] - 4(143) \\[5pt] &= -36 \; J/K \end{align}

Now ΔG = ΔH - TΔS, and our temperature is 298K, which gives this free energy change:

\begin{align} \Delta G &= \Delta H - T \Delta S \\[5pt] &= -144 \; KJ - 298 \, K \left( \frac{-0.036 \, KJ}{K} \right) \\[5pt] &= -133 \; KJ \end{align}

Notice that between the last two steps we realized that 36 J = 0.036 KJ.

The ΔG is negative, so this is a spontaneous reaction at 298K. Notice that at some higher temperatures this reaction is not spontaneous. The point at which ΔG = 0 is

\begin{align} 0 &= -144 - 0.036 T \\[5pt] T &= 4000 \; K \end{align}

So above T = 4000K, the reaction is not spontaneous, and in fact, the reverse reaction is spontaneous.

It's not necessary to use enthalpies and entropies to calculate ΔG like this because standard Gibbs energies of formation have been measured, refined and tabulated, so ΔG can be calculated directly, as the next examples will show.

### Example 2

Calculate the Gibbs energy change for the combustion of octane, $C_8H_{18 \, (g)} + 25 \, O_{2 \, (g)} \, \longrightarrow \, 16 \, CO_{2 \, (g)} + 18 \, H_2O_{(g)}$.

Solution: It's easy to look up the values of G˚ directly, too. They are:

Don't forget to multiply these molar values by the number of moles in the balanced equation. The resulting ΔG is

\begin{align} \Delta G &= 18(-229) + 16(-394) - 2(16.4) \\[5pt] &= -10,459 \; KJ \\[5pt] &= -10.5 \; MJ \end{align}

This is quite a large ΔG. We'll learn later that ΔG is related to the equilibrium constant. Such a large negative ΔG means that the equation as written is not only spontaneous, but that the equilibrium lies largely to the right, as we'd expect for a combustion reaction.

### Example 3

Calculate the temperature at which the reaction   $Cu_2O_{\,(s)} + C_{\,(s)} \, \longrightarrow \, 2 \, Cu_{\,(s)} + CO_{\, (g)}$   becomes spontaneous.

Solution: Approach this problem as though you were calculating ΔG as ΔG = ΔH - TΔS, as we did in example 1. For this reaction (you could calculate them, but I'll give them to you), ΔH = 58 KJ and ΔS = 0.165 KJ/K. Thus we have

$$\Delta G = 58 - 0.165 \, T$$

This is a linear equation that's easy to solve. We want to find out where ΔG is negative, so we'll set ΔG equal to zero and solve for the temperature at the transition from positive to negative:

\begin{align} 0 &= 58 - 0.165 \, T \\[5pt] T &= 352 \, K \\[5pt] &= 78.3˚C \end{align}

To get a feel for what's going on here, we can plot the linear equation highlighted in green:

Notice that the slope of the graph is just ΔS. It's easy to see that for temperatures above 352K, ΔG is negative and the reaction will occur spontaneously.

### Practice problems

1.

Calculate the ΔH, ΔS, and ΔG of the reaction     N2 + O2   →   2 NO     when

(a) T = 25˚C

(b) T = 1000˚C

(c) T = 3000˚C

Solution

$$N_2 + O_2 \longrightarrow 2 \, NO$$

Here are some values you could look up from any table of thermodynamic data:

$\Delta H_f^o \\ KJ/mol$ $S^o \\ J/mol\cdot K$
NO (g) 90.25 210.76
O2 (g) 0* 205.14
N2 (g) 0* 191.61

* Elements in their standard states

Fist calculate ΔH, then ΔS. Then calculate a new &Delta G for each of the three temperatures.

$$\Delta H = 2(90.25) = 180.5 \, KJ/mol$$

\begin{align} \Delta S &= 2(210.76) - 205.14 - 191.61 \\[5pt] &= 24.77 \, J/mol\cdot K \\[5pt] &= 0.02477 KJ/mol \cdot K \end{align}

Now calculate the ΔG's at the three temperatures:

\require{cancel} \begin{align} & 25˚C \\[5pt] &\Delta G = \Delta H - T \Delta S \\[5pt] &= 180.5 - 298 \, \cancel{K}\left(0.02477 \frac{KJ}{mol\cancel{K}} \right) \\[5pt] &= \color{red}{+173 \, KJ/mol} \\[5pt] & 1000˚C \\[5pt] &\Delta G = 180.5 - 1298 \, \cancel{K}\left(0.02477 \frac{KJ}{mol\cancel{K}} \right) \\[5pt] &= \color{red}{+148 \, KJ/mol} \\[5pt] & 3000˚C \\[5pt] &\Delta G = 180.5 - 3298 \, \cancel{K} \left(0.02477 \frac{KJ}{mol \cancel{K}} \right) \\[5pt] &= \color{red}{+99 \, KJ/mol} \end{align}

None of these reactions is spontaneous at the specified temperature. We can go a step further here and calculate the temperature at which the reaction becomes spontaneous. We do that by setting ΔG to zero, the point where it becomes negative. It takes a higher temperature than that for this reaction to be spontaneous.

\begin{align} T &= \frac{\Delta H - \Delta G}{\Delta S} \\[5pt] &= \frac{180.5}{0.02477} = 7287 \, K \end{align}

2.

Consider the reaction 2 SO2 (g) + O2 (g) → 2 SO3 (g). At 298K, ΔGo = -141.6 KJ, ΔHo = -198 KJ and ΔSo = -188 J/K. (a) Is the reaction spontaneous at 25˚C. (b) Use the data to predict how ΔGo will change with increasing T, and (c) Calculate the temperature at which the reaction changes from nonspontaneous to spontaneous.

Solution

$$2 \, SO_{2 \, (g)} + O_{2 \, (g)} \: \longrightarrow \: 2 \, SO_{3 \, (g)}$$

Because both ΔH and ΔS are negative, this reaction is spontaneous at low temperatures, but will become non-spontaneous at some higher temperature. We can find that temperature by setting ΔG = 0.

At 298 K, we have:

\begin{align} \Delta G_o &= -141.6 \, KJ, \\[5pt] \Delta H_o &= -198 \, KJ \\[5pt] \Delta S_o &= -188 \, J/K \end{align}

Then at 25˚C, we have:

\begin{align} \Delta G_{25˚C} &= \Delta H - T \Delta S \\[5pt] &= -198 - (298 \, K)(-0.188) \\[5pt] &= -142 \, KJ/mol \end{align}

The reaction is spontaneous at 25˚C (298 K). The temperature at which it becomes nonspontaneous is

\begin{align} T &= \frac{\Delta H - \cancel{\Delta G}}{\Delta S} \\[5pt] &= \frac{-198}{-0.188} = 1,053 \, K = 780˚C \end{align}

3.

The synthesis of chloroform from methane and chlorine gas goes according to the equation

CH4 (g) + 3Cl2 (g) → CHCl3 (l) + 3HCl (g).

(a) Calculate the Gibbs energy, ΔGorxn of the reaction.

(b) Assuming that ΔHo and ΔSo remain constant, will the reaction be spontaneous at 500K?

Solution

$$CH_{4 \, (g)} + 3 Cl_{2 \, (g)} \: \rightarrow \: CHCl_{3 (l)} + 3 HCl_{(g)}$$

Here are the thermodynamic data you can look up in a table of thermodynamic data:

$\Delta H_f^o \\ KJ/mol$ $S^o \\ J/mol \, K$ $G^o \\ KJ/mol$
CH4 (g) -74.81 185.06 -50.72
Cl2 (g) 0* 223.07 0*
CHCl3 (l) -103.14 295.71 -70.34
HCl(g) -92.31 186.91 -95.3

At 298 K, we have:

\begin{align} \Delta G_o &= 3(-95.3) - 70.34 -[-50.72] \\[5pt] &= -305 \, KJ \; \text{of methane} \\[5pt] \Delta H_o &= -103.14 - 3(92.3) - [-74.81] \\[5pt] &= -305 \, KJ/mol \\[5pt] \Delta S_o &= 295.71 + 3(186.91) \\[5pt] &\phantom{00000}- [185.25 + 3(223.07)] \\[10pt] &= 0.98 \, J/mol \, K \\[5pt] \Delta G_{500˚C} &= \Delta H - T \Delta S \\[5pt] &= -305 - (500)(0.00098) \\[5pt] &= -305.49 \, KJ/mol \end{align}

The entropy change of this reaction is too small to change much as a function of temperature.

X

### State function

A state function is one that depends only on the initial and final conditions of the system, and not at all on how it got from one condition to the other.

Examples of state functions in thermodynamics are enthalpy, entropy, and the Gibbs energy.

xaktly.com by Dr. Jeff Cruzan is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. © 2012-2019, Jeff Cruzan. All text and images on this website not specifically attributed to another source were created by me and I reserve all rights as to their use. Any opinions expressed on this website are entirely mine, and do not necessarily reflect the views of any of my employers. Please feel free to send any questions or comments to jeff.cruzan@verizon.net.