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Free Energy

You may already know that for a reaction to be feasible, the change in enthalpy (∆H) must decrease. A feasible reaction must also show an increase in the change in entropy (∆S). Do you see a problem? What if in a given reaction, the enthalpy change decreased and the entropy change also decreased? Is the reaction feasible or not? How do we…

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Free Energy

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Jetzt kostenlos anmeldenYou may already know that for a reaction to be feasible, the change in **enthalpy **(∆H) must *decrease*. A feasible reaction must also show an *increase* in the change in **entropy **(∆S). Do you see a problem? What if in a given reaction, the enthalpy change decreased and the entropy change also decreased? Is the reaction feasible or not? How do we find out? Keep reading!

Remember:

Enthalpy (H) is the sum of the internal energy (E) and the product of the pressure (P) and the volume (V) of a thermodynamic system.

$H=E+PV$

A decrease in H means that the reaction released heat.

Entropy (S) is the measure of the disorder of a thermodynamic system. An increase in entropy means the system became more disordered (for example, when a solid turns liquid).

- In this article, you will learn how we use the equation for free energy to predict the feasibility of a reaction.
Discover how temperature affects reaction feasibility.

You will learn how to calculate ∆G using ∆Hº and ∆Sº values and using ∆Gº values.

Find out how to determine the temperature at which a reaction becomes feasible.

We cannot predict the feasibility of a reaction by looking *only* at ∆H or ∆S. We need a more accurate way to determine whether a reaction is feasible. **Gibbs free energy** (or simply, **free energy**) helps us predict whether a reaction is feasible by taking both the enthalpy change and entropy change of a reaction into account.

Gibbs free energy (∆G) is the relationship between the change in enthalpy (∆H) and the change in entropy (∆S) to determine the feasibility of a reaction. We say** **it is the amount of energy in a system available to do work.

We express this relationship in the equation:

$\mathbf{\u2206}{\mathbf{G}}^{\mathbf{\Theta}}\mathbf{}\mathbf{=}\mathbf{}\mathbf{\u2206}{\mathbf{H}}^{\mathbf{\Theta}}\mathbf{}\mathbf{-}\mathbf{}\mathbf{T}\mathbf{\u2206}{\mathbf{S}}^{\mathbf{\Theta}}$

- Where ∆G is the change in free energy.
- ∆H is the enthalpy change of the reaction.
- T is the given temperature.
- And, ∆S is the entropy change of the system.

Before we begin, here are a few words on the origin of the equation for Gibbs free energy. You won't be asked about this in your exams, but it adds to your understanding of free energy. Find out more below!

We say that the change in total entropy $(\u2206{\mathrm{S}}_{\left(\mathrm{total}\right)})$ must increase if a reaction is to be feasible. Total entropy is the sum of the entropy change in the system and the surroundings.

$\u2206{S}_{\left(total\right)}=\u2206{S}_{\left(system\right)}+\u2206{S}_{\left(surroundings\right)}$

We can relate the change in the entropy of the surroundings $(\u2206{\mathrm{S}}_{\left(\mathrm{surroundings}\right)})$ to the change in enthalpy (∆H) of a reaction by the following equation:

**$\u2206{\mathrm{S}}_{\left(\mathrm{surroundings}\right)}=-\u2206\mathrm{H}/\mathrm{T}$ **

Where ∆H is the enthalpy change in the reaction and T is the temperature.

If you put the two equations for the change in total entropy and the change in entropy of the surroundings together, you would get an equation that looks a little like this:

**$\u2206{\mathrm{S}}_{\left(\mathrm{total}\right)}=-\u2206\mathrm{H}/\mathrm{T}+\u2206{\mathrm{S}}_{\left(\mathrm{system}\right)}$**

By doing a little rearranging we end up with the following equation:

$\mathrm{T}\u2206{\mathrm{S}}_{\left(\mathrm{total}\right)}=-\u2206\mathrm{H}+\mathrm{T}\u2206{\mathrm{S}}_{\left(\mathrm{system}\right)}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}{\mathbf{.}}{\mathbf{\xb7}}{\mathbf{.}}{\mathbf{}}{\mathbf{-}}{\mathbf{T}}{\mathbf{\u2206}}{{\mathbf{S}}}_{\mathbf{\left(}\mathbf{total}\mathbf{\right)}}{\mathbf{}}{\mathbf{=}}{\mathbf{}}{\mathbf{\u2206}}{\mathbf{H}}{\mathbf{}}{\mathbf{-}}{\mathbf{}}{\mathbf{T}}{\mathbf{\u2206}}{{\mathbf{S}}}_{\mathbf{\left(}\mathbf{system}\mathbf{\right)}}$Does the equation look familiar? It’s the same as the one for free energy - simply replace the term $-\mathrm{T}\u2206{\mathrm{H}}_{\left(\mathrm{total}\right)}$ with ∆G!

${\mathbf{.}}{\mathbf{\xb7}}{\mathbf{.}}{\mathbf{}}{\mathbf{\u2206}}{\mathbf{G}}{\mathbf{}}{\mathbf{=}}{\mathbf{}}{\mathbf{\u2206}}{\mathbf{H}}{\mathbf{}}{\mathbf{-}}{\mathbf{}}{\mathbf{T}}{\mathbf{\u2206}}{{\mathbf{S}}}_{\mathbf{\left(}\mathbf{system}\mathbf{\right)}}$

Notice how the term for ∆G is negative? This means that **for a reaction to be feasible, ∆G must be negative**. Another thing to take note of is how the equation for free energy refers to the **entropy change in a system ∆S(system)**, not total entropy.

You may be wondering why temperature is included in the equation for free energy. The answer is that **temperature affects the feasibility of a reaction**. For example, we say that combustion is a spontaneous process. However, if you were to leave some wood in a room, would it combust? What if you increase the temperature by lighting a fire under the wood? Does the reaction become more or less feasible? Clearly, by increasing the temperature you have increased the feasibility of combustion taking place.

You can see that ∆S is connected to temperature in the equation for free energy. Entropy is limited to the total energy available to a system. This means if you were to add heat to a sample, you would increase its total entropy value. **The entropy of a system increases as its temperature increases** because the particles gain energy and move about more. In other words, the particles become more spread out or disordered. There is a natural tendency towards an increase in entropy.

**All spontaneous processes involve an increase in the total entropy of a system.**

Have a look at the equation for free energy again:

∆Gº = ∆Hº - T∆Sº

We can conclude a few things by considering the effect of temperature on ∆G.

- If ∆H is negative (
*exothermic*, i.e. the reaction releases heat) and ∆S is positive, ∆G will be negative, therefore the reaction is feasible. - If ∆H is negative and ∆S is negative, at low temperatures the reaction is feasible. As the temperature increases, T∆S will also increase enough to overcome ∆H of the reaction. At high temperatures the reaction may not be feasible.
- If ∆H is positive (
*endothermic*, i.e., the reaction absorbs heat) and ∆S is positive, the reaction is feasible at high temperatures. At low temperatures T∆S may not be large enough to overcome ∆H of the reaction. - If ∆H is positive and ∆S is negative, the reaction is not feasible.

That was a lot to take in! The image below will help you put it all together.

There are two ways you can calculate free energy. The first way is by using standard change in enthalpy (∆Hº) and entropy (∆Sº) values. The second is from the free energy (∆Gº) values of the substances in the system.

Before you learn these two methods to calculate free energy, you must remember that Gibbs free energy, like enthalpy and entropy, is a state function and has standard values for compounds. Most of the time, though, we calculate free energy under non-standard conditions. So you often see ∆G without the symbol for standard conditions (º).

The units for the equation ∆Gº = ∆Hº - T∆Sº :

- ∆G in $\mathrm{kJ}{\mathrm{mol}}^{-1}$
- ∆H in $\mathrm{kJ}{\mathrm{mol}}^{-1}$
- T in K
- ∆S in ${\mathrm{JK}}^{-1}{\mathrm{mol}}^{-1}$

Be careful with the **units of entropy** when doing ∆G calculations. Remember to change them to $kJ\mathit{}{K}^{\mathit{-}\mathit{1}}mo{l}^{-1}$by dividing by 1000.

∆Sº here represents the entropy change of a reaction and not total entropy.

1. Calculate ∆Gº for the combustion of methane.

${\mathrm{CH}}_{4\left(\mathrm{g}\right)}+2{\mathrm{O}}_{2\left(\mathrm{g}\right)}\to {\mathrm{CO}}_{2\left(\mathrm{g}\right)}+2{\mathrm{H}}_{2}{\mathrm{O}}_{\left(\mathrm{l}\right)}$

∆Sº = -242.2 ${\mathrm{JK}}^{-1}{\mathrm{mol}}^{-1}$

∆Hº = -890.4 $\mathrm{kJ}{\mathrm{mol}}^{-1}$

Change the value for entropy into $\mathrm{kJ}{\mathrm{mol}}^{-1}$ by dividing by 1000

-242.2 ÷ 1000 = -0.2422 $\mathrm{kJ}{\mathrm{K}}^{-1}{\mathrm{mol}}^{-1}$

Write the equation for Gibbs free energy.

∆G = ∆H - T∆S

Since we are asked for standard values, T = 298K.

Fill in the given values:

∆G = -890 - 298(-0.2422)

∆G = -817.6 $\mathrm{kJ}{\mathrm{mol}}^{-1}$

2. Use the given values for $\u2206{\mathrm{G}}^{\mathrm{\Theta}}$ to find the $\u2206{\mathrm{G}}^{\mathrm{\Theta}}$ of the following reaction

${\mathrm{C}}_{2}{\mathrm{H}}_{5}{\mathrm{OH}}_{\left(\mathrm{l}\right)}+3{\mathrm{O}}_{2\left(\mathrm{g}\right)}\to 2{\mathrm{CO}}_{2\left(\mathrm{g}\right)}+3{\mathrm{H}}_{2}{\mathrm{O}}_{\left(\mathrm{g}\right)}$

Substance | ∆Gº ${\mathbf{kJmol}}^{\mathbf{-}\mathbf{1}}$ |

${\mathrm{C}}_{2}{\mathrm{H}}_{5}{\mathrm{OH}}_{\left(\mathrm{l}\right)}$ | -175 |

${\mathrm{O}}_{2\left(\mathrm{g}\right)}$ | 0 |

${\mathrm{CO}}_{2\left(\mathrm{g}\right)}$ | -395 |

${\mathrm{H}}_{2}{\mathrm{O}}_{\left(\mathrm{g}\right)}$ | -229 |

When we know the ∆G of the substances in the reaction, we can use the equation

$\u2206\mathrm{G}\xba=\sum \u2206{\mathrm{G}}_{\mathrm{products}}\xba-\sum \u2206{\mathrm{G}}_{\mathrm{reactants}}\xba$ to calculate $\u2206{\mathrm{G}}^{\mathrm{\Theta}}$

Fill in the equation using the given values:

∆Gº = [ 2(-394) + 3(-229) ] - [ (-175) + 3(0) ]

∆Gº = -1300 ${\mathrm{kJmol}}^{-1}$

When we talk about spontaneity and feasibility we often mean the same thing. But so we’re on the same page:

- When we say ‘feasible’, we mean the reaction is energetically favourable. In other words, the reaction
*should*take place. - When we say ‘spontaneous’, we usually mean the reaction happens on its own, without outside interference. Spontaneous reactions can be very, very slow!

Some chemists like to describe the combustion of carbon as a spontaneous reaction. However, a piece of carbon left out in the open will not 'spontaneously' combust no matter how long you leave it there, unless you first apply some heat! The reaction is possible, it just has a high activation barrier. That's why some chemists prefer to use the term 'feasible' instead of 'spontaneous'.

Free energy helps us determine the feasibility of a reaction.

- When ∆G is negative the reaction is feasible.
- When ∆G is positive the reaction is not feasible.

By rearranging the equation for free energy we can determine at what temperature a non-spontaneous reaction becomes feasible. A reaction becomes feasible at the point where ∆G = 0. In other words when ∆H = T∆S.

0 = ∆H - T∆S

.˙. ∆H = T∆S

.˙. T = ∆H / ∆S

At what temperature will the reaction between carbon oxide and carbon dioxide become spontaneous?

${\mathrm{CaO}}_{\left(\mathrm{s}\right)}+{\mathrm{CO}}_{2\left(\mathrm{g}\right)}\to {\mathrm{CaCO}}_{3\left(\mathrm{s}\right)}$

∆H = -178 ${\mathrm{kJmol}}^{-1}$

∆S = -161 ${\mathrm{JK}}^{-1}{\mathrm{mol}}^{-1}$

Change units for entropy to $\mathrm{kJ}{\mathrm{K}}^{-1}{\mathrm{mol}}^{-1}$

161 / 1000 = 0.161 kJK^{-1}mol^{-1}

T = ∆H / ∆S

T = 178 / 0.161

T = 1105.59 K

To conclude, let's take a moment to discuss why we call it ‘free’ energy? What's so 'free' about it? Well, we used to call free energy by another name: *available* energy. In other words, free energy is the energy in a system we have available (or free) to do work after a reaction. Think about it: when a chemical reaction takes place, the enthalpy change is the difference between the energy used to break bonds and make new ones. Entropy change is the ‘energy cost’ of the reaction. The energy left over is what we know as free energy!

To explain further, have a look at the following reaction.

${\mathrm{HCl}}_{\left(\mathrm{g}\right)}+{\mathrm{NH}}_{3\left(\mathrm{g}\right)}\to {\mathrm{NH}}_{4}{\mathrm{Cl}}_{\left(\mathrm{s}\right)}$

If you were to calculate ∆H for this reaction you would find that its value is negative (i.e., the reaction is exothermic. It gives out energy). Again, If you were to calculate ∆S (system) you would find that its value is also negative (since energy is less spread out in a solid than in a gas). Why then does the reaction take place? Because of the overall increase in the total entropy change for the system and its surroundings.

Remember the second law of thermodynamics:

In a spontaneous process, the total entropy change for a system and its surroundings is positive." -The second law

This means that the reaction is feasible because the heat energy given out into the surroundings (∆H) increases the entropy of the surroundings ($\u2206{\mathrm{S}}_{\left(\mathrm{surroundings}\right)}$) enough to make up for the decrease in entropy in the system ($\u2206{\mathrm{S}}_{\left(\mathrm{system}\right)}$).

In thermodynamics, a system is a substance or a collection of substances and energy. Everything that is not in the system, we call the surroundings. If a reaction takes place in a jar, then the jar is the system. Everything outside the jar is the surroundings.

Here are the actual ∆H and ∆S values for the above reaction:

∆H = -176 ${\mathrm{kJmol}}^{-1}$

∆S = -284 ${\mathrm{JK}}^{-1}{\mathrm{mol}}^{-1}$

We can conclude that the change in entropy of the surroundings ($\u2206{\mathrm{S}}_{\left(\mathrm{surroundings}\right)}$) must be equal to or greater than +284${\mathrm{JK}}^{-1}{\mathrm{mol}}^{-1}$ (0.284 $\mathrm{kJ}{\mathrm{K}}^{-1}{\mathrm{mol}}^{-1}$). The ∆H of -176 ${\mathrm{kJmol}}^{-1}$ is more than enough to cover this! The value that interests us the most is the amount of the enthalpy change (∆H) we need to make the reaction feasible.

To find this out, we use the relationship between the entropy change of the surroundings $(\u2206{\mathrm{S}}_{\left(\mathrm{surroundings}\right)})$ and the change in enthalpy of a reaction:

$\mathbf{\u2206}{\mathbf{S}}_{\mathbf{\left(}\mathbf{surroundings}\mathbf{\right)}}\mathbf{}\mathbf{=}\mathbf{}\mathbf{-}\mathbf{\u2206}\mathbf{H}\mathbf{}\mathbf{/}\mathbf{}\mathbf{T}$

If we rearrange this expression we get:

$\mathbf{\u2206}\mathbf{H}\mathbf{}\mathbf{=}\mathbf{}\mathbf{-}\mathbf{T}\mathbf{\u2206}{\mathbf{S}}_{\mathbf{\left(}\mathbf{surroundings}\mathbf{\right)}}$

So at room temperature, 298K:

∆H = -(298)(0.284)

∆H = -85 ${\mathrm{kJmol}}^{-1}$

This means, of the -175 ${\mathrm{kJmol}}^{-1}$ of energy produced by the making and breaking of bonds, 86** ${\mathrm{kJmol}}^{-1}$** of it *must* be released into the surroundings as heat for the reaction to be feasible. The rest of it is 'free' energy, and is available to be used as other forms of energy!

- Gibbs free energy is the amount of energy available in a system to do work.
- The equation for free energy shows the relationship between entropy and enthalpy as ∆G = ∆H - T∆S.
- For a reaction to be feasible ∆G must be negative or equal to zero.
- Temperature affects the feasibility of a reaction.
- Entropy increases with temperature because the particles become more spread out and disordered.

1. Calculate ∆Gº for the combustion of methane.

∆Sº = -242.2

∆Hº = -890.4

Change the value for entropy into by dividing by 1000

-242.2 ÷ 1000 = -0.2422

Write the equation for Gibbs free energy:

∆G = ∆H - T∆S

Since we are asked for standard values, T = 298K.

Fill in the given values:

∆G = -890 - 298(-0.2422)

∆G = -817.6

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