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, and American scientist who made numerous important contributions to chemistry and physics, particularly in thermodynamics.

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.


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, ∆G,

∆G = ∆H - T∆S

∆G 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.

Conditions for spontaneity

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

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

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

1. ∆H < 0, ∆S < 0

When this is true, the sign of ∆G 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 ∆H and T·∆S in order to determine spontaneity.

2. ∆H < 0, ∆S > 0

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

3. ∆H > 0, ∆S < 0

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

4. ∆H > 0, ∆S > 0

This is like case 1. The size of ∆S 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

∆G = ∆H - T∆S

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

Conditions for spontaneity

∆H ∆S ∆G Spontaneity
negative negative ? Depends on the T and relative sizes of ∆H, ∆S
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 ∆H, ∆S

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 ∆H < 0 and ∆S < 0, there is a good chance that the free energy change will be negative at low temperatures, where the ∆S term in the Gibbs energy doesn't mean as much.

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

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.

Remember that the unit of enthalpy is the Joule/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.


Example 1

Calculate the Gibbs energy change for the reaction

4 KClO3 (s) → 3 KClO4 (s) + KCl(s) at T = 298K


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

First the ∆H:

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

Now the ∆S:

The entropy change is

Now ∆G = ∆H - T∆S, and our temperature is 298K, which gives this free energy change:

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

The ∆G 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 ∆G = 0 is

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 ∆G like this because standard Gibbs energies of formation have been measured, refined and tabulated, so ∆G can be calculated directly, as the next examples will show.


Example 2

Calculate the Gibbs energy change for the combustion of octane,

C8H18 (g) + 25 O2 (g) → 16 CO2 (g) + 18 H2O(g)


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

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

This is quite a large ∆G. We'll learn later that ∆G is related to the equilibrium constant. Such a large negative ∆G 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

Cu2O(s) + C(s) → 2 Cu(s) + CO(g)

becomes spontaneous.


Solution: Approach this problem as though you were calculating ∆G as ∆G = ∆H - T∆S, as we did in example 1. For this reaction (you could calculate them, but I'll give them to you), ∆H = 58 KJ and ∆S = 0.165 KJ/K. Thus we have

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

 

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 ∆S. It's easy to see that for temperatures above 352K, ∆G is negative and the reaction will occur spontaneously.


Practice problems


  1. Calculate the ∆H, ∆S, and ∆G of the reaction     N2 + O2 → 2 NO     when

        (a) T = 25˚C

        (b) T = 1000˚C

        (c) T = 3000˚C


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

  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, ∆Gorxn of the reaction.

    (b) Assuming that ∆Ho and ∆So remain constant, will the reaction be spontaneous at 500K?

Calculate the entropy change, ∆S, for each of these reactions. See if you can predict ahead of your calculation whether the change will be positive or negative.

  1. CaH2 (s) + H2O (l) → Ca(OH)2 (s) + H2 (g)

  2. 2 HCl(g) + Br2 (l) → 2 HBr(g) + Cl2 (g)

  3. 4 Fe(s) + 3 O2 (g) → 2 Fe2O3 (s)

  4. 2 NO2 (g) → N2O4 (g)

 

Solutions are on their way ... hang tight. (12/15/2016) Download solutions

Creative Commons License   optimized for firefox

xaktly.com by Dr. Jeff Cruzan is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. © 2016, 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.