20.2: Spontaneous Reactions and Free Energy
Lesson Objectives
 Define Gibbs free energy, and be able to calculate the change in Gibbs free energy for a given process when provided with the appropriate data.
 Use the value for Gibbs free energy to predict whether a reaction will occur spontaneously.
Vocabulary
 Gibbs free energy: A thermodynamic quantity that combines enthalpy and entropy into a single value in order to predict whether or not a process is spontaneous.
Check Your Understanding
 What must be true for a reaction to be considered spontaneous?
Gibbs Free Energy
We have learned that a spontaneous reaction must increase the total entropy in the universe: ΔS_{univ} > 0. To analyze the entropic effects of a given process, we generally break this quantity down into two separate components. This is expressed with the following equation:


 ΔS_{univ} = ΔS_{sys} + ΔS_{surr} > 0

In general, the entropy change for a chemical reaction or phase change can be easily determined from standard entropy values. Additionally, we saw in the previous lesson that the following relationship is true:



ΔS_{surr} =  \begin{align*}\mathrm{\frac{\Delta H_{sys}}{T}}\end{align*}
ΔHsysT .

ΔS_{surr} =  \begin{align*}\mathrm{\frac{\Delta H_{sys}}{T}}\end{align*}

Substituting this into the above equation, we get the following:



ΔS_{univ} = ΔS_{sys}  \begin{align*}\mathrm{\frac{\Delta H_{sys}}{T}}\end{align*}
ΔHsysT > 0

ΔS_{univ} = ΔS_{sys}  \begin{align*}\mathrm{\frac{\Delta H_{sys}}{T}}\end{align*}

This equation can then be rearranged as follows:


 TΔS_{univ} =  ΔH_{sys} + TΔS_{sys} > 0
  TΔS_{univ} = ΔH_{sys}  TΔS_{sys} < 0

Now we can talk about the entropy changes in the universe simply by looking at data about the system of interest. If the results of the equation are less than zero, the reaction will be spontaneous. The American physicist Josiah Gibbs took this idea and introduced a thermodynamic quantity that combines enthalpy and entropy into a single value that has been named after him. Gibbs free energy is defined by the following equation:


 ΔG = ΔH_{sys}  TΔS_{sys}

where ΔG is the change in free energy, ΔH is the change in enthalpy, T is the temperature (in Kelvin), and ΔS is the change in entropy. As we can see by comparing this to the derived equations above, the value of ΔG must be less than zero for a spontaneous process. Stated another way, systems have a natural tendency to move towards a minimum amount of free energy.
Gibbs postulated that a system progressed towards the equilibrium direction which exhibited the highest entropy.
Standard Free Energy Change
Much like the corresponding equations for ΔH°_{rxn} and ΔS°_{rxn}, ΔG°_{rxn} for a given reaction can be calculated from the corresponding ΔG°_{f} values, which are tabulated for a wide variety of substances.


 ΔG°_{rxn} = ΣnΔG°_{f}(products) – ΣnΔG°_{f}(reactants)

As with the standard enthalpy of formation values, there is no absolute zero for Gibbs free energy, so each ΔG°_{f} value is the change in free energy for a compound when prepared from its constituent elements in their standard states. Also like the enthalpy of formation values, ΔG°_{f} for any element in its most stable form at 25°C is zero.
Example 20.3
Calculate the value of ΔG°_{rxn} for the combustion of methane:


 CH_{4}(g) + 2O_{2}(g) → CO_{2}(g) + 2H_{2}O(l)

ΔG°_{f} values can be found at the following website: http://chemed.chem.wisc.edu/chempaths/GenChemTextbook/TheFreeEnergy629.html
Answer:
Multiply each ΔG°_{f} value by the coefficient from the balanced equation, and subtract the free energy of the reactants from that of the products. Note that water is present as a liquid, so make sure to use the ΔG°_{f} value for the correct state. Because O_{2}(g) is the standard form of oxygen, it has a ΔG°_{f} value of zero.


 ΔG°_{rxn} = ΣnΔG°_{f}(products) – ΣnΔG°_{f}(reactants)
 ΔG°_{rxn} = [(394.36 kJ/mol) + 2(237.13 kJ/mol)] – [(50.72 kJ/mol + 2(0 kJ/mol)]
 ΔG°_{rxn} = 868.62 kJ/mol – (50.72 kJ/mol)
 ΔG°_{rxn} = 817.90 kJ/mol

The large negative value for ΔG°_{rxn} indicates that the forward reaction is heavily favored under standard conditions. This reaction proceeds spontaneously in the forward direction.
Gibbs Free Energy and NonStandard Conditions
The sign of ΔG° predicts the behavior of a chemical reaction at the standard conditions of 25°C and 1 atm of pressure. If ΔG° is negative, the reaction will proceed spontaneously, if ΔG° is zero, the reaction is at equilibrium, and if ΔG° is positive, the reaction will proceed spontaneously in the reverse direction. However, this does not tell us what will happen with a given reaction under nonstandard conditions.
Consider the following reaction between ammonia and hydrogen chloride gas:


 NH_{3}(g) + HCl(g) → NH_{4}Cl(s)

We could calculate ΔG°_{rxn} for this reaction using ΔG°_{f} values, as we did in the previous example problem, but that would only tell us the value of ΔG at 25°C. Alternatively, we could calculate ΔG at other temperatures using the following equation:


 ΔG_{rxn} = ΔH_{rxn}  TΔS_{rxn}

Unlike ΔG, ΔH and ΔS do not vary much with changes in temperature. For simplicity, we will assume that for a specified reaction the ΔH_{rxn} and ΔS_{rxn} values are the same at any temperature. By calculating ΔH_{rxn} and ΔS_{rxn} from the relevant standard enthalpy and entropy values, we can calculate the value of ΔG at any temperature. Let's start by finding ΔH_{rxn}:


 ΔH°_{rxn} = ΣnΔH°_{f}(products) – ΣnΔH°_{f}(reactants)
 ΔH°_{rxn} = 315.39 kJ/mol – [(46.3 kJ/mol) + (92.3 kJ/mol)]
 ΔH°_{rxn} = 176.70 kJ/mol

The large negative value indicates that this is a highly exothermic reaction. Then, find ΔS_{rxn}:


 ΔS°_{rxn} = ΣnS°_{f}(products) – ΣnS°_{f}(reactants)
 ΔS°_{rxn} = 94.6 J/K•mol – [193.0 J/K•mol + 187.0 J/K•mol]
 ΔS°_{rxn} = 285.4 J/K•mol

A negative value here indicates a decrease in entropy over the course of the reaction. This is consistent with what we might expect for a reaction in which two gases combine to make a single solid product.
Now, let's calculate the value of ΔG_{rxn} for this reaction at a couple of different temperatures. First, let's look at what we would get using the standard temperature of 25°C (298 K). Before plugging any values into our equation, we first need to make sure that our units match. Our enthalpy value is written in terms of kilojoules (kJ), but the value for entropy is written in terms of joules. One of these must be changed in order to add the two quantities together. Because free energy changes are usually written in units of kJ/mol, like enthalpy changes, we should convert the value of ΔS°_{rxn} from 285.4 J/K•mol to 0.2854 kJ/K•mol by dividing by 1,000. Now, we can plug values into the following equation:


 ΔG_{rxn} = ΔH_{rxn}  TΔS_{rxn}
 ΔG_{rxn} = 176.70 kJ/mol  (298 K)(0.2854 kJ/K•mol)
 ΔG_{rxn} = 176.70 kJ/mol  (85.05 kJ/mol)
 ΔG_{rxn} = 91.65 kJ/mol

ΔG_{rxn} < 0, so this reaction would proceed spontaneously at a temperature of 25°C. It should also be noted that, because we are using the standard temperature at which thermodynamic data is generally measured, this would be the approximate value obtained if we had calculated ΔG_{rxn} using ΔG°_{f} values.
What if we increase the reaction temperature to 500°C (773 K)? The same calculation can be performed at the new temperature:


 ΔG_{rxn} = ΔH_{rxn}  TΔS_{rxn}
 ΔG_{rxn} = 176.70 kJ/mol  (773 K)(0.2854 kJ/K•mol)
 ΔG_{rxn} = 176.70 kJ/mol  (220.61 kJ/mol)
 ΔG_{rxn} = 43.91 kJ/mol

At this temperature, the reaction would not be spontaneous in the forward direction. However, the reverse reaction in which ΔG_{rxn} = 43.91 kJ/mol would be spontaneous at 500°C.
Lesson Summary
 Gibbs free energy combines enthalpy and entropy into a single thermodynamic variable that can be used to predict whether a given reaction will occur spontaneously.
 The sign of ΔG predicts the behavior of a chemical reaction at a constant temperature and pressure. If ΔG is negative, the forward reaction will proceed spontaneously, if ΔG is zero, the reaction is at equilibrium, and if ΔG is positive, the reverse reaction will be spontaneous.
 ΔG_{rxn} can be calculated at 25°C using ΔG°_{f} values.
 ΔG_{rxn} can be calculated at other temperatures from the values of ΔH_{rxn} and ΔS_{rxn}.
Lesson Review Questions
 What thermodynamic quantities are used to define Gibbs free energy?
 Define each component of the equation describing Gibbs free energy.
 What is true for all spontaneous processes regarding Gibbs free energy equation?
 What is the change in Gibbs free energy for the formation of any element in its most stable state at 25degC?
 What can be inferred from a very large ΔG_{rxn}? From a very small ΔG_{rxn}?
 What does a ΔG value of zero imply?
 Describe how it is possible that a reaction is spontaneous at some temperatures but not at others.
 Use the table referenced by the URL below to calculate ΔG°_{rxn} for the following reactions at 25°C:
 CaCO_{3}(s) → CaO(s) + CO_{2}(g)
 2Mg(s) + O_{2}(g) → 2MgO(s)
 2SO_{2}(g) + O_{2}(g) → 2SO_{3}(g)
 2C_{2}H_{6}(g) + 7O_{2}(g) → 4CO_{2}(g) + 6H_{2}O(l)
http://chemed.chem.wisc.edu/chempaths/GenChemTextbook/TheFreeEnergy629.html
Further Reading/Supplementary Links
 Gibbs free energy: http://chemed.chem.purdue.edu/genchem/topicreview/bp/ch21/gibbs.php
 ΔG values: http://chemed.chem.wisc.edu/chempaths/GenChemTextbook/TheFreeEnergy629.html
Points to Consider
 How does the equilibrium of a reaction affect the free energy?
 Can we use free energy calculations to determine equilibrium constants?
 Can we use equilibrium constants to determine free energy changes?
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