1.1 What is Chemistry?

20.2. Spontaneous Reactions and Free Energy http://www.ck12.org

Unlike∆G,∆H and∆S do not vary much with changes in temperature. For simplicity, we will assume that for a
specified reaction the∆Hrxnand∆Srxnvalues are the same at any temperature. By calculating∆Hrxnand∆Srxnfrom
the relevant standard enthalpy and entropy values, we can calculate the value of∆G at any temperature. Let’s start
by finding∆Hrxn:

∆H°rxn=Σn∆H°f(products) –Σn∆H°f(reactants)

∆H°rxn= -315.39 kJ/mol –[(-46.3 kJ/mol) + (-92.3 kJ/mol)]

∆H°rxn= -176.70 kJ/mol

The large negative value indicates that this is a highly exothermic reaction. Then, find∆Srxn:

∆S°rxn=ΣnS°f(products) –ΣnS°f(reactants)

∆S°rxn= 94.6 J/K•mol –[193.0 J/K•mol + 187.0 J/K•mol]

∆S°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∆Grxnfor 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∆S°rxnfrom -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:

∆Grxn=∆Hrxn- T∆Srxn

∆Grxn= -176.70 kJ/mol - (298 K)(-0.2854 kJ/K•mol)

∆Grxn= -176.70 kJ/mol - (-85.05 kJ/mol)

∆Grxn= -91.65 kJ/mol

∆Grxn<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∆Grxnusing∆G°fvalues.

What if we increase the reaction temperature to 500°C (773 K)? The same calculation can be performed at the new
temperature:

∆Grxn=∆Hrxn- T∆Srxn

∆Grxn= -176.70 kJ/mol - (773 K)(-0.2854 kJ/K•mol)

∆Grxn= -176.70 kJ/mol - (-220.61 kJ/mol)

∆Grxn= 43.91 kJ/mol

At this temperature, the reaction wouldnotbe spontaneous in the forward direction. However, the reverse reaction
in which∆Grxn= -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.