20.5: Calculating Free Energy Change \(\left( \Delta G^\text{o} \right)\)

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When you are baking something, you heat the oven to the temperature indicated in the recipe. Then you mix all the ingredients, put them in the proper baking dish, and place them in the oven for a specified amount of time. If you had mixed the ingredients and left them out at room temperature, not much would change. The materials need to be heated to a given temperature, for a set time, in order for the ingredients to react with one another and produce a delicious final product.

Calculating Free Energy \(\left( \Delta G^\text{o} \right)\)

The free energy change of a reaction can be calculated using the following expression:

\[\Delta G^\text{o} = \Delta H^\text{o} - T \Delta S^\text{o}\nonumber \]

where \(\Delta G =\) free energy change \(\left( \text{kJ/mol} \right)\)

\(\Delta H =\) change in enthalpy \(\left( \text{kJ/mol} \right)\)

\(\Delta S =\) change in entropy \(\left( \text{J/K} \cdot \text{mol} \right)\)

\(T =\) temperature (Kelvin)

Note that all values are for substances in their standard state. In performing calculations, it is necessary to change the units for \(\Delta S\) to \(\text{kJ/K} \cdot \text{mol}\), so that the calculation of \(\Delta G\) is in \(\text{kJ/mol}\).

Example \(\PageIndex{1}\)

Methane gas reacts with water vapor to produce a mixture of carbon monoxide and hydrogen, according to the balanced equation below.

\[\ce{CH_4} \left( g \right) + \ce{H_2O} \left( g \right) \rightarrow \ce{CO} \left( g \right) + 3 \ce{H_2} \left( g \right)\nonumber \]

The \(\Delta H^\text{o}\) for the reaction is \(+206.1 \: \text{kJ/mol}\), while the \(\Delta S^\text{o}\) is \(+215 \: \text{J/K} \cdot \text{mol}\). Calculate the \(\Delta G^\text{o}\) at \(25^\text{o} \text{C}\) and determine if the reaction is spontaneous at that temperature.

Solution:

Step 1: List the known values and plan the problem.

Known

\(\Delta H^\text{o} = 206.1 \: \text{kJ/mol}\)
\(\Delta S^\text{o} = 215 \: \text{J/K} \cdot \text{mol} = 0.215 \: \text{kJ/K} \cdot \text{mol}\)
\(T = 25^\text{o} \text{C} = 298 \: \text{K}\)

Unknown

Prior to substitution into the Gibbs free energy equation, the entropy change is converted to \(\text{kJ/K} \cdot \text{mol}\) and the temperature to Kelvins.

Step 2: Solve.

\[\Delta G^\text{o} = \Delta H^\text{o} - T \Delta S^\text{o} = 206.1 \: \text{kJ/mol} - 298 \: \text{K} \left( 0.215 \: \text{kJ/K} \cdot \text{mol} \right) = +142.0 \: \text{kJ/mol}\nonumber \]

The resulting positive value of \(\Delta G\) indicates that the reaction is not spontaneous at \(25^\text{o} \text{C}\).

Step 3: Think about your result.

The unfavorable driving force of increasing enthalpy outweighed the favorable increase in entropy. The reaction will be spontaneous only at some elevated temperature.

Available values for enthalpy and entropy changes are generally measured at the standard conditions of \(25^\text{o} \text{C}\) and \(1 \: \text{atm}\) pressure. The values are slightly temperature dependent and so we must use caution when calculating specific \(\Delta G\) values at temperatures other than \(25^\text{o} \text{C}\). However, since the values for \(\Delta H\) and \(\Delta S\) do not change a great deal, the tabulated values can safely be used when making general predictions about the spontaneity of a reaction at various temperatures.