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20.8: Calculations of Free Energy and \(K_\text{eq}\)

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  • Formation of stalactites (pointing down) and stalagmites (pointing up) is a complex process. Solutions of minerals drip down and absorb carbon dioxide as water flows through the cave. Calcium carbonate dissolves in this liquid and redeposits on the rock as the carbon dioxide is dissipated into the environment.

    Equilibrium Constant and \(\Delta G\)

    At equilibrium the \(\Delta G\) for a reversible reaction is equal to zero. \(K_\text{eq}\) relates the concentrations of all substances in the reaction at equilibrium. Therefore we can write (through a more advanced treatment of thermodynamics) the following equation:

    \[\Delta G^\text{o} = -RT \: \text{ln} \: K_\text{eq}\]

    The variable \(R\) is the ideal gas constant \(\left( 8.314 \: \text{J/K} \cdot \text{mol} \right)\), \(T\) is the Kelvin temperature, and \(\text{ln} \: K_\text{eq}\) is the natural logarithm of the equilibrium constant.

    When \(K_\text{eq}\) is large, the products of the reaction are favored and the negative sign in the equation means that the \(\Delta G^\text{o}\) is negative. When \(K_\text{eq}\) is small, the reactants of the reaction are favored. The natural logarithm of a number less than one is negative and so the sign of \(\Delta G^\text{o}\) is positive. The table below summarizes the relationship of \(\Delta G^\text{o}\) to \(K_\text{eq}\):

    Table \(\PageIndex{1}\): Relationship of \(\Delta G^\text{o}\) and \(K_\text{eq}\)
    \(K_\text{eq}\) \(\text{ln} \: K_\text{eq}\) \(\Delta G^\text{o}\) Description
    >1 positive negative Products are favored at equilibrium.
    1 0 0 Reactants and products are equally favored.
    <1 negative positive Reactants are favored at equilibrium.

    Knowledge of either the standard free energy change or the equilibrium constant for a reaction allows for the calculation of the other. The following two sample problems illustrate each case.

    Example \(\PageIndex{1}\)

    The formation of nitrogen monoxide from nitrogen and oxygen gases is a reaction that strongly favors the reactants at \(25^\text{o} \text{C}\).

    \[\ce{N_2} \left( g \right) + \ce{O_2} \left( g \right) \rightleftharpoons 2 \ce{NO} \left( g \right)\]

    The actual concentrations of each gas would be difficult to measure, and so the \(K_\text{eq}\) for the reaction can more easily be calculated from the \(\Delta G^\text{o}\), which is equal to \(173.4 \: \text{kJ/mol}\). Find the \(K_\text{eq}\).


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


    • \(\Delta G^\text{o} = +173.4 \: \text{kJ/mol}\)
    • \(R = 8.314 \: \text{J/K} \cdot \text{mol}\)
    • \(T = 25^\text{o} \text{C} = 298 \: \text{K}\)


    • \(K_\text{eq} = ?\)

    In order to make the units agree, the value of \(\Delta G^\text{o}\) will need to be converted to \(\text{J/mol}\) \(\left( 173,400 \: \text{J/mol} \right)\). To solve for \(K_\text{eq}\), the inverse of the natural logarithm, \(e^x\), will be used.

    Step 2: Solve.

    \[\begin{align} \Delta G^\text{o} &= -RT \: \text{ln} \: K_\text{eq} \\ \text{ln} \: K_\text{eq} &= \frac{-\Delta G^\text{o}}{RT} \\ K_\text{eq} &= e^{\frac{-\Delta G^\text{o}}{RT}} = e^{\frac{-173,400 \: \text{J/mol}}{8.314 \: \text{J/K} \cdot \text{mol} \left( 298 \: \text{K} \right)}} = 4.0 \times 10^{-31} \end{align}\]

    Step 3: Think about your result.

    The large positive free energy change leads to a \(K_\text{eq}\) that is extremely small. Both lead to the conclusion that the reactants are highly favored and very few product molecules are present at equilibrium.

    Example \(\PageIndex{2}\)

    The solubility product constant \(\left( K_\text{sp} \right)\) of lead (II) iodide is \(1.4 \times 10^{-8}\) at \(25^\text{o} \text{C}\). Calculate \(\Delta G^\text{o}\) for the dissociation of lead (II) iodide in water.

    \[\ce{PbI_2} \left( s \right) \rightleftharpoons \ce{Pb^{2+}} \left( aq \right) + 2 \ce{I^-} \left( aq \right)\]


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


    • \(K_\text{eq} = K_\text{sp} = 1.4 \times 10^{-8}\)
    • \(R = 8.314 \: \text{J/K} \cdot \text{mol}\)
    • \(T = 25^\text{o} \text{C} = 298 \: \text{K}\)


    • \(\Delta G^\text{o} = ? \: \text{kJ/mol}\)

    The equation relating \(\Delta G^\text{o}\) to \(K_\text{eq}\) can be solved directly.

    Step 2: Solve.

    \[\begin{align} \Delta G^\text{o} &= -RT \: \text{ln} \: K_\text{eq} \\ &= -8.314 \: \text{J/K} \cdot \text{mol} \left( 298 \: \text{K} \right) \: \text{ln} \left( 1.4 \times 10^{-8} \right) \\ &= 45,000 \: \text{J/mol} \\ &= 45 \: \text{kJ/mol} \end{align}\]

    Step 3: Think about your result.

    The large, positive \(\Delta G^\text{o}\) indicates that the solid lead (II) iodide is nearly insoluble and so very little of the solid is dissociated at equilibrium.


    • The relationship between \(\Delta G\) and \(K_\text{eq}\) is described.
    • Calculations involving these two parameters are shown.

    Contributors and Attributions

    • CK-12 Foundation by Sharon Bewick, Richard Parsons, Therese Forsythe, Shonna Robinson, and Jean Dupon.