# Gibbs Free Energy

Gibbs free energy, ΔG, for a process (at constant T and P) is an alternative way to predict the spontaneity of a process since measuring the entropy change of the universe is difficult. It can be expressed as

$\Delta G^{\circ} = \Delta H^{\circ} - T\Delta S^{\circ}$ where every term is related to the system.

Free energy is a measure of the maximum amount of work that can be extracted from a system based on the

• energy produced by a process, ΔH
• amount of energy lost to the surroundings, TΔS

Free energy is another good definition for spontaneity.

• ΔG < 0; spontaneous
• ΔG > 0; nonspontaneous
• ΔG = 0; at equilibrium

## Determining state function quantities

ΔG, ΔH, and ΔS, are state functions; they only depend on the final (or products) and initial (or reactants) states of the system.

$\Delta X^{\circ} = \sum \nu \Delta X^{\circ} (\mathrm{products}) - \sum \nu \Delta X^{\circ} (\mathrm{reactants})$

### Example: Determine the standard enthalpy and entropy for the following process

$\mathrm{H_2O}(l) \longrightarrow \mathrm{H_2O}(g)$

Looking at the standard thermodynamic quantities given in Appendix G, we see

\begin{align*} \mathrm{H_2O}(l) \quad \Delta H^{\circ} &= -285.83~\mathrm{kJ~mol^{-1}} \\ S^{\circ} &= 70.0~\mathrm{J~mol^{-1}~K^{-1}} \end{align*}

\begin{align*} \mathrm{H_2O}(g) \quad \Delta H^{\circ} &= -241.82~\mathrm{kJ~mol^{-1}} \\ S^{\circ} &= 188.8~\mathrm{J~mol^{-1}~K^{-1}} \end{align*}

Solve

\begin{align*} \Delta H^{\circ} &= (-241.82) - (-285.83)\\ &= 44.01~\mathrm{kJ}\\ \Delta S^{\circ} &= (188.8) - (70.0)\\ &= 118.8~\mathrm{J~K^{-1}} \end{align*}

### Example: Determine the standard enthalpy, entropy, and free energy changes (at 25 °C)

$\require{color}$ $\mathrm{\textcolor{red}{3}H_2}(g) + \mathrm{N_2}(g) \longrightarrow \mathrm{\textcolor{blue}{2}NH_3}(g)$

Get thermodynamic data Appendix G

\begin{align*} & \qquad \Delta H^{\circ} (\mathrm{kJ~mol^{-1}}) && S^{\circ} (\mathrm{J~mol^{-1}~K^{-1}})\\ \mathrm{H_2}(g) & \qquad 0 && 130.7 \\ \mathrm{N_2}(g) & \qquad 0 && 191.6 \\ \mathrm{NH_3}(g) & \qquad -45.9 && 192.8 \end{align*}

Solve for ΔG

\begin{align*} \Delta H^{\circ} &= [\textcolor{blue}{2}(-45.9)] - [\textcolor{red}{3}(0) + (0)] = -91.8~\mathrm{kJ}\\[1.5ex] \Delta S^{\circ} &= [\textcolor{blue}{2}(192.8)] - [\textcolor{red}{3}(130.7) + (191.6)] = -198.1~\mathrm{J~K^{-1}}\\[1.5ex] \Delta G^{\circ} &= \Delta H^{\circ} - T\Delta S^{\circ}\\ &= (-91.8~\mathrm{kJ}) - (298.15~\mathrm{K})(-0.1981~\mathrm{kJ~K^{-1}}) \\ &= -32.7~\mathrm{kJ~mol^{-1}} \end{align*}

## Spontaneity and Temperature Dependence

Changing the temperature of a process can sometimes affect its spontaneity (i.e. cause a sign change in ΔG).

$\Delta G = \Delta H - T\Delta S$ For processes that are non/spontaneous at low/high temperature, you can determine the temperature at which the process “switches” in spontaneity by setting ΔG = 0 and solving for T.

$T = \dfrac{\Delta H}{\Delta S}$

### Example: Temperature Dependence on Spontaneity

What is the temperature (in °C) at which the following process switches between being nonspontaneous and spontaneous?

$\mathrm{H_2O}(l) \longrightarrow \mathrm{H_2O}(g)$

Get thermodynamic data Appendix G

\begin{align*} \mathrm{H_2O}(l) \quad \Delta H^{\circ} &= -285.83~\mathrm{kJ~mol^{-1}} \\ S^{\circ} &= 70.0~\mathrm{J~mol^{-1}~K^{-1}} \end{align*}

\begin{align*} \mathrm{H_2O}(g) \quad \Delta H^{\circ} &= -241.82~\mathrm{kJ~mol^{-1}} \\ S^{\circ} &= 188.8~\mathrm{J~mol^{-1}~K^{-1}} \end{align*}

Solve for ΔH and ΔS

\begin{align*} \Delta H^{\circ} &= (-241.82) - (-285.83)\\ &= 44.01~\mathrm{kJ}\\ \Delta S^{\circ} &= (188.8) - (70.0)\\ &= 118.8~\mathrm{J~K^{-1}} \end{align*}

Solve for T

\begin{align*} T &= \dfrac{\Delta H}{\Delta S} \\ &= \dfrac{44.01~\mathrm{kJ}}{0.1188~\mathrm{J~K^{-1}}}\\[1.5ex] &= 370.5~\mathrm{K} = 97.3~\mathrm{^{\circ}C} \end{align*}

Recall that the thermodynamic values used are obtained at 298.15 K (25 °C). The answer here is in reasonable agreement with the actual boiling point of water (100 °C).

## Free Energy and Equilibrium

Free energy, Q, and K are related. Free energy can be considered to be the “driving force” behind why a reaction proceeds in the direction that it does.

• ΔG < 0 ;   Q < K  -   reaction proceeds right
• ΔG > 0 ;   Q > K  -   reaction proceeds left
• ΔG = 0 ;   Q = K  -   reaction is at equilibrium

Most reactions are carried out under nonstandard conditions. Standard free energy can be related to a nonstandard free energy for a reaction

$\Delta G = \Delta G^{\circ} + RT \ln Q \qquad R = 8.315~\mathrm{J~mol^{-1}~K^{-1}}$

If the system is at equilibrium, ΔG = 0 and Q = K to give

$\Delta G^{\circ} = -RT \ln K \qquad R = 8.315~\mathrm{J~mol^{-1}~K^{-1}}$

and ΔG° and K can be used to predict reactant or product favorability for a reaction

• ΔG° < 0 ;   K > 1  -   products are more abundant at equilibrium
• ΔG° > 0 ;   K < 1  -   reactants are more abundant at equilibrium
• ΔG° = 0 ;   K = 1  -   reactants and products are comparably abundant at equilibrium

### Example: ΔG, Q, and K

Determine ΔG for the following process at 25 °C. Will the reaction proceed to the right or left?

$\mathrm{3H_2}(g) + \mathrm{N_2}(g) \rightleftharpoons \mathrm{2NH_3}(g)$

• ΔG° = –32.7 kJ mol–1
• Q = 8.00 × 105

\begin{align*} \Delta G &= \Delta G^{\circ} + RT \ln Q \\ &= (-32.7~\mathrm{kJ~mol^{-1}}) + (0.008315~\mathrm{kJ~mol^{-1}~K^{-1}})(298.15~\mathrm{K})\ln(8.00 \times 10^{5}) \\ &= 1.00~\mathrm{kJ} \end{align*}

Since ΔG is positive, the forward reaction is nonspontaneous and Q > K. Now determine K.

$\Delta G^{\circ} = -RT \ln K$

\begin{align*} K &= e^{-\frac{\Delta G^{\circ}}{RT}}\\ &= e^{-\frac{-32.7~\mathrm{kJ~mol^{-1}}}{(0.008315~\mathrm{kJ~mol^{-1}~K^{-1}})(298.15~\mathrm{K})}} \\ &= 5.35\times 10^5 \end{align*}

We see that Q was greater than K. The reaction will not spontaneously continue forward. However, it will spontaneously move in the reverse direction to approach equilibrium (K).

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