Вільна енергія Гіббса та рівновага

Let's explore the relationship between the Gibbs free energy ($\Delta G$) and the equilibrium constant (K)!

First, let's recall the three criteria for whether a reaction will be spontaneous:

(Remember that a spontaneous reaction favors the formation of the products at the given conditions without the addition of any external energy. These processes are also exothermic)

• $\Delta S_{univ}>0$

• $\Delta G_{sys}<0$

• K > Q

If a reaction does not meet these qualifications, then it must occur either non-spontaneously or be at equilibrium. Reactions that are non-spontaneous occur in the reverse direction and require an input of external energy for the reaction to proceed. These reactions must also be endothermic.

Non-spontaneous reactions will meet the following criteria:

• $\Delta S_{univ}<0$

• $\Delta G_{sys}>0$

• K < Q

When a system is at equilibrium, no net reaction occurs. In this instance:

• $\Delta S_{univ}=0$

• $\Delta G_{sys}0$

• K = Q

Note: Q $\rightarrow$ Reaction Quotient

$\Delta S_{univ} \rightarrow$ Entropy of Universe

K $\rightarrow$ Equilibrium Constant

Since all three criteria address the spontaneity of the reaction, we can mathematically explore the relationship between the standard free energy of a reaction ($\Delta G^{o}$) and the equilibrium constant (K) through the equation:

$\Delta G^{o} = -RT ln K_{eq}$

This reaction can be rearranged to solve for the equilibrium constant, as well:

$K_{eq}$ = $e^{-\Delta G^{o}/ RT}$ or $ln K_{eq}$ = $\frac{-\Delta G^{o}}{R T}$

• R is the gas constant ($8.314 J K^{-1} mol^{-1}$ or $0.008314 kJ mol^{-1} K^{-1}$)

• T is the temperature of the reaction (in Kelvin)

• $K_{eq}$ is the equilibrium constant of the reaction

• Occurs under standard conditions:

All reactants and products present in unit concentrations; 1 atm of pressure; temperature at 273.15 K

Note: If $\Delta G^{o}<0$, then K>1 and the products are favored at equilibrium (spontaneous forward), but if $\Delta G^{o}>0$, then K<1 and the reactants are favored (spontaneous in reverse direction). However, if $\Delta G^{o}=0$, then K=1 and the reaction does not take place, neither the products or reactants are favored.

Let's practice using the different forms of the equation to solve for either $\Delta G^{o}$ or K!

Example #1:

Nitrogen and oxygen gases will react together to form nitrogen monoxide. At $25^{o}$, the reaction will strongly favor the reactants. What is the the equilibrium constant ($K_{eq}$) when $\Delta G^{o}$ is equal to 184.6 kJ/mol?

$N_{2(g)} + 0_{2(g)} \Leftrightarrow 2NO_{(g)}$

First, let's think about the information given:

• T= $25^{o}C$ or 298 K

• $\Delta G^{o}$= +173.4 kJ/mol

• R=8.314 J/K*mol

And then what we need to find:

• $K_{eq}$= ?

The first thing that you should notice is that the units of our known values do not agree-- $\Delta G^{o}$ is in kJ/mol while our gas constant (K) is in J/K*mol. The value of $\Delta G^{o}$ will need be converted to J/mol (173,400 J/mol) for our equation to work properly.

Next, we will choose which form of our equation to use, plug in our known values, and solve!

$K_{eq}$ = $e^{-\Delta G^{o}/ RT}$

$K_{eq}$ =$e^{\frac{-173,400 J/mol}{8.314 J/K*mol (298 K)}}$

$K_{eq}$ = 4.0 x $10^{-31}$

$K_{eq}$ is a unitless value, so this is our final answer!

Lastly, let's think about our result:

The large, positive Gibbs free energy change results in a $K_{eq}$ value that is extremely small (<1). If we look back to our earlier criteria for spontaneity, if $\Delta G^{o} > 0$ and K<1, then the reactants are favored, which our problem said was the case!

Now, let's try solving for $\Delta G^{o}$ from a given $K_{eq}$ value.

Example #2:

For a given reaction taking place at 300 K, the equilibrium constant is 2.7 x $10^{8}$. Solve for $\Delta G^{o}$ and determine if the products or reactants are favored at this temperature.

$3H_{2(g)} + N_{2(g)} \Leftrightarrow 2NH_{3(g)}$

First, what are the known values?

• T= 300K

• $K_{eq}= 2.7 x 10^{8}$

• R=8.314 J/K*mol

And what do we need to find?

• $\Delta G^{o} = ?$

Since the equilibrium constant is unit-less, our units all agree and no conversion is necessary!

Next, choose an equation, plug in the values, and solve!

$\Delta G^{o} = -RT ln K_{eq}$

$\Delta G^{o} = -(8.314 J/K*mol)(300 K) ln(2.7 x 10^{8})$

$\Delta G^{o}$ = -48422.2 J/mol = -48.4 kJ/mol

Lastly, let's think about our result to answer the second part of the question.

Since we have a large, negative Gibbs free energy change and K>1, if we look back at the criteria previously listed, we can see that this reaction will favor formation of the products at 300 K.

Now, try a problem on your own!

(Video) Free Energy and the Equilibrium Constant

by Bozeman Science

In this video, Mr. Andersen explains the connections between Gibbs Free Energy and the equilibrium constant. He explains that we can consider chemical reactions from a thermodynamic or equilibrium perspective. But, we can also connect them. He shows you the equation and explains it conceptually.