In the previous post, we talked about the Gibbs free energy for reactions at nonstandard-state conditions,** Δ***G*. It is related to the free energy change of the reaction at standard-state conditions, Δ*G*^{o, }and the reaction quotient, *Q* with the following equation:

Remember, the reaction quotient is used to predict the direction of the equilibrium depending on the initial quantities of the reactants and products:

*Q < K**Reaction tends to***form more products**.*Q > K**Reaction tends to***form more reactants**.*Q = K**Reaction is already at***equilibrium**.

Because the system always tries to reach an equilibrium, *Q* tends to become equal to *K*, and when that happens, an equilibrium is established, *K* = *Q*.

Now, by definition, when the **equilibrium** is established, the free energy of the reaction becomes a zero, **Δ***G = ***0**. So, if we replace *Q* with K, and Δ*G *with 0, we obtain a very important equation in chemistry that correlates the equilibrium constant the Δ*G ^{o}:*

0 = Δ*G*° + *RT* ln *K, *therefore,

We can apply this equation to reactions in solutions, *K*_{c} and the ones involving gases in which case, *K _{P} *is used. The equation can also be used for the solubility constant,

*K*

_{sp}.

**How do we interpret this equation?**

The key link here is the **negative sign** which tells us the larger the *K*, the more negative Δ*G*° is. Remember, a *larger K indicates that the products are favored* which is essentially to say that the reaction is spontaneous – it tends to proceed forward producing more products. A smaller equilibrium constant, on the other hand, makes the Δ*G*° larger and if it becomes a positive value, the reaction is not spontaneous anymore. This happens when ** K < 1**, because the

*RT*ln

*K*term becomes negative, and the minus sign makes the

**Δ**. Again,

*G*° positive*K*< 1 means that the equilibrium mixture is mainly reactants, and therefore, the forward reaction is

**nonspontaneous**.

To summarize the correlation between the equilibrium constant the Δ*G*°, we can write that:

♦ When ** K < 1**, ln

*K*is negative and Δ

*G*^{o}

_{rxn}**> 0**. The reaction is not

**spontaneous.**

♦ When

**, ln**

*K*> 1*K*is negative, and Δ

*G*^{o}

_{rxn}**< 0**. The reaction is

**spontaneous.**

♦ When

**, ln**

*K*= 1*K*= 0, Δ

*G*^{o}

_{rxn}**= 0**. The reaction is at

**equilibrium.**

** **

The significance of the equation *Δ**G**° **= -RT *ln* K* is evident when, for example, the equilibrium constant of the reaction cannot be determined experimentally. What we do then is calculate the Δ*G*° and use it to determine the equilibrium constant.

**For example**, using the data from an appendix for thermodynamic data, calculate the equilibrium constants of the following reaction at 25 °C:

NO(*g*) + O_{3}(*g*) ⇆ NO_{2}(*g*) + O_{2}(*g*)

**Solution**: The first is to determine the Δ*G ^{o }*from the standard free energies of formation using the following equation:

Δ*G*°_{rxn} = Σ*n*_{p}Δ*G*^{o}_{f} (products) – Σ*n*_{r}Δ*G*^{o}_{f }(reactants)

NO(*g*) + O_{3}(*g*) ⇆ NO_{2}(*g*) + O_{2}(*g*)

Δ*G*° = [Δ*G*^{o}_{f} (NO_{2}(*g*)) + Δ*G*^{o}_{f} (O_{2}(*g*))] – [Δ*G*^{o}_{f} (NO(*g*)) + Δ*G*^{o}_{f} (O_{3}(*g*))]

Δ*G*° = [(52 + 0) – (87 + 163)] = -198 kJ

Δ*G*° = –*RT* ln *K*

* * \[\ln \,K\; = \;\frac{{ – \Delta G^\circ }}{{RT}}\]

Taking antilog to get an expression for K:

\[K\; = \;{e^{\frac{{ – \Delta G^\circ }}{{RT}}}}\]

You can either calculate the exponent term separately or use the numbers in this equation.

\[K\; = \;{e^{\frac{{{\rm{ – }}\left( {{\rm{ – 198}}\,{\rm{ \times }}\,{\rm{1}}{{\rm{0}}^{\rm{3}}}\,{\rm{J}}} \right)}}{{{\rm{8}}{\rm{.314}}\,{\rm{J/K}} \cdot {\rm{mol}}\, \cdot {\rm{298}}\;{\rm{K}}}}}}\, = \,{e^{79.9}}\; = \;5.01\, \times \,{10^{34}}\,\]

The **equilibrium constant is very large**, and this is in agreement with a highly **negative Δ G°**.