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Chemistry 351

\( \Delta G^o \) and \( K_p \)

The Connection Between \(\Delta G^\circ\) and the Equilibrium Constant

In previous chapters, we learned that the Gibbs energy determines the direction of spontaneous change. A reaction proceeds spontaneously when

\[ \Delta G < 0 \]

and reaches equilibrium when

\[ \Delta G = 0 \]

The Gibbs energy change for a reaction mixture can be expressed as

\[ \Delta G = \Delta G^\circ + RT\ln Q \]

where \(Q\) is the reaction quotient. At equilibrium,

\[ Q=K \]

and

\[ \Delta G = 0 \]

Substituting these equilibrium conditions gives

\[ 0 = \Delta G^\circ + RT\ln K \]

Rearranging yields one of the most important equations in chemical thermodynamics:

\[ \boxed{\Delta G^\circ=-RT\ln K} \]

This equation provides a direct connection between a thermodynamic quantity (\(\Delta G^\circ\)) and an experimentally measurable quantity (\(K\)).

Interpreting the Relationship

The sign and magnitude of \(\Delta G^\circ\) determine the size of the equilibrium constant.

\(\Delta G^\circ\) \(K\) Equilibrium Position
Negative \(K>1\) Products favored
Zero \(K=1\) Neither side favored
Positive \(K<1\) Reactants favored

A large negative value of \(\Delta G^\circ\) corresponds to a very large equilibrium constant and strong product formation. Conversely, a large positive value of \(\Delta G^\circ\) corresponds to a very small equilibrium constant and strong reactant preference.

\(K_p\), \(K_c\), and Thermodynamic Equilibrium Constants

Equilibrium constants can be written using different measurable quantities.

The thermodynamically rigorous equilibrium constant is written in terms of activities and fugacities:

\[ K = \prod_i a_i^{\nu_i} \]

Since activities and fugacities are dimensionless, the thermodynamic equilibrium constant is also dimensionless.

In many systems, especially dilute solutions and low-pressure gases, activities can be approximated by concentrations and fugacities can be approximated by partial pressures. Under these conditions, \(K_c\) and \(K_p\) provide excellent approximations to the true thermodynamic equilibrium constant.

What Does the Equilibrium Constant Really Tell Us?

Students often interpret a large equilibrium constant as meaning that a reaction "goes to completion." This is not quite correct.

The equilibrium constant describes the composition of a system at equilibrium. It does not determine how rapidly equilibrium is reached, nor does it by itself determine the direction a reaction will shift from a particular starting composition.

To predict the direction of spontaneous change, we must compare the reaction quotient \(Q\) to the equilibrium constant \(K\), a topic we will examine in the next section.

Big picture: The relationship \[ \Delta G^\circ=-RT\ln K \] connects thermodynamics and chemical equilibrium. The equilibrium constant is a measurable manifestation of the Gibbs energy change for a reaction, allowing equilibrium compositions to be predicted directly from thermodynamic data.

Worked examples

Worked Example: Calculating \(K_p\) for the Formation of HI(g)

Calculate \(K_p\) at 298 K for the gas-phase reaction

\[ \text{H}_2(g)+\text{I}_2(g)\rightleftharpoons 2\text{HI}(g) \]

using the following standard Gibbs energies of formation:

Species \(\Delta G_f^\circ\) (kJ mol\(^{-1}\))
\(\text{H}_2(g)\) 0
\(\text{I}_2(g)\) 19.36
\(\text{HI}(g)\) 1.70

Solution

First calculate the standard Gibbs energy change for the reaction:

\[ \Delta G^\circ_{\text{rxn}} = \sum \nu \Delta G_f^\circ(\text{products}) - \sum \nu \Delta G_f^\circ(\text{reactants}) \] \[ \Delta G^\circ_{\text{rxn}} = 2(1.70) - \left[ 0+19.36 \right] \] \[ \Delta G^\circ_{\text{rxn}} = -15.96\ \text{kJ mol}^{-1} \]

Now use the relationship between \(\Delta G^\circ\) and the equilibrium constant:

\[ \Delta G^\circ=-RT\ln K_p \] \[ \ln K_p = -\frac{\Delta G^\circ}{RT} \] \[ \ln K_p = -\frac{-15.96\times10^3\ \text{J mol}^{-1}} {(8.314\ \text{J mol}^{-1}\text{K}^{-1})(298\ \text{K})} \] \[ \ln K_p = 6.44 \] \[ K_p=e^{6.44}=6.3\times10^2 \]

Therefore,

\[ \boxed{K_p\approx 6.3\times10^2} \]

Physical interpretation: Because \(\Delta G^\circ_{\text{rxn}}\) is negative, the equilibrium constant is greater than one. At 298 K, the gas-phase reaction strongly favors formation of HI relative to \(\text{H}_2\) and \(\text{I}_2\).

Practice

Problem 1
Calculate \(K_p\) at 298 K for the reaction \[ \text{N}_2(g)+3\text{H}_2(g)\rightleftharpoons2\text{NH}_3(g) \] given the following standard Gibbs energies of formation:
Species \(\Delta G_f^\circ\) (kJ mol\(^{-1}\))
\(\text{N}_2(g)\) 0
\(\text{H}_2(g)\) 0
\(\text{NH}_3(g)\) -16.45
Show Answer \[ \Delta G^\circ_{\text{rxn}} = 2(-16.45)-0-0 = -32.9\ \text{kJ mol}^{-1} \] \[ \ln K_p = -\frac{\Delta G^\circ}{RT} = \frac{32900} {(8.314)(298)} = 13.3 \] \[ K_p=e^{13.3} = 5.8\times10^5 \] \[ \boxed{K_p\approx5.8\times10^5} \]

The large value of \(K_p\) indicates that ammonia formation is strongly favored at equilibrium.


Problem 2
Calculate \(K_p\) at 298 K for the reaction \[ 2\text{NO}(g)+\text{O}_2(g) \rightleftharpoons 2\text{NO}_2(g) \] using the following standard Gibbs energies of formation:
Species \(\Delta G_f^\circ\) (kJ mol\(^{-1}\))
\(\text{NO}(g)\) 86.55
\(\text{NO}_2(g)\) 51.53
\(\text{O}_2(g)\) 0
Show Answer \[ \Delta G^\circ_{\text{rxn}} = 2(51.53)-2(86.55)-0 \] \[ \Delta G^\circ_{\text{rxn}} = -70.04\ \text{kJ mol}^{-1} \] \[ \ln K_p = \frac{70040} {(8.314)(298)} = 28.3 \] \[ K_p=e^{28.3} = 1.9\times10^{12} \] \[ \boxed{K_p\approx1.9\times10^{12}} \]

This enormous equilibrium constant indicates that the reaction strongly favors formation of \(\text{NO}_2\).


Problem 3
Calculate \(K_p\) at 298 K for the reaction \[ \text{CO}(g)+\text{H}_2\text{O}(g) \rightleftharpoons \text{CO}_2(g)+\text{H}_2(g) \] using the following standard Gibbs energies of formation:
Species \(\Delta G_f^\circ\) (kJ mol\(^{-1}\))
\(\text{CO}(g)\) -137.2
\(\text{H}_2\text{O}(g)\) -228.6
\(\text{CO}_2(g)\) -394.4
\(\text{H}_2(g)\) 0
Show Answer \[ \Delta G^\circ_{\text{rxn}} = (-394.4)+0 - [(-137.2)+(-228.6)] \] \[ \Delta G^\circ_{\text{rxn}} = -28.6\ \text{kJ mol}^{-1} \] \[ \ln K_p = \frac{28600} {(8.314)(298)} = 11.5 \] \[ K_p=e^{11.5} = 9.9\times10^4 \] \[ \boxed{K_p\approx1.0\times10^5} \]

Since \(\Delta G^\circ_{\text{rxn}}\) is negative, the equilibrium favors formation of \(\text{CO}_2\) and \(\text{H}_2\).


Key points (one glance)

Big picture: The relationship \[ \Delta G^\circ=-RT\ln K \] provides a direct connection between thermodynamics and chemical equilibrium. It shows that equilibrium constants are simply another way of expressing the Gibbs energy change for a reaction and allows equilibrium behavior to be predicted directly from thermodynamic data.