DAT General Chemistry: Free Energy, the Reaction Quotient, and the Equilibrium Constant

DAT High-Yield Concepts

Nonstandard Free Energy ( $\Delta G$ ): Determines instantaneous reaction spontaneity under nonstandard concentrations or pressures: $\Delta G = \Delta G^\circ + RT \ln Q$ .
Standard Free Energy ( $\Delta G^\circ$ ): Quantifies the thermodynamic driving force under standard conditions and directly determines the equilibrium constant: $\Delta G^\circ = -RT \ln K$ .
Equilibrium State: At dynamic equilibrium, free energy is minimized, yielding $\Delta G = 0$ and $Q = K$ .
Sign of $\Delta G^\circ$ and Magnitude of $K$ : A negative $\Delta G^\circ$ corresponds to $K > 1$ (products favored), a positive $\Delta G^\circ$ corresponds to $K < 1$ (reactants favored), and $\Delta G^\circ = 0$ means $K = 1$ .

Key DAT Distinction: Do not confuse

\Delta G^\circ

with

\Delta G

. The standard reaction free energy

\Delta G^\circ

is a single, constant reference value at a specified temperature (where all reactants and products are at standard states:

\pu{1 atm}

for gases,

\pu{1 M}

for solutions). Nonstandard free energy

\Delta G

changes continuously as reactants turn into products until equilibrium is reached. Gibbs free energy concepts are covered in depth in their own dedicated thermodynamic lessons.

Nonstandard Free Energy and the Reaction Quotient

Chemical reactions rarely occur with all species at standard state. Under nonstandard conditions, the instantaneous driving force $\Delta G$ depends on the standard free energy change $\Delta G^\circ$ and the reaction quotient $Q$ under isothermal (constant temperature) and isobaric (constant pressure) conditions:

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

Where $R = \pu{8.314 J/(mol K)}$ , $T$ is absolute temperature in kelvins, and $Q$ is the reaction quotient calculated from nonequilibrium pressures or concentrations. When a system reaches dynamic equilibrium, net chemical change ceases, meaning $\Delta G = 0$ and $Q = K$ :

0 = \Delta G^\circ + RT \ln K \implies \Delta G^\circ = -RT \ln K

Worked Example: Calculating Nonstandard Free Energy

Consider the gas-phase decomposition reaction $\ce{N2O4(g) <=> 2NO2(g)}$ at $\pu{298 K}$ , where $\Delta G^\circ = \pu{4.8 kJ/mol}$ . Calculate $\Delta G$ when the partial pressures in a reaction vessel are $P_{\ce{N2O4}} = \pu{2.0 atm}$ and $P_{\ce{NO2}} = \pu{0.10 atm}$ .

Step-by-Step Solution:

Write the expression for the reaction quotient $Q_p$ :
$Q_p = \frac{(P_{\ce{NO2}})^2}{P_{\ce{N2O4}}} = \frac{(0.10)^2}{2.0} = 0.0050$
Calculate $RT \ln Q_p$ using dimensional analysis:
$RT \ln Q_p = (\pu{8.314 J/(mol K)})(\pu{298 K}) \ln(0.0050) = (\pu{2477.57 J/mol})(-5.2983) = -\pu{13127 J/mol} = -\pu{13.1 kJ/mol}$
Combine $\Delta G^\circ$ and $RT \ln Q_p$ :
$\Delta G = \pu{4.8 kJ/mol} + (-\pu{13.1 kJ/mol}) = -\pu{8.3 kJ/mol}$

Because $\Delta G < 0$ , the reaction proceeds spontaneously in the forward direction under these nonstandard initial conditions.

Common Exam Trap: Mixing units during addition. $\Delta G^\circ$ is typically tabulated in $\pu{kJ/mol}$ , whereas $RT \ln Q$ evaluates to $\pu{J/mol}$ . Always convert $RT \ln Q$ to $\pu{kJ/mol}$ before adding it to $\Delta G^\circ$ .

Standard Reaction Free Energy and the Equilibrium Constant K

The standard free energy change $\Delta G^\circ$ dictates the equilibrium constant $K$ for any reversible reaction:

\Delta G^\circ = -RT \ln K \quad \implies \quad K = e^{-\frac{\Delta G^\circ}{RT}}

As demonstrated by the industrial water-gas shift reaction $\ce{CO(g) + H2O(g) <=> CO2(g) + H2(g)}$ , temperature changes alter $\Delta G^\circ$ , exponentially impacting the magnitude of $K$ . Because the reaction is exothermic, increasing temperature causes $\Delta G^\circ$ to become less negative, decreasing $K$ .

Worked Example: Calculating K from Standard Free Energy

Calculate the equilibrium constant $K_p$ for $\ce{N2O4(g) <=> 2NO2(g)}$ at $\pu{298 K}$ given $\Delta G^\circ = \pu{4.80 kJ/mol}$ .

Step-by-Step Solution:

Convert $\Delta G^\circ$ to $\pu{J/mol}$ : $\Delta G^\circ = \pu{4800 J/mol}$ .
Rearrange the equation to isolate $\ln K$ :
$\ln K = \frac{-\Delta G^\circ}{RT} = \frac{-\pu{4800 J/mol}}{(\pu{8.314 J/(mol K)})(\pu{298 K})} = \frac{-4800}{2477.57} = -1.937$
Solve for $K$ using the exponential function:
$K_p = e^{-1.937} = 0.144$

Common Exam Trap: Assuming $\Delta G^\circ > 0$ implies a reaction cannot occur. A positive $\Delta G^\circ$ simply indicates $K < 1$ (reactants favored at equilibrium under standard states), but nonstandard mixtures with small $Q$ values can still yield spontaneous forward reaction ( $\Delta G < 0$ ).

Predicting K from the Sign of Standard Free Energy

The sign of $\Delta G^\circ$ directly indicates whether products or reactants dominate the mixture once equilibrium is established.

Standard Free Energy ( $\Delta G^\circ$ )	Logarithmic Value ( $\ln K$ )	Equilibrium Constant ( $K$ )	Equilibrium Mixture Composition
$\Delta G^\circ < 0$ (Negative)	$\ln K > 0$	$K > 1$	Products strongly favored at equilibrium
$\Delta G^\circ = 0$	$\ln K = 0$	$K = 1$	Reactants and products equally favored
$\Delta G^\circ > 0$ (Positive)	$\ln K < 0$	$K < 1$	Reactants strongly favored at equilibrium

DAT Mnemonic: "Negative $\Delta G^\circ$ , $K$ is Grand ( $K > 1$ ); Positive $\Delta G^\circ$ , $K$ is Petite ( $K < 1$ )."

Critical Thinking Questions

A reaction has $\Delta G^\circ = \pu{+15.0 kJ/mol}$ at $\pu{298 K}$ . What reaction quotient $Q$ is required for the reaction to become spontaneous in the forward direction ( $\Delta G < 0$ )?
Why is $\Delta G = 0$ at dynamic equilibrium even though $\Delta G^\circ$ is non-zero for most chemical reactions?
For an endothermic reaction ( $\Delta H^\circ > 0$ ), how does raising the temperature under isothermal conditions affect the sign of $\Delta G^\circ$ and the magnitude of $K$ ?

References

Sources returned by Google Search grounding during lesson verification.

Water gas shift equilibria via the NIST Webbook - Matlab in Chemical Engineering at CMU — kitchinhub.cheme.cmu.edu
Water–gas shift reaction - Wikipedia — en.wikipedia.org

Free Energy, the Reaction Quotient, and the Equilibrium Constant

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