Gibbs Free Energy and Reaction Spontaneity
Definition
Gibbs free energy
Gibbs free energy ($\Delta G$) is a thermodynamic property that determines whether a chemical reaction is spontaneous under constant pressure and temperature.
- A reaction is spontaneous if it proceeds toward completion or equilibrium without requiring external energy.
- The relationship between Gibbs free energy and other thermodynamic properties is expressed by the equation: $$\Delta G^\circ = \Delta H^\circ - T\Delta S^\circ$$ where:
- $\Delta G^\circ$: Standard Gibbs free energy change ($kJ \, mol^{-1}$)
- $\Delta H^\circ$: Standard enthalpy change ($kJ \, mol^{-1}$)
- $T$: Temperature in Kelvin (K)
- $\Delta S^\circ$: Standard entropy change ($J \, K^{-1} mol^{-1}$)
Let’s break this down:
- Enthalpy ($\Delta H^\circ$):
- Represents the heat energy absorbed or released during a reaction.
- Exothermic reactions ($\Delta H^\circ < 0$) release heat, while endothermic reactions ($\Delta H^\circ > 0$) absorb heat.
- Entropy ($\Delta S^\circ$):
- Measures the disorder or randomness in a system.
- Reactions that increase disorder ($\Delta S^\circ > 0$) are more likely to be spontaneous.
- Temperature ($T$):
- Higher temperatures amplify the influence of entropy on Gibbs free energy.
Hint
- Temperature must always be in Kelvin.
- To convert from °C to K, add 273.15.
When Is a Reaction Spontaneous?
The sign of $\Delta G$ determines whether a reaction is spontaneous:
- $\Delta G< 0$: The reaction is spontaneous.
- $\Delta G = 0$: The reaction is at equilibrium.
- $\Delta G >0$: The reaction is non-spontaneous.
Note
- The interplay between $\Delta H^\circ$, $\Delta S^\circ$, and $T$ influences $\Delta G^\circ$.
- We will recap this aspect in the next article.
Calculating Gibbs Free Energy
To calculate $\Delta G^\circ$, you need thermodynamic data for $\Delta H^\circ$ and $\Delta S^\circ$, which are available in the IB Chemistry data booklet.
Example
Combustion of Propane ($\text{C}_3\text{H}_8$)
Consider the combustion of propane:
$$\text{C}_3\text{H}_8(g) + 5\text{O}_2(g) \rightarrow 3\text{CO}_2(g) + 4\text{H}_2\text{O}(g)$$
Given data:
- $\Delta H^\circ = -2045 \, \text{kJ mol}^{-1}$
- $\Delta S^\circ = +103 \, \text{J K}^{-1}\ \text{mol}^{-1}$
- Temperature: $T = 298 \, \text{K}$
- Step 1: Convert $\Delta S^\circ$ to $kJ \, K^{-1} \, mol^{-1}$:
$$\Delta S^\circ = \frac{103}{1000} = 0.103 /, \text{kJ K}^{-1} \text{mol}^{-1}$$ - Step 2: Substitute values into the Gibbs free energy equation:
$$\Delta G^\circ = \Delta H^\circ - T\Delta S^\circ$$
$$\Delta G^\circ = -2045 - (298 \times 0.103)$$
$$\Delta G^\circ = -2045 - 30.7 = -2075.7 \, \text{kJ mol}^{-1}$$ - Step 3: Interpret the result:
Since $\Delta G^\circ< 0$, the reaction is spontaneous under standard conditions.
Hint
- Always check your units!
- Entropy values in the IB data booklet are often in $\text{J K}^{-1} \text{mol}^{-1}$, so convert them to $\text{kJ K}^{-1}\ \text{mol}^{-1}$ by dividing by 1000.
Temperature and Spontaneity: When Does $\Delta G = 0$?
- For some reactions, spontaneity depends on temperature.
- At the temperature where $\Delta G = 0$, the reaction is at the boundary of becoming spontaneous.
- Rearranging the Gibbs free energy equation gives:
$$T = \frac{\Delta H^\circ}{\Delta S^\circ}$$
Example
Dissociation of Ammonium Chloride
The decomposition of ammonium chloride ($\text{NH}_4\text{Cl}$) is:
$$\text{NH}_4\text{Cl}(s) \rightarrow \text{NH}_3(g) + \text{HCl}(g)$$
Given data:
- $\Delta H^\circ = +176 \, \text{kJ mol}^{-1}$
- $\Delta S^\circ = +285 \, \text{J K}^{-1} \text{mol}^{-1}$
- Step 1: Convert $\Delta S^\circ$ to $\text{kJ K}^{-1} \text{mol}^{-1}$:
$$\Delta S^\circ = \frac{285}{1000} = 0.285 \, \text{kJ K}^{-1} \text{mol}^{-1}$$ - Step 2: Calculate the temperature:
$$T = \frac{\Delta H^\circ}{\Delta S^\circ}$$
$$T = \frac{176}{0.285} = 617.5 \, \text{K}$$ - Step 3: Interpret the result:
The reaction becomes spontaneous above 617.5 K.
Using Thermodynamic Data to Calculate $\Delta G^\circ$
To calculate $\Delta G^\circ$ for a reaction using data booklet values:
- Find $\Delta H^\circ$ and $\Delta S^\circ$ for each species in the reaction.
- Use the equations:
$$\Delta H^\circ = \sum \Delta H^\circ_{\text{products}} - \sum \Delta H^\circ_{\text{reactants}}$$
$$\Delta S^\circ = \sum S^\circ_{\text{products}} - \sum S^\circ_{\text{reactants}}$$ - Substitute into $\Delta G^\circ = \Delta H^\circ - T\Delta S^\circ$.
Common Mistake
- Many students forget to convert entropy from $\text{J K}^{-1} \text{mol}^{-1}$ to $\text{kJ K}^{-1} \text{mol}^{-1}$ or temperature from °C to K.
- Always double-check your units!
Self review
- What happens to $\Delta G^\circ$ if both $\Delta H^\circ$ and $\Delta S^\circ$ are negative?
- Under what conditions would the reaction be spontaneous?
