Spontaneity of Reactions - Chemistry

1. Fundamental Concepts

Definition: A spontaneous reaction is one that occurs without the need for an external energy source, and it tends to proceed in the direction of increasing entropy and decreasing Gibbs free energy.
Entropy (S): A measure of disorder or randomness in a system. The second law of thermodynamics states that the total entropy of a closed system must always increase over time.
Gibbs Free Energy (G): A thermodynamic potential that measures the maximum reversible work that a system can perform at constant temperature and pressure. It is defined as $$ G = H - TS $$, where $$ H $$ is enthalpy, $$ T $$ is temperature, and $$ S $$ is entropy.

2. Key Concepts

Gibbs Free Energy Change (ΔG): $$\Delta G = \Delta H - T\Delta S$$

Spontaneity Criteria:

If $$ \Delta G < 0 $$, the reaction is spontaneous.
If $$ \Delta G > 0 $$, the reaction is non-spontaneous.
If $$ \Delta G = 0 $$, the system is at equilibrium.

Application: Used to predict the spontaneity of chemical reactions and to understand the feasibility of processes in chemistry and biology.

3. Examples

Example 1 (Basic)

Problem: Determine if the following reaction is spontaneous at 298 K: $$\text{N}_2\text{O}_4(g) \rightarrow 2\text{NO}_2(g)$$ Given: $$ \Delta H = 57.2 \, \text{kJ/mol} $$, $$ \Delta S = 176.6 \, \text{J/(mol·K)} $$

Step-by-Step Solution:

Convert $$ \Delta S $$ to kJ: $$ \Delta S = 0.1766 \, \text{kJ/(mol·K)} $$
Calculate $$ \Delta G $$: $$\Delta G = \Delta H - T\Delta S = 57.2 \, \text{kJ/mol} - (298 \, \text{K} \times 0.1766 \, \text{kJ/(mol·K)})$$ $$\Delta G = 57.2 \, \text{kJ/mol} - 52.8 \, \text{kJ/mol} = 4.4 \, \text{kJ/mol}$$

Validation: Since $$ \Delta G > 0 $$, the reaction is non-spontaneous at 298 K.

Example 2 (Intermediate)

Problem: Calculate the temperature at which the following reaction becomes spontaneous: $$\text{CaCO}_3(s) \rightarrow \text{CaO}(s) + \text{CO}_2(g)$$ Given: $$ \Delta H = 178.3 \, \text{kJ/mol} $$, $$ \Delta S = 160.2 \, \text{J/(mol·K)} $$

Step-by-Step Solution:

Set $$ \Delta G = 0 $$ for the transition point: $$\Delta G = \Delta H - T\Delta S = 0$$
Solve for $$ T $$: $$T = \frac{\Delta H}{\Delta S} = \frac{178.3 \, \text{kJ/mol}}{0.1602 \, \text{kJ/(mol·K)}} = 1113 \, \text{K}$$

Validation: The reaction becomes spontaneous above 1113 K.

4. Problem-Solving Techniques

Sign Analysis: Always check the signs of $$ \Delta H $$ and $$ \Delta S $$ to determine the general behavior of $$ \Delta G $$.
Temperature Dependence: Use the relationship $$ \Delta G = \Delta H - T\Delta S $$ to find the temperature at which a reaction changes from non-spontaneous to spontaneous.
Dimensional Consistency: Ensure that all units are consistent, especially when converting between kJ and J.