Feasible Reactions

The Fundamental Equation: The relationship between enthalpy, entropy, and temperature is given by: $\Delta G = \Delta H - T\Delta S$ where $\Delta H$ is the enthalpy change, $T$ is the absolute temperature in Kelvin, and $\Delta S$ is the entropy change of the system.

Enthalpy-Entropy Balance: The term $-T\Delta S$ represents the energy 'lost' to the surroundings to satisfy the second law of thermodynamics. For a reaction to be feasible, the combination of heat release (negative $\Delta H$) and disorder increase (positive $\Delta S$) must result in a negative $\Delta G$.

Unit Consistency: A critical principle in applying this equation is that $\Delta H$ is usually measured in $kJ\,mol^{-1}$, while $\Delta S$ is measured in $J\,K^{-1}\,mol^{-1}$. To calculate $\Delta G$ in $kJ\,mol^{-1}$, the entropy value must be divided by $1000$.

Case 1: Always Feasible: If $\Delta H$ is negative (exothermic) and $\Delta S$ is positive (increasing disorder), $\Delta G$ will be negative at all temperatures. These reactions are thermodynamically favored under any conditions.

Case 2: Never Feasible: If $\Delta H$ is positive (endothermic) and $\Delta S$ is negative (decreasing disorder), $\Delta G$ will always be positive. Such reactions cannot occur spontaneously without external energy input.

Case 3: Temperature Dependent: If both $\Delta H$ and $\Delta S$ have the same sign, feasibility depends on the magnitude of $T$. For endothermic reactions with increasing entropy, feasibility occurs only at high temperatures where the $T\Delta S$ term outweighs $\Delta H$.

Calculating the Threshold: To find the exact temperature at which a reaction becomes feasible, set $\Delta G = 0$ and solve for $T$: $T = \frac{\Delta H}{\Delta S}$

Thermodynamic Stability: Refers to a system in its lowest energy state (most negative $\Delta G$). A system is stable if no feasible reaction exists to change it.

Kinetic Stability: Refers to a system that should react according to $\Delta G$ but is prevented from doing so by a high energy barrier. These substances are often called 'metastable'.

The 1000 Rule: Always check your units for $\Delta S$. Exams frequently provide $\Delta H$ in $kJ$ and $\Delta S$ in $J$. Forgetting to divide $\Delta S$ by $1000$ is the most common cause of lost marks.

Sign Conventions: Be meticulous with signs. A negative gradient on a $\Delta G$ vs $T$ graph implies a positive $\Delta S$ (since $m = -\Delta S$). If the line slopes upward, the entropy change is negative.

Standard Conditions: Remember that $\Delta G^\theta$ refers to standard conditions ($298\,K$, $100\,kPa$). If the question asks about feasibility at a different temperature, you must use the Gibbs equation with that specific $T$ value.

Sanity Check: If a reaction involves the production of a gas from solids (like thermal decomposition), $\Delta S$ must be positive. If your calculation shows otherwise, re-evaluate your steps.

The van't Hoff Link: Gibbs Free Energy is directly related to the equilibrium constant ($K$) via the equation: $\Delta G = -RT \ln K$ where $R$ is the gas constant ($8.31\,J\,K^{-1}\,mol^{-1}$).

Interpreting K: If $\Delta G$ is very negative, $K$ will be very large ($K > 1$), meaning the equilibrium position lies far to the right (products favored). If $\Delta G$ is positive, $K$ is small ($K < 1$), and reactants are favored.

Dynamic Equilibrium: At the exact point of feasibility transition ($\Delta G = 0$), the equilibrium constant $K = 1$, meaning neither reactants nor products are favored over the other.