Entropy and thermodynamic feasibility

What you’ll learn

• Calculate system, surroundings and total entropy changes.
• Predict how temperature affects feasibility.
• Distinguish thermodynamic feasibility from reaction speed.

Understand the idea

Entropy describes the dispersal of energy among possible arrangements of particles and energy levels. Gases generally have greater molar entropy than comparable liquids or solids, and raising temperature increases entropy. A perfect crystal has zero entropy at zero kelvin. For a reaction, multiply each standard molar entropy by its equation coefficient before subtracting the reactant total from the product total.

The natural direction of change increases total entropy: system plus surroundings. An exothermic reaction transfers energy to the surroundings and usually increases their entropy. An endothermic reaction can still be feasible if the system's entropy gain outweighs the surroundings' loss. A positive total entropy change says nothing about whether an activation barrier makes the reaction too slow to observe.

Key formulae

$$\Delta S_{\mathrm{system}}=\sum\nu S^\circ_{\mathrm{products}}-\sum\nu S^\circ_{\mathrm{reactants}}$$

$$\Delta S_{\mathrm{surroundings}}=-\frac{\Delta H}{T}$$

$$\Delta S_{\mathrm{total}}=\Delta S_{\mathrm{system}}+\Delta S_{\mathrm{surroundings}}$$

$$\Delta S_{\mathrm{total}}>0\quad\text{for a thermodynamically feasible direction}$$

Worked example

A reaction has $\Delta H=+40.0\,\mathrm{kJ\,mol^{-1}}$ and $\Delta S_{\mathrm{system}}=+150\,\mathrm{J\,mol^{-1}\,K^{-1}}$. Is it feasible at $300\,\mathrm K$?

1. Convert the enthalpy change to $+40000\,\mathrm{J\,mol^{-1}}$.
2. The surroundings change is $-40000/300=-133.3\,\mathrm{J\,mol^{-1}\,K^{-1}}$.
3. The total change is $150-133.3=+16.7\,\mathrm{J\,mol^{-1}\,K^{-1}}$.

Answer: Yes: the total entropy change is positive, although the reaction may still be slow.