5.2.2(d)-(f) - Free energy, feasibility and kinetic limits

The Gibbs balance

Feasibility is about whether a process is thermodynamically allowed under the stated conditions. OCR uses the sign of the free energy change, $\Delta G$, to make that judgement.

A process is feasible when $\Delta G$ has a negative value. If $\Delta G$ is positive, the process is not thermodynamically feasible under those stated conditions. If $\Delta G = 0$, the process is at the boundary where feasibility changes.

The terms in the equation have to be in compatible units:

The most common method is to convert $\Delta S$ from $\,\mathrm{J} \,\mathrm{K}^{-1}$ $\,\mathrm{mol}^{-1}$ to $\,\mathrm{kJ} \,\mathrm{K}^{-1}$ $\,\mathrm{mol}^{-1}$ by dividing by 1000. Then $T\Delta S$ has units of $\,\mathrm{kJ} \,\mathrm{mol}^{-1}$, matching $\Delta H$.

Do not read "feasible" as "fast". Negative $\Delta G$ means the energy balance favours the process; it does not yet say whether particles can get over the activation-energy barrier.

Calculating free energy change

A free energy calculation is usually a careful substitution exercise. The chemistry is in the signs and units.

Worked example:

A reaction has:

• $\Delta H = -74.8\ \mathrm{kJ\ mol^{-1}}$
• $\Delta S = -80.7\ \mathrm{J\ K^{-1}\ mol^{-1}}$
• $T = 298\ \mathrm{K}$

Calculate $\Delta G$ and state whether the reaction is feasible at 298 $\mathrm K$.

First convert entropy into $\,\mathrm{kJ} \,\mathrm{K}^{-1}$ $\,\mathrm{mol}^{-1}$:

Now substitute into the Gibbs equation:

Estimate before pressing on: $300 \times 0.08 \approx 24$, so the entropy term will add about $+24\ \mathrm{kJ\ mol^{-1}}$ because subtracting a negative gives a positive contribution.

The value is negative, so the reaction is feasible at 298 $\mathrm K$.

If a question gives entropies of individual substances instead of $\Delta S$, use the previous entropy method first:

Then use the resulting $\Delta S$ in the Gibbs equation. Coefficients in the balanced equation still matter when calculating $\Delta S$.

Temperature and sign reasoning

The term $T\Delta S$ explains why temperature can change feasibility. As temperature increases, the entropy term has a bigger effect on $\Delta G$.

[DIAGRAM: asset_name: Lesson 5.2.2d-f: Free energy, feasibility and kinetic limits - diagram 01; asset_slug: 05_02_02b_free_energy_feasibility_and_kinetic_limits__diagram_01; recommended_method: drawn_chem; description: Deterministic sign map for Delta H and Delta S combinations, showing always feasible, never feasible, high-temperature feasible and low-temperature feasible cases under the Gibbs equation.]

The sign of $\Delta H$ tells you whether the enthalpy term helps or opposes feasibility. A negative $\Delta H$ helps make $\Delta G$ negative; a positive $\Delta H$ makes $\Delta G$ more positive.

The sign of $\Delta S$ has to be read through the term $-T\Delta S$:

• if $\Delta S$ is positive, $-T\Delta S$ is negative, so increasing temperature helps feasibility
• if $\Delta S$ is negative, $-T\Delta S$ is positive, so increasing temperature works against feasibility

This gives four useful cases:

These patterns assume $\Delta H$ and $\Delta S$ are approximately constant over the temperature range considered. They are not a replacement for calculation. If numbers are given, calculate $\Delta G$.

Worked example:

A reaction has $\Delta H = +85.0\ \mathrm{kJ\ mol^{-1}}$ and $\Delta S = +170\ \mathrm{J\ K^{-1}\ mol^{-1}}$. Both signs are positive, so the reaction is only feasible at high enough temperature.

Convert entropy:

At the feasibility boundary, $\Delta G = 0$:

Rearrange:

Substitute:

Because $\Delta H$ and $\Delta S$ are both positive, higher temperature makes $-T\Delta S$ more negative. The reaction becomes feasible above 500 $\mathrm K$.