13B.18-13B.22 - Gibbs Energy And Feasibility

Enthalpy, Entropy And Temperature

Some reactions release heat and proceed readily, but enthalpy alone is not enough to decide feasibility. An endothermic process can be feasible if it gives a large enough increase in entropy, and an exothermic process can become less favourable at high temperature if it causes a large enough decrease in entropy.

Gibbs free energy is the quantity that combines those two effects:

The sign of $\Delta G$ gives the thermodynamic prediction for the reaction as written under the stated conditions.

The equation is closely related to total entropy. From the previous lesson, a feasible change has a positive total entropy change. Since the surroundings entropy term depends on the enthalpy change, Gibbs free energy packages the balance between system entropy and enthalpy into one value: negative $\Delta G$ corresponds to the entropy balance favouring the reaction.

Common-error contrast: "feasible" does not mean "fast". Gibbs free energy tells you whether the final state is thermodynamically favoured; it does not tell you whether particles can get over the activation energy barrier quickly enough to observe the reaction.

Using the Gibbs Equation

Most errors in Gibbs calculations are unit errors. $T$ must be in kelvin, and $\Delta H$ and $T \Delta S_{\mathrm{system}}$ must use the same energy unit before they are subtracted.

Two safe routes are possible:

Worked example: predict feasibility at 298 K.

A reaction has $\Delta H = +45.0\,\mathrm{kJ\,mol^{-1}}$, $\Delta S_{\mathrm{system}} = +160\,\mathrm{J\,K^{-1}\,mol^{-1}}$ and $T = 298\,\mathrm{K}$.

Use the kJ route:

$\Delta G = +45.0 - (298\times 0.160)$

$\Delta G = +45.0 - 47.7$

$\Delta G = -2.7 \mathrm{kJ\,mol^{-1}}$

The prediction is thermodynamically feasible at $298\,\mathrm{K}$ because $\Delta G$ is negative. The chemical interpretation matters: the enthalpy term is unfavourable because the reaction is endothermic, but the positive entropy term is large enough at this temperature to outweigh it.

Finding the Feasible Temperature

When a question asks for the temperature at which a reaction becomes feasible, set $\Delta G = 0$. That gives the boundary temperature:

[DIAGRAM: asset_name: Free energy against temperature for positive enthalpy and entropy; asset_slug: edexcel_a_level_chemistry_l063_gibbs_temperature; recommended_method: deterministic_chart; description: Free energy against temperature for positive enthalpy and entropy.]

This calculation only makes sense when $\Delta H$ and $\Delta S_{\mathrm{system}}$ are in matching units. The answer is in kelvin.

Worked example: determine the temperature boundary.

A decomposition has:

$\Delta H = +178 \mathrm{kJ\,mol^{-1}}$

Convert entropy to kJ:

At the boundary:

$T = 178 / 0.161$

$T = 1106\,\mathrm{K}$

Because both $\Delta H$ and $\Delta S_{\mathrm{system}}$ are positive, the reaction becomes feasible above this temperature. Below $1106\,\mathrm{K}$, the $T \Delta S_{\mathrm{system}}$ term is too small to overcome the positive enthalpy change. Above $1106\,\mathrm{K}$, the entropy contribution is large enough to make $\Delta G$ negative.

The sign pattern is a useful sense check:

Guided practice: a reaction has:

$\Delta H = -92.0 \mathrm{kJ\,mol^{-1}}$

Determine the temperature boundary and state whether the reaction is feasible above or below it.

Expected answer:

$T = (-92.0) / (-0.199)$

$T = 462\,\mathrm{K}$

The reaction is feasible below $462\,\mathrm{K}$. For example, at $298\,\mathrm{K}$, $\Delta G = -92.0 - (298\times -0.199) = -32.7 \mathrm{kJ\,mol^{-1}}$, but at $600\,\mathrm{K}$, $\Delta G = -92.0 - (600\times -0.199) = +27.4 \mathrm{kJ\,mol^{-1}}$.

Gibbs Energy and Equilibrium

The standard reaction Gibbs free energy, $\Delta G^{\circ}$, is linked to the dimensionless equilibrium constant $K$. The specification often abbreviates this symbol to $\Delta G$ in this relationship:

The signs explain the chemistry:

Worked example: show the link for a large equilibrium constant.

At $298\,\mathrm{K}$, a reaction has:

$K = 2.0\times 10^{5}$

Calculate:

$\ln K = \ln(2.0\times 10^{5}) = 12.2$

$\Delta G^{\circ} = -RT\ln K$

$\Delta G^{\circ} = -(8.31)(298)(12.2)$