Gibbs Energy Change

Why a new function is needed. The second law says a process is spontaneous only if $\Delta S_{total} = \Delta S_{system} + \Delta S_{surroundings} \geq 0$. That is exact, but $\Delta S_{surroundings}$ is awkward to measure directly for a chemical system. Gibbs energy repackages the same physics using only system properties.

The definition. For any system at constant $T$:

where: $G$ is Gibbs (free) energy, $H$ is enthalpy, $T$ is absolute temperature (kelvin), and $S$ is entropy — all properties of the system alone.

The working equation. At constant temperature and pressure, the change in $G$ between two states is:

where: $\Delta H$ is the enthalpy change (kJ mol⁻¹), $\Delta S$ is the entropy change (usually given in J K⁻¹ mol⁻¹, so it must be converted to kJ before use), and $T$ is the constant absolute temperature at which the process occurs.

The spontaneity criterion. This single quantity replaces the need to track the surroundings:

• $\Delta G < 0$: the process is spontaneous (exergonic).
• $\Delta G > 0$: the process is non-spontaneous in the forward direction — the reverse process is spontaneous instead (endergonic).
• $\Delta G = 0$: the system is at equilibrium; forward and reverse rates balance and there is no net driving force either way.

How $\Delta H$ and $\Delta S$ interact. Because $\Delta G$ is a sum of one term independent of $T$ and one term proportional to $T$, the four sign combinations behave very differently:

For the two temperature-dependent rows, there is a crossover temperature where the reaction switches character. Setting $\Delta G = 0$ gives $\Delta H = T\Delta S$, so:

Gibbs energy and the equilibrium constant. At standard conditions, $\Delta G^\circ$ links directly to the equilibrium constant $K$:

where: $R = 8.314\ \text{J K}^{-1}\text{mol}^{-1}$ is the gas constant, $T$ is absolute temperature, and $K$ is the equilibrium constant. A large negative $\Delta G^\circ$ gives a large $K$ (products favoured at equilibrium); a large positive $\Delta G^\circ$ gives a small $K$ (reactants favoured). This is only an introductory, qualitative-to-basic-quantitative link — full treatment of $K$ and the reaction quotient $Q$ belongs to the Equilibrium chapter.

Problem: For a reaction at 298 K, $\Delta H = -35.5\ \text{kJ mol}^{-1}$ and $\Delta S = -83.6\ \text{J K}^{-1}\text{mol}^{-1}$. Calculate $\Delta G$ at 298 K, comment on spontaneity, and find the temperature at which the reaction switches from spontaneous to non-spontaneous.

Step 1. Match units before combining. $\Delta H$ is in kJ mol⁻¹ but $\Delta S$ is in J K⁻¹mol⁻¹, so convert: $\Delta S = -83.6\ \text{J K}^{-1}\text{mol}^{-1} = -0.0836\ \text{kJ K}^{-1}\text{mol}^{-1}$.

Step 2. Apply $\Delta G = \Delta H - T\Delta S$ at $T = 298\ \text{K}$:

Since $\Delta G < 0$, the reaction is spontaneous at 298 K.

Step 3. Check whether spontaneity depends on $T$. Both $\Delta H$ and $\Delta S$ are negative here, so from the sign table this reaction is spontaneous only at sufficiently low temperature. Find the crossover temperature by setting $\Delta G = 0$, i.e., $\Delta H = T\Delta S$:

The reaction stays spontaneous for $T < 424.6\ \text{K}$ and becomes non-spontaneous above it — consistent with 298 K (well below 424.6 K) giving a negative $\Delta G$.

Competitive stretch

Problem: A reaction has equilibrium constant $K = 1.8\times10^{-5}$ at 298 K. Find $\Delta G^\circ$ and state whether the reaction favours reactants or products at equilibrium.

Step 1. Compute $\ln K$: since $K = 1.8\times10^{-5}$, $\ln K = \ln(1.8) + \ln(10^{-5}) = 0.588 - 11.513 = -10.925$.

Step 2. Apply $\Delta G^\circ = -RT\ln K$ with $R = 8.314\ \text{J K}^{-1}\text{mol}^{-1}$, $T = 298\ \text{K}$:

Since $\Delta G^\circ$ is positive and $K \ll 1$, the reaction strongly favours reactants at equilibrium — the forward reaction is non-spontaneous under standard conditions, exactly as the small $K$ suggests.

• Unit mismatch: $\Delta H$ is usually in kJ mol⁻¹ while $\Delta S$ is in J K⁻¹mol⁻¹ — forgetting to convert before combining them in $\Delta G = \Delta H - T\Delta S$ is the single most common numerical error.
• Reading $\Delta G < 0$ as "the reaction will happen fast." Gibbs energy tests thermodynamic feasibility only — reaction rate is a separate (kinetics) question.
• Applying the sign table by checking only one of $\Delta H$ or $\Delta S$ — both signs are needed to know whether spontaneity is universal, never, or temperature-dependent.
• Treating $\Delta G^\circ = -RT\ln K$ as the full equilibrium story at this level — here it is only an introductory link; $K$ and reaction quotient $Q$ are developed properly in the Equilibrium chapter.

Ask the tutor: want to work through more $\Delta G$-versus-temperature problems or see how $\Delta G^\circ = -RT\ln K$ connects to the equilibrium constant in more detail?