Free Energy (College Board AP® Chemistry): Flashcards

Increasing temperature makes the reaction less thermodynamically favorable because the −TΔS° term becomes more positive as T rises (ΔS° is negative), and at high enough temperatures this outweighs the negative ΔH°, making ΔG° positive. The reaction is therefore only favorable at low temperatures.

A thermodynamically favorable reaction has a negative value of ΔG°. If ΔH° is positive and ΔS° is negative, the reaction is thermodynamically unfavorable at all temperatures.

ΔG° = ΔH° − TΔS°

• ΔG°: kJ mol^-1

• ΔH°: kJ mol^-1

• T: K

• ΔS°: J K^-1 mol^-1 (must be converted to kJ K^-1 mol^-1 by dividing by 1000)

True.

ΔH° is in kJ mol^-1 but ΔS° is typically given in J K^-1 mol^-1. Dividing ΔS° by 1000 converts it to kJ K^-1 mol^-1 so that units are consistent throughout the calculation.

Set ΔG° = 0 in the equation ΔG° = ΔH° − TΔS° and solve for T:

T = ΔH° / ΔS°

Below this temperature the reaction is favorable if both ΔH° and ΔS° are negative; above it if both are positive. This is the crossover temperature.

False.

The reaction converts two solids and a gas into two solids. Removing the gas phase drastically reduces disorder, so ΔS° is negative — more moles of product does not override the loss of a gaseous species.

Standard entropy values (S°) are tabulated in J K^-1 mol^-1 while enthalpy values are in kJ mol^-1, so ΔS° must be divided by 1000 before substituting into the Gibbs equation. Mixing units without this conversion would give an incorrect ΔG°.

When using ΔG° = ΔH° − TΔS°, ΔS° must be converted from J K^-1 mol^-1 to kJ K^-1 mol^-1 by dividing by 1000. The crossover temperature at which a reaction changes favorability is found by setting ΔG° = 0 and solving for T.

A reaction is under kinetic control when it is thermodynamically favorable (ΔG° < 0) but proceeds at a very slow or unmeasurable rate at a given temperature because the activation energy (Ea) is very high.

False.

A negative ΔG° indicates the reaction is energetically downhill, but says nothing about the rate. If the activation energy is high, the reaction may be immeasurably slow — it is under kinetic control.

The decomposition of H₂O₂ has a very high activation energy, making it kinetically controlled at room temperature. MnO₂ acts as a catalyst, providing an alternative reaction pathway with a lower activation energy. This allows the reaction to proceed at a measurable rate without changing ΔG°.

False.

A catalyst lowers the activation energy and increases reaction rate, but it does not change ΔG°. A thermodynamically unfavorable reaction (ΔG° > 0) remains unfavorable in the presence of a catalyst.

Thermodynamic control relates to ΔG° — whether the reaction is energetically favorable (products lower in free energy than reactants).

Kinetic control relates to activation energy (Ea) — whether the reaction is fast enough to occur at a measurable rate.

A reaction can be thermodynamically favorable yet kinetically controlled (slow) if Ea is high.

A reaction is under kinetic control when it is thermodynamically favorable but proceeds at a very slow or immeasurable rate due to a high activation energy. A catalyst speeds up such reactions by providing an alternative pathway with a lower activation energy.

The two are related by:

ΔG° = −RT ln K

where R is the ideal gas constant (8.314 J mol^-1 K^-1) and T is temperature in Kelvin.

Equivalently: K = e^{(−ΔG°/RT)}

True.

Substituting ΔG° = 0 into ΔG° = −RT ln K gives ln K = 0, so K = 1. Products and reactants are present in equal amounts at equilibrium. Note: ΔG° = 0 does not mean the system is currently at equilibrium — it is a standard-state property that tells us K = 1.

From ΔG° = −RT ln K, a more negative ΔG° gives a more positive ln K, and therefore a larger K. A large K means products are strongly favored at equilibrium — the reaction proceeds almost to completion in the forward direction.

True.

A positive ΔG° means the reverse reaction is thermodynamically favored. Substituting into ΔG° = −RT ln K gives a negative ln K, so K < 1 — reactants are favored at equilibrium.

For a favorable reaction (ΔG° < 0): the minimum free energy (equilibrium position) is closer to the product side, and the slope of the curve is negative toward the product side.

For an unfavorable reaction (ΔG° > 0): the minimum is closer to the reactant side, and the slope is positive — indicating the system moves toward reactants.

The relationship between Gibbs free energy and the equilibrium constant is given by: ΔG° = −RT ln K. When ΔG° < 0, K is greater than 1, and when ΔG° > 0, K is less than 1.

Coupled reactions are two reactions that share a common intermediate. A thermodynamically favorable reaction (ΔG° < 0) is linked to an unfavorable one (ΔG° > 0) so that the overall combined ΔG° is negative, making the net process thermodynamically favorable.

False.

Electrolysis drives an unfavorable reaction using an external electrical energy source — not by coupling to a favorable reaction. Coupling reactions specifically involves sharing a common intermediate between two chemical reactions.

Cu₂S → 2Cu + S has ΔG° = +85.1 kJ mol^-1 (unfavorable).\n\nS + O₂ → SO₂ has ΔG° = −301.4 kJ mol^-1 (favorable).\n\nSulfur is the shared intermediate. The favorable second reaction consumes sulfur as it forms, shifting the first reaction's equilibrium toward products (Le Chatelier's Principle). Combined ΔG° = −216.3 kJ mol^-1, so overall copper production is thermodynamically favorable.

The condensation of amino acids (e.g., alanine + glycine → alanylglycine + H₂O) has ΔG° = +31 kJ mol^-1 (unfavorable).\n\nATP + H₂O → ADP + Pi has ΔG° = −33 kJ mol^-1 (favorable). The water molecules cancel as the common intermediate, giving a combined ΔG° = −2 kJ mol^-1 — the peptide bond formation is now thermodynamically favorable.

True.

Because ΔG° is a state function, the free energy changes are additive. If the sum of the two ΔG° values is negative, the coupled process is thermodynamically favorable overall.

In coupled reactions, a thermodynamically favorable reaction drives a thermodynamically unfavorable reaction by sharing a common intermediate. The overall ΔG° is found by adding the individual ΔG° values.

The free energy of dissolution (ΔG°_diss) is the thermodynamic parameter that determines the spontaneity of a dissolution process:

ΔG°_diss = ΔH°_diss − TΔS°_diss

If negative, dissolution is thermodynamically favorable; if positive, it is not.

True.

Although Step 3 (ion-dipole interaction formation) has a negative entropy contribution, the positive entropy gained from breaking the lattice (Step 1) and reorganizing water (Step 2) outweighs it, giving a net positive ΔS°_diss.

Mg²⁺ is smaller and carries a higher charge (+2) than Na⁺ (+1). This means it forms stronger ion-dipole interactions with water molecules in Step 3 of dissolution. More energy is released when these interactions form, making Step 3 more exothermic and the overall dissolution more exothermic for MgCl₂.

False.

For dissolution with a positive ΔS°_diss (more disorder), increasing T makes TΔS°_diss larger, reducing ΔG°_diss and making dissolution more thermodynamically favorable.

Step 1 — Breaking the lattice: Endothermic (ΔH₁ > 0), entropy increases (ΔS₁ > 0) as ions separate.

Step 2 — Preparing solvent: Endothermic (ΔH₂ > 0), entropy increases (ΔS₂ > 0) as water molecules move apart to make room.

Step 3 — Ion-dipole formation: Exothermic (ΔH₃ < 0), entropy decreases (ΔS₃ < 0) as water molecules are organized around ions.

The free energy of dissolution is calculated using ΔG°_diss = ΔH°_diss − TΔS°_diss. If ΔG°_diss is negative, dissolution is thermodynamically favorable. Increasing temperature makes dissolution more favorable when ΔS°_diss is positive.