9.5 ΔG° & K

Free Energy and Equilibrium

Adjust concentrations and T for 4 reactions; a beaker of particles and a ΔG vs ln Q line show ΔG = ΔG° + RT ln Q flipping sign as Q crosses K.

ΔG° = −RT ln KΔG = ΔG° + RT ln QQ vs K4 Reactions
Topic 9.5

Free Energy and Equilibrium

Explain whether a process is thermodynamically favored using the relationships between K, ΔG°, and T.

"Thermodynamically favored" (ΔG° < 0) means that the products are favored at equilibrium (K > 1) under standard conditions. Free energy and the equilibrium constant are two descriptions of the same fact.

The relationships, with the second on the equation sheet:

K = e−ΔG°/RT and ΔG° = −RT ln K

EK 9.5.A.3 emphasizes qualitative estimation: when ΔG° is near zero, K is close to 1. When ΔG° is much larger or much smaller than RT, K deviates strongly from 1. At 298 K, RT ≈ 2.5 kJ/mol — a useful benchmark.

EK 9.5.A.4 summarizes: processes with ΔG° < 0 favor products (K > 1); those with ΔG° > 0 favor reactants (K < 1).

ΔG°KAt equilibrium
< 0> 1products favored
= 0= 1comparable amounts
> 0< 1reactants favored

ΔG versus ΔG° — a distinction worth getting straight:

  • ΔG° is a fixed property of the reaction at a given temperature, defined for standard conditions (Q = 1).
  • ΔG is the driving force under the current conditions. It changes as the reaction proceeds and reaches zero at equilibrium.

A reaction with ΔG° > 0 can still proceed forward if Q is small enough — it simply will not get far.

Units: use R = 8.314 J·mol⁻¹·K⁻¹ with ΔG° in joules, or divide by 1000 for kJ.

Key points

  • ΔG° < 0 ⇔ K > 1 ⇔ products favored. The three statements are equivalent.
  • ΔG° = 0 corresponds to K = 1, not to equilibrium in general.
  • ΔG (not ΔG°) equals zero at equilibrium.
  • RT ≈ 2.5 kJ/mol at 298 K — the scale against which to judge whether K is near 1.

Equations

  • on the exam sheet
    • 8.314 J mol⁻¹ K⁻¹
    • temperature in kelvin
    • equilibrium constant
  • not on the sheetThe same relationship solved for K; cited in EK 9.5.A.2.

Common mistakes

  • Confusing ΔG° with ΔG. ΔG° is fixed; ΔG goes to zero at equilibrium.
  • Unit mismatch with R. 8.314 J·mol⁻¹·K⁻¹ requires ΔG° in joules.
  • Using log instead of ln.
  • Thinking ΔG° = 0 means the reaction is at equilibrium. It means K = 1.

Worked example

At 298 K a reaction has ΔG° = −28.5 kJ/mol. (a) Calculate K. (b) A second reaction has K = 4.5 × 10⁻⁵. Find its ΔG°. (c) Which reaction is more product-favored?

(a) K from ΔG°.
ΔG° = −RT ln K → ln K = −ΔG°/(RT)
ln K = −(−28 500 J/mol) / [(8.314)(298)] = 28 500 / 2477.6 = 11.50
K = e11.50 = 9.9 × 10⁴

K is much greater than 1, consistent with a negative ΔG°.

(b) ΔG° from K.
ΔG° = −(8.314)(298) ln(4.5 × 10⁻⁵)
ln(4.5 × 10⁻⁵) = −10.01
ΔG° = −(2477.6)(−10.01) = 24 800 J/mol = +24.8 kJ/mol

Positive, consistent with K < 1.

(c) The first reaction is far more product-favored: K ≈ 10⁵ versus 10⁻⁵, a difference of ten orders of magnitude. Note that this says nothing about which reaction is faster — that is a kinetics question (9.4).

Full notes for topic 9.5 →