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Chemistry · Intermediate · 35 min · Chapter 25 of 28

Equilibrium, energy, and rates

The equilibrium constant, the free-energy test for spontaneity, and the rate laws that say how fast you get there.

Read first: Amounts, stoichiometry, and gases, Functions, exponentials, and logarithms

The idea

A reaction at equilibrium has not stopped. The forward and reverse rates are equal, so concentrations stay constant. The equilibrium constant K expresses that condition. For aA + bB ⇌ cC + dD, K is [C]ᶜ[D]ᵈ / [A]ᵃ[B]ᵇ, with each concentration raised to its coefficient. Pure solids and the solvent in a dilute solution are omitted, because their activities are essentially 1. K depends on temperature and not, directly, on how you mix the starting materials. The reaction quotient Q has the same form using the current concentrations. If Q < K the reaction runs forward; if Q > K, backward.

Enthalpy H is the heat of reaction at constant pressure. Entropy S measures the number of ways the energy can be arranged. The quantity that actually decides spontaneity at constant temperature and pressure is the Gibbs free energy, G = H − TS, and the change ΔG = ΔH − TΔS. Negative ΔG means the forward direction increases the entropy of the universe, which is the second law restated in a convenient potential. The standard free-energy change is tied to the equilibrium constant by ΔG° = −RT ln K. A reaction can be spontaneous and still immeasurably slow. Thermodynamics does not mention time.

Kinetics does. A rate law such as rate = k[A]ᵐ[B]ⁿ is determined by experiment. The exponents are the orders, and their sum is the overall order. For a first-order process, concentration decays as e^{−kt} and the half-life is ln 2 / k, independent of how much you started with. Temperature dependence of k is summarized by the Arrhenius equation: a plot of ln k against 1/T is a straight line whose slope is −Eₐ/R. A catalyst lowers the activation energy and is not consumed. It does not change K.

Hold onto this

  1. Equilibrium is a balance of rates, not a situation in which nothing happens.
  2. K is a ratio of concentrations (or pressures) raised to stoichiometric powers. Large K favors products.
  3. ΔG < 0 means the forward direction is spontaneous as written. ΔG° = −RT ln K connects that to the equilibrium constant.
  4. A rate law is experimental. The orders are not the stoichiometric coefficients, unless you have shown they are.

Formulas

Equilibrium constant

For aA + bB ⇌ cC + dD. Concentrations at equilibrium, each to the power of its coefficient.

[ ]
Concentration, or partial pressure in Kₚ

Gibbs free energy

The test for spontaneity at constant T and P. Negative means spontaneous in the forward direction.

ΔH
Enthalpy change
ΔS
Entropy change of the system
T
Absolute temperature

Free energy and K

The standard free-energy change fixes the equilibrium constant, and conversely.

ΔG°
Standard free-energy change
K
The thermodynamic equilibrium constant

First-order decay

Concentration falls exponentially. The half-life does not depend on the starting amount.

k
Rate constant
t₁/₂
Half-life

Arrhenius equation

Rate constants grow with temperature. Eₐ is the activation energy.

A
Pre-exponential factor
Eₐ
Activation energy

Worked examples

1. Reading Q against K

For H₂ + I₂ ⇌ 2 HI, K = 50 at the temperature of interest. You mix equal concentrations [H₂] = [I₂] = [HI] = 0.10 M. Which way does the reaction go?

  1. Q = [HI]² / ([H₂][I₂]) = (0.10)² / (0.10 × 0.10) = 1.
  2. Q = 1 and K = 50, so Q < K.
  3. The reaction runs forward, making more HI, until Q rises to 50.

Forward. Products are under-represented relative to equilibrium.

2. A half-life

A first-order decomposition has k = 0.020 per minute. What fraction remains after 1.0 hour?

  1. t = 60 min. kt = 0.020 × 60 = 1.20.
  2. Fraction left: e^{−1.20} ≈ 0.301.
  3. Half-life is 0.693/0.020 = 34.7 min, so 60 min is a bit under two half-lives, and (1/2)² = 0.25, close to 0.30.

About 30% remains.

Try these

ΔH = −40 kJ/mol and ΔS = −80 J/(mol·K). At what temperature does ΔG change sign?

If K = 1, what is ΔG°?

Common mistakes

  • K has no unit in the thermodynamic definition because each concentration is divided by a standard concentration. In beginner calculations the units are often carried and then quietly dropped. Be consistent.
  • A large K does not mean a fast reaction. Diamond turning into graphite is spontaneous and extraordinarily slow.
  • Orders in a rate law are not the coefficients in the balanced equation, unless the reaction is elementary and you have said so.