10.2 Equilibrium constants and rate constants

Chemical reactions are all about balance and speed. Equilibrium constants tell us where reactions end up, while rate constants show how fast they get there. These two constants are closely linked, helping us understand and predict chemical behavior.

By comparing equilibrium and rate constants, we can figure out which direction a reaction will go and how quickly. This knowledge is crucial for controlling reactions in labs and industry, making it a key part of chemical kinetics.

Equilibrium Constants and Rate Constants

Equilibrium and rate constants

• Equilibrium constant ($K$) represents the ratio of product concentrations to reactant concentrations at equilibrium, indicating the extent to which a reversible reaction proceeds at equilibrium and determining the equilibrium composition of a reaction mixture without depending on the initial concentrations of reactants or products
• Rate constant ($k$) quantifies the speed of a chemical reaction, determining how quickly reactants are converted into products, and depends on factors such as temperature, activation energy, and the presence of catalysts, being specific to each elementary step in a complex reaction mechanism

Relationship between constants

• Consider a general reversible reaction: $aA + bB \rightleftharpoons cC + dD$
• The rate law for the forward reaction: $\text{Rate}_f = k_f[A]^a[B]^b$
• The rate law for the reverse reaction: $\text{Rate}_r = k_r[C]^c[D]^d$
• At equilibrium, the forward and reverse rates are equal: $k_f[A]^a[B]^b = k_r[C]^c[D]^d$
• Rearrange the equation to obtain the equilibrium constant expression: $K = \frac{[C]^c[D]^d}{[A]^a[B]^b} = \frac{k_f}{k_r}$, showing that the equilibrium constant is equal to the ratio of the forward and reverse rate constants

Calculations with constants

• Given the rate constants $k_f$ and $k_r$, calculate the equilibrium constant: $K = \frac{k_f}{k_r}$ (example: if $k_f = 2.5 \times 10^{-3}$ M$^{-1}$s$^{-1}$ and $k_r = 5.0 \times 10^{-4}$ s$^{-1}$, then $K = \frac{2.5 \times 10^{-3}}{5.0 \times 10^{-4}} = 5.0$ M$^{-1}$)
• Given the equilibrium constant $K$ and one of the rate constants, calculate the other rate constant:
  1. If $K$ and $k_f$ are known, calculate $k_r$: $k_r = \frac{k_f}{K}$
  2. If $K$ and $k_r$ are known, calculate $k_f$: $k_f = K \times k_r$

Reaction direction from constants

• The magnitude of the equilibrium constant indicates the relative concentrations of products and reactants at equilibrium:
  • If $K > 1$, the equilibrium favors the products (forward reaction is favored)
  • If $K < 1$, the equilibrium favors the reactants (reverse reaction is favored)
  • If $K = 1$, the concentrations of products and reactants are equal at equilibrium
• The direction of a reaction can be predicted by comparing the reaction quotient ($Q$) to the equilibrium constant ($K$):
  1. If $Q < K$, the reaction will proceed in the forward direction to reach equilibrium
  2. If $Q > K$, the reaction will proceed in the reverse direction to reach equilibrium
  3. If $Q = K$, the reaction is at equilibrium, and no net change in concentrations will occur