10.2 Equilibrium constants and rate constants

Equilibrium Constants and Rate Constants

Equilibrium constants and rate constants are two ways of describing the same reaction from different angles. The equilibrium constant tells you where a reaction ends up, while rate constants tell you how fast it gets there. The powerful connection between them is that one can be derived directly from the other.

Equilibrium and Rate Constants Defined

The equilibrium constant ($K$) is the ratio of product concentrations to reactant concentrations at equilibrium, each raised to their stoichiometric coefficients. It tells you the extent to which a reversible reaction proceeds. A key feature of $K$: it depends only on temperature, not on the initial concentrations you start with.

The rate constant ($k$) quantifies the speed of a chemical reaction. It determines how quickly reactants convert into products (or vice versa) for a given elementary step. Unlike $K$, the rate constant depends on temperature, activation energy, and the presence of catalysts. Each elementary step in a mechanism has its own rate constant.

Relationship Between Constants

This is the core idea of the section: you can connect $K$ to the forward and reverse rate constants. Here's how the derivation works for a general reversible reaction:

$aA + bB \rightleftharpoons cC + dD$

1. Write the rate law for the forward reaction: $\text{Rate}_f = k_f[A]^a[B]^b$
2. Write the rate law for the reverse reaction: $\text{Rate}_r = k_r[C]^c[D]^d$
3. At equilibrium, the forward and reverse rates are equal, so set them equal: $k_f[A]^a[B]^b = k_r[C]^c[D]^d$
4. Rearrange by dividing both sides by $k_r$ and by $[A]^a[B]^b$:

$K = \frac{[C]^c[D]^d}{[A]^a[B]^b} = \frac{k_f}{k_r}$

This result is important: the equilibrium constant equals the ratio of the forward rate constant to the reverse rate constant. A large $k_f$ relative to $k_r$ means the forward reaction is faster, so products accumulate and $K$ is large. The reverse logic holds too.

Calculations with Constants

Finding $K$ from rate constants:

$K = \frac{k_f}{k_r}$

For example, if $k_f = 2.5 \times 10^{-3}$ M$^{-1}$s$^{-1}$ and $k_r = 5.0 \times 10^{-4}$ s$^{-1}$:

$K = \frac{2.5 \times 10^{-3}}{5.0 \times 10^{-4}} = 5.0 \text{ M}^{-1}$

Notice that $K$ carries units here because $k_f$ and $k_r$ have different units (which happens when the forward and reverse reactions have different molecularities).

Finding a missing rate constant:

• If you know $K$ and $k_f$, solve for the reverse: $k_r = \frac{k_f}{K}$
• If you know $K$ and $k_r$, solve for the forward: $k_f = K \times k_r$

These rearrangements come straight from the same equation, so there's nothing new to memorize.

Predicting Reaction Direction from Constants

The magnitude of $K$ tells you the equilibrium composition:

• $K > 1$: Products are favored at equilibrium. The forward reaction "wins."
• $K < 1$: Reactants are favored at equilibrium. The reverse reaction "wins."
• $K = 1$: Neither side is strongly favored; product and reactant concentrations are comparable.

To predict which way a reaction will shift from a non-equilibrium state, compare the reaction quotient ($Q$) to $K$. The reaction quotient has the same formula as $K$, but you plug in the current concentrations instead of the equilibrium ones.

1. $Q < K$: The system has too few products relative to equilibrium. The reaction proceeds forward.
2. $Q > K$: The system has too many products relative to equilibrium. The reaction proceeds in reverse.
3. $Q = K$: The system is already at equilibrium. No net change occurs.

Think of it this way: the reaction always shifts in whatever direction pushes $Q$ toward $K$.