Dynamic Equilibrium & the Equilibrium Constant

You’ll be able to

Describe chemical equilibrium as a dynamic balance of forward and reverse rates
Write the equilibrium-constant expression (Kc and Kp) for a reaction
Interpret the size of K and use the reaction quotient Q to predict shift direction

Equilibrium is dynamic, not dead

When a reversible reaction reaches equilibrium, the reaction has not stopped. The forward and reverse reactions are both still running — they have simply reached equal rates, so every molecule consumed is replaced just as fast. The result is that the concentrations stay constant even though molecules are constantly interconverting. The double arrow ⇌ is the signal: the reaction runs both ways at once.

The equilibrium constant captures the balance

At a given temperature, the ratio of products to reactants at equilibrium is always the same number, no matter what amounts you started with. That number is the equilibrium constant, Kc — products over reactants, each concentration raised to the power of its coefficient. Pure solids and pure liquids are left out (their concentration is effectively constant), so only gases and aqueous species appear.

Equilibrium-constant expression

For aA + bB ⇌ cC + dD: Kc = [C]ᶜ[D]ᵈ / ([A]ᵃ[B]ᵇ)

Products on top, reactants on the bottom, each raised to its coefficient. Omit any pure solid (s) or pure liquid (l).

Kp for gaseous equilibria

Kp = Kc (RT)^Δn, where Δn = (moles gas product) − (moles gas reactant)

Kp uses partial pressures instead of concentrations. If the number of gas moles is unchanged (Δn = 0), then Kp = Kc.

What the size of K tells you

K is a scorecard for how far a reaction goes. A large K (K ≫ 1) means the top of the fraction dominates — products are strongly favored and the mixture is mostly product at equilibrium. A small K (K ≪ 1) means reactants are favored and little product forms. K near 1 means comparable amounts of both. K says nothing about speed — only about the final balance.

Worked example

For N₂O₄(g) ⇌ 2NO₂(g), an equilibrium mixture has [N₂O₄] = 0.40 M and [NO₂] = 0.20 M. Write Kc and calculate its value.

1.Products over reactants, each raised to its coefficient. NO₂ has coefficient 2, N₂O₄ has coefficient 1.
2.Kc = [NO₂]² / [N₂O₄].
3.Substitute: Kc = (0.20)² / (0.40) = 0.040 / 0.40.
4.Kc = 0.10.

Answer: Kc = [NO₂]² / [N₂O₄] = 0.10. Because K < 1, reactant (N₂O₄) is favored at this temperature.

Checkpoint

Which is the correct Kc expression for N₂(g) + 3H₂(g) ⇌ 2NH₃(g)?

Checkpoint

For H₂(g) + I₂(g) ⇌ 2HI(g), Kc = 64. A mixture is measured and its reaction quotient is Q = 100. Which way does the reaction shift to reach equilibrium?

Reaction quotient Q vs. K

Q < K → shift right (forward); Q = K → at equilibrium; Q > K → shift left (reverse)

Q has the exact same form as K but uses the current (not necessarily equilibrium) concentrations. Comparing Q to K predicts the direction of change.

Tip

Q and K are computed with the identical formula. The only difference is the concentrations you plug in: Q takes whatever you have right now, K takes the values at equilibrium. A reaction always moves in the direction that pushes Q toward K.

On the exam

On the exam, state the reason with the comparison. "Q < K, so the reaction proceeds forward" earns the point; "it goes forward" alone often does not. Also remember: reversing a reaction inverts K (K becomes 1/K), and multiplying an equation through by n raises K to the nth power.

Answer the 2 checkpoints as you read.