Chemical equilibrium
The idea
Chemical equilibrium is a dynamic standoff: forward and reverse reactions keep running, but at equal rates, so concentrations stop changing. The law of mass action assigns each reaction a constant, K = products over reactants (each raised to its coefficient, pure solids and liquids omitted), fixed at a given temperature. The reaction quotient Q uses the same expression with current concentrations: Q < K means the system shifts forward, Q > K backward, Q = K means it has arrived. You already know reactions can be reversible; here you make the balance point quantitative.
The standard computational tool is the ICE table — Initial, Change, Equilibrium — which encodes stoichiometry into one unknown x and turns the K expression into algebra. Look for structure before grinding: perfect-square cases collapse under a square root, and small-K systems justify the approximation that x is negligible against initial amounts (check it afterward). Le Chatelier's principle then predicts qualitatively how the position responds to added species, pressure changes, or temperature.
Strike the misconception that equilibrium means equal concentrations of reactants and products, or that the reaction has stopped. K can be enormous or tiny — the balance point can sit far to either side — and molecules continue converting in both directions forever; only the net change is zero.
Worked example
At 448 °C, Kc = 50.5 for H₂(g) + I₂(g) ⇌ 2 HI(g). If 1.00 mol of H₂ and 1.00 mol of I₂ are sealed in a 1.00 L flask, find all three equilibrium concentrations.
- Set up the ICE table with x mol/L of each reactant consumed: [H₂] = [I₂] = 1.00 − x and [HI] = 2x, since each reaction event makes two HI.
- Write the equilibrium condition: Kc = [HI]²/([H₂][I₂]) = (2x)²/(1.00 − x)² = 50.5.
- Notice the right side is a perfect square — take the square root of both sides instead of expanding a quadratic: 2x/(1.00 − x) = √50.5 = 7.11.
- Solve the linear equation: 2x = 7.11 − 7.11x, so 9.11x = 7.11 and x = 0.780.
- Back-substitute: [H₂] = [I₂] = 1.00 − 0.780 = 0.220 M and [HI] = 2 × 0.780 = 1.56 M.
- Verify: (1.56)²/(0.220 × 0.220) = (7.09)² ≈ 50.3, matching Kc within rounding — and with K well above 1 it is sensible that products dominate at equilibrium.
Answer. [HI] ≈ 1.56 M and [H₂] = [I₂] ≈ 0.220 M at equilibrium.
Check your understanding
- Why does a catalyst change how fast equilibrium is reached but not where it lies?
- How do you decide when the small-x approximation is safe, and what do you do when the check fails?
- What happens to Kc itself, versus the equilibrium position, when you compress the flask or when you raise the temperature of an exothermic reaction?
- How would you explain dynamic equilibrium to a friend using a two-way escalator or crowd analogy, and where does the analogy break down?
Build the foundations first
Chemical equilibrium builds on these concepts. If any feel shaky, start there.