Chemical Equilibrium

A reversible reaction can proceed in both the forward and reverse directions. When the rates of the forward and reverse reactions become equal, the system reaches chemical equilibrium — concentrations of reactants and products remain constant over time even though both reactions continue at the molecular level. This is why equilibrium is described as dynamic, not static.

Consider N₂O₄(g) ⇌ 2 NO₂(g). Initially only N₂O₄ is present, so the forward rate is nonzero and the reverse rate is zero. As NO₂ accumulates, the reverse rate increases while the forward rate decreases until they match. At that point, the composition no longer changes and the system is at equilibrium.

Equilibrium can be reached from either direction — starting with all reactants, all products, or any mixture — and the same equilibrium composition results at a given temperature.

The equilibrium constant expression for a general reaction aA + bB ⇌ cC + dD is written as:

K_c = [C]^c[D]^d / [A]^a[B]^b

where brackets denote molar concentrations at equilibrium, and the exponents are the stoichiometric coefficients. Key rules for writing K expressions:

• Only aqueous and gaseous species appear in the expression.
• Pure solids and pure liquids are omitted because their effective concentrations (activities) equal 1.
• Coefficients become exponents — doubling an equation squares K.
• Reversing a reaction inverts K: K_reverse = 1/K_forward.

For gas-phase equilibria, K_p uses partial pressures: K_p = (P_C)^c(P_D)^d / (P_A)^a(P_B)^b. The two constants are related by K_p = K_c(RT)^Δn, where Δn equals moles of gaseous products minus moles of gaseous reactants, R = 0.08206 L·atm/(mol·K), and T is in kelvins.

The numerical value of the equilibrium constant tells you the relative amounts of products and reactants at equilibrium:

• K ≫ 1 — Products are favored. Most of the reactant has been converted to product by the time equilibrium is established.
• K ≪ 1 — Reactants are favored. Only a small fraction converts to product.
• K ≈ 1 — Appreciable amounts of both reactants and products coexist.

In this course the working thresholds are: K > 10³ → favors products; K < 10⁻³ → favors reactants; anything between 10⁻³ and 10³ counts as comparable amounts — even K = 160 sits inside the comparable window (leaning products).

Importantly, K says nothing about how fast equilibrium is reached. A reaction can have a very large K yet proceed extremely slowly without a catalyst. K depends only on temperature; changing concentration, pressure, or adding a catalyst does not alter its value.

When combining equations, K values combine algebraically: adding two reactions multiplies their K values, and multiplying an equation’s coefficients by a factor n raises K to the nth power.

An ICE table (Initial, Change, Equilibrium) organises the algebra needed to solve for unknown equilibrium concentrations:

1. Write the balanced equation and the K_c expression.
2. I — Enter the initial concentrations of every species.
3. C — Define changes in terms of a variable x using stoichiometric ratios. Reactants that are consumed change by −x (or −2x, etc.); products formed change by +x (or +2x, etc.).
4. E — Write each equilibrium concentration as Initial + Change.
5. Substitute the equilibrium expressions into K_c and solve for x.

When K is very small, the small-x approximation simplifies the math: assume x is negligible compared to the initial concentration. After solving, verify that x is less than 5% of the initial value. If it exceeds 5%, discard the approximation and solve the full quadratic (or cubic) equation. Always substitute your answer back into the K expression to confirm the result.

The reaction quotient (Q) has exactly the same mathematical form as K, but it uses current (non-equilibrium) concentrations or pressures. Comparing Q to K predicts which direction a net reaction will occur:

• Q < K — The ratio of products to reactants is too small. The reaction proceeds forward (toward products) to reach equilibrium.
• Q > K — The product-to-reactant ratio is too large. The reaction proceeds in reverse (toward reactants).
• Q = K — The system is already at equilibrium and no net change occurs.

This comparison is a powerful diagnostic tool. You can use it to predict whether a precipitate will form (Q vs. K_sp), whether a buffer will shift, or whether an industrial reactor has reached steady state. When Q and K are far apart, the shift is large; when they are close, only a small adjustment is needed.

Le Châtelier’s principle states that if a system at equilibrium is subjected to a stress, it will shift in the direction that partially counteracts that stress:

• Concentration change — Adding a reactant (or removing a product) increases the forward rate and shifts the equilibrium toward products. Removing a reactant (or adding a product) shifts it toward reactants.
• Volume/pressure change (gases only) — Decreasing volume increases pressure and shifts toward the side with fewer moles of gas. If both sides have equal moles of gas, no shift occurs.
• Temperature change — Treat ΔH as a “reactant” (endothermic) or “product” (exothermic). Raising temperature shifts the equilibrium in the endothermic direction and changes the value of K. This is the only common stress that alters K.
• Catalyst — Speeds both forward and reverse reactions equally. Equilibrium is reached faster, but the position (and K) are unchanged.

The industrial synthesis of ammonia illustrates how equilibrium principles guide real-world chemistry. The reaction N₂(g) + 3 H₂(g) ⇌ 2 NH₃(g) is exothermic (ΔH < 0), so equilibrium is strongly product-favored at low temperature (large K_p), while K_p decreases as temperature increases.

To maximise yield, the Haber–Bosch process manipulates every Le Châtelier lever:

• High pressure (~150–250 atm) — Four moles of gas on the left versus two on the right, so high pressure favours products.
• Moderate temperature (~400–500 °C) — Lower temperature would increase K (exothermic), but the rate becomes impractically slow. A compromise temperature is chosen alongside an iron catalyst to achieve acceptable rates.
• Product removal — NH₃ is continuously condensed and removed, keeping Q below K and driving the reaction forward.

This case shows that optimising a chemical process often involves balancing thermodynamic favourability (K) against kinetic feasibility (rate), a theme that recurs throughout chemistry.

Use this decision framework for equilibrium problems:

1. Write the balanced equation and the K expression. Include only aqueous and gaseous species; pure solids and liquids do not appear in the expression.
2. Determine what you know: K value, initial concentrations, or equilibrium concentrations?
3. If you need equilibrium concentrations: set up an ICE table. Define x as the change, express all equilibrium concentrations in terms of x, and substitute into the K expression.
4. If you need to predict direction: calculate Q from current concentrations. If Q < K, the reaction proceeds forward. If Q > K, it proceeds in reverse. If Q = K, the system is at equilibrium.
5. For Le Châtelier perturbations: identify the stress (concentration change, pressure change, or temperature change) and predict which direction offsets it. Remember: only temperature changes alter the value of K.

Common mistakes: including pure solids or liquids in the equilibrium expression, confusing Q (any point) with K (equilibrium only), assuming that adding a catalyst shifts equilibrium (it does not—it only speeds attainment of equilibrium), misidentifying the exothermic or endothermic direction when predicting the temperature effect, and forgetting that pressure changes only affect equilibria involving gases with unequal moles on each side.