Equilibrium and Le Chatelier's principle: Kc, the ICE table solved line by line, Ksp, and which way the system shifts and why

A reaction at equilibrium is still running, forward and back at the same rate, so the concentrations hold steady at values fixed by the equilibrium constant Kc. To find those values you set up an ICE table (initial, change, equilibrium), write Kc in terms of the change x, and solve. Le Chatelier's principle then tells you which way the position moves when you add a substance, change the pressure or change the temperature: toward whatever undoes the change. Below, one equilibrium is solved to the end, the shift rules are given with their reasons, and Ksp is worked once.

Dynamic equilibrium and the Kc expression

Dynamic means both reactions run. At equilibrium the forward reaction has not stopped; it is making products exactly as fast as the reverse reaction is unmaking them, so nothing you can measure changes. Add a drop of dye that only one side carries and it spreads to both sides, because molecules are still crossing.

The equilibrium constant describes where that balance sits. For a reaction at a fixed temperature, Kc is the product concentrations over the reactant concentrations, each raised to the power of its coefficient in the balanced equation, all in mol/L. Solids and pure liquids are left out, because their concentrations do not change.

For H₂(g) + I₂(g) ⇌ 2 HI(g):

Kc = [HI]² ÷ ([H₂][I₂])

A large Kc means the products are favoured at equilibrium; a small one means the reactants are. Kc has a value only at one temperature, and the problem states it.

The common wrong line. Coefficients as multipliers instead of exponents: Kc = 2[HI] ÷ ([H₂][I₂]). The 2 in front of HI becomes a power, because [HI] is multiplied by itself once for each HI the equation makes. The same error returns in the ICE table's change row, where the 2 is a multiplier, and keeping the two uses apart is most of this unit.

In SCH4U this is three ideas of strand E: Dynamic equilibrium and the equilibrium constant, Le Chatelier's principle and Solubility equilibria and Ksp; the SCH4U course map places them.

Worked example: the ICE table, solved to the end

The problem. 1.00 mol of H₂ and 1.00 mol of I₂ are placed in a 1.00 L flask at a temperature where Kc = 50.0 for H₂(g) + I₂(g) ⇌ 2 HI(g). The 50.0 is a given of this problem, not a value to look up. Find every concentration at equilibrium.

The table has three rows and one column per species.

	[H₂]	[I₂]	[HI]
I (initial)	1.00	1.00	0
C (change)	−x	−x	+2x
E (equilibrium)	1.00 − x	1.00 − x	2x

The initial concentrations are 1.00 mol ÷ 1.00 L = 1.00 mol/L each. The change row is the balanced equation written down a column: the reactants lose x each, and HI gains 2x because the equation makes two HI for every H₂ that reacts. The coefficient enters here, as a multiplier.

• Kc = (2x)² ÷ (1.00 − x)² = 50.0 (substituting the E row; the 2 is squared with the x)
• The whole left side is a perfect square, so take the square root of both sides: 2x ÷ (1.00 − x) = 7.071 (√50.0 = 7.071; the positive root, since concentrations are positive)
• 2x = 7.071 − 7.071x (multiplying both sides by 1.00 − x)
• 9.071x = 7.071 (collecting the x terms)
• x = 0.7795 mol/L (dividing)
• [HI] = 2 × 0.7795 = 1.559 → 1.56 mol/L (the E row with x in)
• [H₂] = [I₂] = 1.00 − 0.7795 = 0.2205 → 0.220 mol/L

Three significant figures, from the 1.00 and the 50.0.

Check. 1.559² ÷ 0.2205² = 2.431 ÷ 0.04862 = 50.0. The numbers go back into the correct Kc expression and give the Kc you were given, so the table was right.

The common wrong line. Writing the change in HI as +x instead of +2x. The table then gives Kc = x² ÷ (1 − x)² = 50.0, x = 0.876, and [HI] = 0.876 mol/L, well short of the right 1.56 mol/L. The trap is that a check with the same wrong expression passes: 0.876² ÷ 0.124² ≈ 50 (49.9 after rounding). A check must use the Kc expression built from the balanced equation, not the expression you happened to solve, or it only confirms your own algebra. The second wrong line is squaring the numerator and not the denominator; the exponent on (1.00 − x) comes from the two reactants, one each, multiplied.

When x is small: the approximation and the 5 % rule

When Kc is very small, hardly any reactant converts, so x is tiny next to the initial concentration and (initial − x) can be replaced by (initial). Solve the simpler equation, then check: if x ÷ initial is under 5 %, the approximation stands; if not, go back and solve the quadratic. The weak-acid case on the pH page uses this every time, with Ka in Kc's place, and the sibling at the end of this page shows a case where the approximation fails.

Ksp: the equilibrium of a sparingly soluble salt

A salt that barely dissolves still reaches an equilibrium between the solid and its ions:

AgCl(s) ⇌ Ag⁺(aq) + Cl⁻(aq), with Ksp = [Ag⁺][Cl⁻]

The solid is left out, as every solid is. Ksp, the solubility product, is just Kc for this kind of reaction.

The problem. The Ksp of AgCl at 25 °C is given as 1.6 × 10⁻¹⁰ (the value in the open textbook's solubility-product table; other tables list 1.8 × 10⁻¹⁰, which rounds to the same answer below). Find its molar solubility.

• Let s be the moles of AgCl that dissolve per litre; then [Ag⁺] = s and [Cl⁻] = s (one of each ion per formula unit)
• Ksp = s × s = s² = 1.6 × 10⁻¹⁰ (substituting)
• s = √(1.6 × 10⁻¹⁰) = 1.26 × 10⁻⁵ → 1.3 × 10⁻⁵ mol/L (taking the root; two significant figures, from the Ksp)

For a 1 : 2 salt the coefficient appears twice, once as a multiplier and once as an exponent. PbCl₂(s) ⇌ Pb²⁺(aq) + 2 Cl⁻(aq): [Pb²⁺] = s, [Cl⁻] = 2s, and Ksp = s × (2s)² = 4s³. Given Ksp, s is the cube root of Ksp ÷ 4.

The common wrong line. Ksp = s² for every salt, which is only true for 1 : 1 salts; or s = Ksp, with no root taken at all. A second ion already in the solution lowers the solubility (the common-ion effect), and comparing the ion product of a mixture with Ksp predicts whether a precipitate forms; both are one line each in this course.

Le Chatelier's principle: which way, and why

Le Chatelier's principle: when a system at equilibrium is disturbed, the position of equilibrium shifts in the direction that partly undoes the disturbance. Take N₂(g) + 3 H₂(g) ⇌ 2 NH₃(g), with ΔH negative, so the forward direction is exothermic. Each change, its direction and its reason:

Change	Shift	Why
Add N₂	Right	Uses up some of the added reactant; the ratio of concentrations is below K until it does
Remove NH₃	Right	Makes more of what was removed
Increase the pressure by shrinking the volume	Right	Toward fewer moles of gas: 4 on the left, 2 on the right
Raise the temperature	Left	Toward the endothermic direction, which absorbs the added heat; K itself changes
Add a catalyst	None	Both rates rise equally; equilibrium is reached faster, in the same place
Add an inert gas at constant volume	None	No concentration changes, so nothing is disturbed

Only temperature changes K. Every other change moves the position of equilibrium and leaves K where it was; the system shifts until the concentrations satisfy the same K again. Say "the position of equilibrium shifts" for those, and "K changes" only for temperature.

The common wrong line. "Adding heat is like adding a product." That is the right instinct for an exothermic reaction, where heat is on the product side, and backwards for an endothermic one. Write the heat into the equation first, on the left for endothermic and on the right for exothermic, and then treat it like any other substance.

Equilibrium on one screen

• The Kc expression. Products over reactants, each to the power of its coefficient, solids and liquids left out.
• The ICE table. Three rows; the change row is the balanced equation down a column, with each coefficient as a multiplier on x.
• The perfect-square trick. When both sides of Kc are squares, take the square root and solve a linear equation.
• Ksp. 1 : 1 salt, Ksp = s²; 1 : 2 salt, Ksp = 4s³.
• The shifts. Add a reactant, right; remove a product, right; squeeze, toward fewer gas moles; heat, toward the endothermic direction and K changes; catalyst or inert gas, no shift.

Attempting the problem below before reading its answer is chapter 3 of the free playbook.

Try one cold: an ICE table that needs the quadratic

Cover the page above. PCl₅(g) ⇌ PCl₃(g) + Cl₂(g) has Kc = 0.0410 at the temperature of the problem (a given). The initial [PCl₅] is 0.500 mol/L and nothing else is present. Find every concentration at equilibrium, and say whether the small-x approximation would have been allowed.

Check

• E row: [PCl₅] = 0.500 − x, [PCl₃] = x, [Cl₂] = x.
• Kc = x² ÷ (0.500 − x) = 0.0410. Not a perfect square, so multiply out: x² = 0.0205 − 0.0410x, so x² + 0.0410x − 0.0205 = 0.
• Quadratic formula: x = [−0.0410 + √(0.0410² + 4 × 0.0205)] ÷ 2 = (−0.0410 + √0.08368) ÷ 2 = (−0.0410 + 0.2893) ÷ 2 = 0.124 mol/L. The negative root is discarded.
• [PCl₃] = [Cl₂] = 0.124 mol/L; [PCl₅] = 0.500 − 0.124 = 0.376 mol/L.
• Check: 0.124² ÷ 0.376 = 0.0409, which is 0.0410 to the precision of the rounding.
• The approximation would have given x = √(0.0410 × 0.500) = 0.143, and 0.143 ÷ 0.500 = 29 %, far above 5 %. The quadratic was required.

If your x was 0.143, you skipped the 5 % check. If your [PCl₅] was 0.124, you forgot to subtract x from the initial value in the E row.

Where this sits in the course

These are ideas 16, 17 and 18 of 24 in SCH4U, Dynamic equilibrium and the equilibrium constant, Le Chatelier's principle and Solubility equilibria and Ksp, the first three ideas of strand E. They follow strand D's Activation energy, catalysts and reaction mechanisms, and they build on Rate laws and reaction order, because equilibrium is the state where the forward and reverse rates are equal; the rate laws page is where those rates come from.

The ideas after them are 19 and 20, Acid–base theories, pH and strong/weak acids and Buffers and acid–base titration. Those are this page's method with Ka and Kb in place of Kc, the ICE table and the 5 % rule included, and nothing else new in the arithmetic.

The ICE table itself is the mole ratio from the stoichiometry page written down a column, with x in place of a known amount. Every idea of the three science courses has a page like this one.

How to practise it

Mix the kinds in one set: an ICE table with a perfect square, one without, a Ksp for a 1 : 1 salt and one for a 1 : 2 salt, and three shift questions. Rohrer and Taylor (2007) found that students who practised problem types shuffled together did worse during practice and much better on a test a week later than students who practised each type in its own block, because the shuffled set makes you choose the method, and the choice is what a test asks. Sweller and Cooper (1985) found that a worked example followed by a matching problem teaches a procedure faster than problems alone, which is why each idea here is one example and one sibling.

Keep the error log by row: I, C or E. An I error is a concentration not divided by the volume; a C error is a missing coefficient; an E error is a sign. Mixed practice has the schedule, and how to study chemistry treats this strand as a procedure with a short list of lines.