Lesson 3 · 25 min

Equilibrium calculations: ICE tables

Initial, change, equilibrium: three rows turn any equilibrium problem into one equation in x. See how exact root finding solves it — and when the shortcut fails.

In this lesson

Set up an equation in the extent of reaction x with Initial, Change and Equilibrium rows
Solve quadratic and perfect-square problems, and check the answer
Decide whether the “x is small” shortcut is safe (the 5 % rule)
See why higher-order equations need numerical root finding, and why the root is unique

In 1862 Berthelot and Péan de Saint-Gilles mixed equal amounts of acetic acid and ethanol and found that the reaction always stopped at about two-thirds: however long they waited, there was no more ethyl acetate — and starting from ethyl acetate and water, the reaction stopped at the same point. It was one of the first reversible reactions to be studied quantitatively, with an equilibrium constant of about 4. Once K is known, the end point of any starting mixture can be calculated. The tool is the ICE table.

Predict first

Mix 1.00 mol of acetic acid with 1.00 mol of ethanol (K = 4.0; the volume cancels). How much ethyl acetate is there at equilibrium?

Pick an answer and lock it in before you look. A wrong guess teaches more than no guess.

1.00 mol (complete) 0.80 mol 0.67 mol 0.50 mol

Three rows, one unknown

I (Initial): write the starting concentration (or partial pressure) of each species.
C (Change): let the extent of reaction be x. Each change is coefficient × x — negative for reactants, positive for products.
E (Equilibrium): equilibrium = initial + change.
Substitute the E row into the K expression to get an equation in x alone; solve for x and then every equilibrium concentration.
Check: the Q you compute must equal K, and no concentration may be negative.

c_i = c_i^{\,0} + \nu_i x,\qquad K = \prod_i \left(c_i^{\,0} + \nu_i x\right)^{\nu_i}

One equation covers every ICE table. Within the allowed range (every cᵢ ≥ 0) the right-hand side rises steadily with x, so there is exactly one root.

$c_i^{\,0}$: initial concentration of species i
$\nu_i$: signed coefficient: negative for reactants, positive for products
$x$: extent of reaction (mol/L); x < 0 means the reaction ran backwards

Perfect squares: when both sides are squares you can skip the quadratic. At 430 °C for $\ce{H2} + \ce{I2} \mathrel{\ce{<=>}} \ce{2HI}$ ( $K_c = 54.3$ ), starting from 0.500 M each of $\ce{H2}$ and $\ce{I2}$ :

\frac{(2x)^2}{(0.500-x)^2} = 54.3\ \Rightarrow\ \frac{2x}{0.500-x} = 7.37\ \Rightarrow\ x = 0.393

so $[\ce{HI}] = 0.786$ M and $[\ce{H2}] = [\ce{I2}] = 0.107$ M. Unequal starting amounts give a quadratic; a reaction like $\ce{N2} + \ce{3H2} \mathrel{\ce{<=>}} \ce{2NH3}$ gives a quartic that has to be solved numerically — the solver above uses Brent’s method, guaranteed to find the single root in the allowed range.

Worked example

Dissociation of N₂O₄: is the shortcut safe?

At 25 °C, $K_c = 5.99\times10^{-3}$ for $\ce{N2O4} \mathrel{\ce{<=>}} \ce{2NO2}$ . Starting from $[\ce{N2O4}] = 0.100$ M, find the equilibrium concentrations.

Initial · Change · Equilibrium → exact root

ICE-table solver

Reaction

\ce{N2O4} \mathrel{\ce{<=>}} \ce{2NO2}

K = 0.00599 · Kc at 25 °C (from Kp = 0.148)

	$\ce{N2O4}$	$\ce{2NO2}$
I initial / M

① Write the initial concentrations.

Change the starting N₂O₄ to 1.00 M or 0.010 M: the lower the concentration, the larger the fraction that dissociates and the worse the shortcut. Then try N₂ + 3H₂ ⇌ 2NH₃ — a quartic, and the solver still finds the root at once.

Concept check

For $\ce{2SO2} + \ce{O2} \mathrel{\ce{<=>}} \ce{2SO3}$ starting from $\ce{SO2}$ and $\ce{O2}$ , the C row is:

−x, −x, +x −2x, −x, +2x −2x, −2x, +2x +2x, +x, −2x

Try the calculation

At 430 °C, 0.200 mol of $\ce{H2}$ and 0.200 mol of $\ce{I2}$ are placed in a 1.00 L vessel ( $K_c = 54.3$ ). What is $[\ce{HI}]$ at equilibrium?

An excess of one reagent raises the conversion: double the ethanol (1.00 mol acid + 2.00 mol alcohol) and with the same K = 4, $\dfrac{x^2}{(1-x)(2-x)} = 4$ gives $x = 0.845$ — the conversion of acid rises from 67 % to 85 %. Chemists use a cheap reagent in excess to drive a valuable one further for exactly this reason. It is also a first example of Le Chatelier’s principle, the subject of the next lesson.

Key takeaways

ICE tables: initial + change (coefficient × x) = equilibrium; substitute into K for an equation in x.
The equation has exactly one root where all concentrations are non-negative; higher-order cases are solved numerically.
For small K try the “x is small” shortcut, but always test it with the 5 % rule.
The last step is always to check that Q = K.

Progress is stored only in this browser and never uploaded.

Data and references

M. Berthelot, L. Péan de Saint-Gilles, “Recherches sur les affinités”, Annales de Chimie et de Physique (1862–1863).
P. Atkins, J. de Paula, J. Keeler, Atkins’ Physical Chemistry, 11th ed. (2018), Topic 6A.
D. D. Wagman et al., J. Phys. Chem. Ref. Data 11, Suppl. 2 (1982) (N₂O₄ ⇌ 2NO₂).