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The ICE table: initial amounts, changes in the ratio of the coefficients, and equilibrium amounts that can go into the constant.
Paper packet. Every task here also exists on screen, where it is checked automatically; answers written on paper are not assessed by Nydus. When you are back at a device, enter your answers there.
By the end of this lesson you will be able to set up an ICE table for a reaction, fill its Change row from one measured quantity and the coefficients, complete the Equilibrium row, convert amounts to concentrations and calculate $K_c$. You will also be able to work backwards from $K_c$ and the starting amounts to the equilibrium mixture when the expression is a perfect square.
In the last lesson you wrote the expression for $K_c$ and put equilibrium concentrations into it. That lesson handed you every concentration. Real measurements rarely do: a chemist knows what was weighed into the flask and can usually measure one substance at the end — by its colour, by titration, by the pressure. This lesson is the bookkeeping that gets from those two facts to all the others, and it uses nothing but the mole ratios you met in Chemistry 1.
An ICE table has three rows — Initial, Change and Equilibrium — and one column for each substance in the equation. The Initial row is what was put in before any reaction. The Change row is how much of each substance was used up (negative) or formed (positive) on the way to equilibrium. The Equilibrium row is initial plus change. The unknown change is often called $x$, and every other change is a multiple of it fixed by the coefficients.
Take the Haber process, $\mathrm{N_2 + 3H_2 \rightleftharpoons 2NH_3}$, starting with 2.0 mol of nitrogen, 6.0 mol of hydrogen and no ammonia. At equilibrium 1.2 mol of ammonia is measured.
| $\mathrm{N_2}$ | $\mathrm{H_2}$ | $\mathrm{NH_3}$ | |
|---|---|---|---|
| Initial | 2.0 | 6.0 | 0 |
| Change | $-0.6$ | $-1.8$ | $+1.2$ |
| Equilibrium | 1.4 | 4.2 | 1.2 |
The ammonia column is the only one with both ends known, so its change, $+1.2$, is read straight off. The equation then dictates the rest. Two moles of ammonia need one mole of nitrogen, so 1.2 mol of ammonia used 0.6 mol of nitrogen; the same two moles need three of hydrogen, so 1.8 mol of hydrogen went. Reactants go down, products go up, and the changes are always in the ratio of the coefficients — $0.6 : 1.8 : 1.2$ is $1 : 3 : 2$.
The Initial and Equilibrium rows are not in that ratio, and never need to be. What went in is the chemist's choice; only what changed is governed by the equation.
The table works in moles. Before any number goes into $K_c$ it must be divided by the volume to become a concentration — in a 2.0 L vessel the ammonia above is 0.60 mol/L, not 1.2.
Another way: picture
Think of a shop's stock sheet. Opening stock is the Initial row; sales and deliveries are the Change row; closing stock is the Equilibrium row. A recipe links the sales — every cake sold uses two eggs and one bag of flour — so knowing how many cakes went tells you how many eggs went. Nobody adds opening stock to closing stock, and nobody assumes the eggs went down by the same number as the cakes.
Another way: steps
To fill an ICE table:
Sometimes nothing is measured at equilibrium: the question gives the starting amounts and $K_c$, and asks what the mixture ends up as. The Change row is then written with an unknown.
For $\mathrm{H_2 + I_2 \rightleftharpoons 2HI}$ starting with 0.50 mol of each reactant and no product, at a temperature where $K_c = 64$:
| $\mathrm{H_2}$ | $\mathrm{I_2}$ | $\mathrm{HI}$ | |
|---|---|---|---|
| Initial | 0.50 | 0.50 | 0 |
| Change | $-x$ | $-x$ | $+2x$ |
| Equilibrium | $0.50 - x$ | $0.50 - x$ | $2x$ |
$$K_c = \frac{(2x)^2}{(0.50 - x)^2} = 64$$
The volume has cancelled — two moles of gas on each side — and both top and bottom are perfect squares. Taking the square root of both sides leaves $2x \div (0.50 - x) = 8$, so $2x = 4 - 8x$, $x = 0.40$ mol, and there are 0.80 mol of hydrogen iodide. A check: $0.80^2 \div 0.10^2 = 64$.
Not every case is this tidy. When the starting amounts differ, the expression is a quadratic, and the answer that makes any equilibrium amount negative is thrown away. The items in this lesson are chosen so the square root does the work.
The ICE table looks like an extra step, and it is there because the tempting shortcuts are all wrong in ways that produce believable numbers.
If you put initial amounts into $K_c$ you have calculated a reaction quotient for a mixture that has not reacted yet — a real number, and the wrong one. If you put amounts rather than concentrations in, every term is off by the volume, and because the terms carry different powers the errors do not cancel unless the moles of gas on the two sides happen to match. If you copy one change into every column, the equilibrium row describes a reaction that does not conserve atoms.
Each of those gives a number with the right look. Only the table, filled row by row, guarantees that the number describes the mixture actually in the flask.
Adding initial and equilibrium amounts. They are two readings of the same column at two times, not two quantities to be combined. The link between them is the change.
One change for every column. A mole of ammonia forming does not mean a mole of hydrogen used; the coefficients say one and a half. The Change row is in the ratio of the coefficients and nothing else is.
Forgetting that reactants go down. A reactant's change is negative. A table whose equilibrium hydrogen is larger than its initial hydrogen, with no hydrogen made anywhere, has a sign wrong.
Moles into the expression. $K_c$ is written in concentrations. Divide by the volume first — it matters most exactly when the two sides of the equation have different numbers of moles.
Expecting the Initial row to match the equation. Nobody has to start with nitrogen and hydrogen in a 1 : 3 ratio. Only the changes are bound to the coefficients.
$\mathrm{2SO_2 + O_2 \rightleftharpoons 2SO_3}$ starts with 5.0 mol of sulfur dioxide and 3.0 mol of oxygen in a 2.0 L vessel. At equilibrium there are 4.0 mol of sulfur trioxide.
Sulfur trioxide is the column with both ends known: 0 to 4.0 mol.
Changes: sulfur trioxide $+4.0$, sulfur dioxide $-4.0$, oxygen $-2.0$ mol.
Coefficients $2 : 1 : 2$, and the reactants go down.
Equilibrium: sulfur dioxide 1.0 mol, oxygen 1.0 mol, sulfur trioxide 4.0 mol; as concentrations 0.50, 0.50 and 2.0 mol/L.
Initial plus change, then divide by 2.0 L.
$K_c = 2.0^2 \div (0.50^2 \times 0.50) = 4.0 \div 0.125 = 32$.
Concentrations only, each raised to its coefficient.
Ethanoic acid and ethanol, 0.90 mol of each, make ethyl ethanoate and water with $K_c = 4$.
Equal starting amounts and a perfect-square constant: the root will do the work.
Equilibrium row: acid and ethanol $0.90 - x$, ester and water $x$. So $x^2 \div (0.90 - x)^2 = 4$.
Two moles on each side, so the volume cancels.
Square roots: $x \div (0.90 - x) = 2$, so $x = 1.8 - 2x$ and $x = 0.60$ mol of ester.
A straight line in $x$, solved in one step.
Ammonia changed by $+2.0$ mol, so nitrogen changed by $-1.0$ mol.
Two ammonia for every one nitrogen.
Hydrogen changed by three times the nitrogen change, which is $\ldots$
…$-3.0$ mol, so the hydrogen left is $9.0 - 3.0 = 6.0$ mol at equilibrium.
Sulfur dioxide and oxygen are sealed in a vessel with a catalyst: $\mathrm{2SO_2(g) + O_2(g) \rightleftharpoons 2SO_3(g)}$. It starts with $8$ mol of sulfur dioxide, $3$ mol of oxygen and no sulfur trioxide. At equilibrium there are $4$ mol of sulfur trioxide. Complete the Change and Equilibrium rows, in mol.
| sulfur dioxide | oxygen | sulfur trioxide | |
|---|---|---|---|
| Initial, in mol | 8 | 3 | 0 |
| Change, in mol | |||
| Equilibrium, in mol | 4 |
$0.4$ mol of dinitrogen tetroxide is sealed in a $2.0$ L flask and allowed to reach equilibrium: $\mathrm{N_2O_4(g) \rightleftharpoons 2NO_2(g)}$. At equilibrium the flask holds $0.4$ mol of nitrogen dioxide. What is $K_c$ at this temperature?
Answer: mol/L
Hydrogen and iodine vapour react in a sealed flask: $\mathrm{H_2(g) + I_2(g) \rightleftharpoons 2HI(g)}$. The flask starts with $0.4$ mol of each and no hydrogen iodide, at a temperature where $K_c = 4$. How much hydrogen iodide is there at equilibrium?
Answer: unit: mol / mmol
In a Haber process vessel, $\mathrm{N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g)}$ starts with $6$ mol of nitrogen, $17$ mol of hydrogen and no ammonia. At equilibrium there are $4$ mol of ammonia. How much hydrogen is left?
A chemist knows what went into a flask and has measured one substance at equilibrium. Put the steps for finding $K_c$ in order.
Number the steps in order (write the number in the box):
Dinitrogen tetroxide at a starting concentration of $0.3$ mol/L comes to equilibrium: $\mathrm{N_2O_4(g) \rightleftharpoons 2NO_2(g)}$. Let $x$ be the concentration of dinitrogen tetroxide that reacts. Write $K_c$ as an expression in $x$.
Answer:
A flavour chemist makes ethyl ethanoate, the pear-drop smell, by mixing $0.9$ mol of ethanoic acid with $0.9$ mol of ethanol and a little acid catalyst: $\mathrm{CH_3COOH + C_2H_5OH \rightleftharpoons CH_3COOC_2H_5 + H_2O}$. At this temperature $K_c = 4$. How many moles of ethyl ethanoate are there at equilibrium?
Answer: mol
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
A $1.0$ L vessel starts with $0.7$ mol of sulfur dioxide and $0.4$ mol of oxygen: $\mathrm{2SO_2(g) + O_2(g) \rightleftharpoons 2SO_3(g)}$. At equilibrium it holds $0.6$ mol of sulfur trioxide. What is $K_c$?
Answer:
You can keep initial, change and equilibrium amounts apart and let the coefficients fix the changes. Say out loud why, when ammonia goes up by two moles, hydrogen goes down by three. Next: disturbing an equilibrium and predicting which way it moves.
10. Your turn: $\mathrm{N_2 + 3H_2 \rightleftharpoons 2NH_3}$ starts with 3.0 mol of nitrogen and 9.0 mol of hydrogen, and 2.0 mol of ammonia is measured at equilibrium. How much hydrogen is left?, step 3