How to Convert Moles into Concentration for an ICE Table

Put 0.0200 mol of a gas into a 2.00 L vessel. Another student puts exactly the same amount into a 0.500 L vessel. Would you enter 0.0200 in the initial row of both ICE tables? It might seem reasonable because the number of moles is identical. But the second vessel has four times the concentration.

Before making an ICE table, you need to convert the initial amount of each substance into its concentration. The calculation is simple:

\[
\boxed{c=\frac{n}{V}}
\]

Divide the number of moles by the volume of the actual reaction mixture, measured in litres. Getting this step right gives you the correct starting point for every calculation that follows.

01Converting moles into concentration

Moles tell you how much of a substance is present. Concentration tells you how much of that substance is present per litre of mixture.

Think of putting the same number of students into two classrooms. The smaller classroom has more students per square metre, even though the total number of students is unchanged.

Concentration works in the same way. The volume matters just as much as the number of moles.

The formula is:

\[
c=\frac{n}{V}
\]

where:

• \(c\) = concentration, in mol L^-1
• \(n\) = amount of substance, in mol
• \(V\) = volume of the mixture, in L

For example, suppose a reaction vessel contains 0.0200 mol of a substance in 2.00 L.

\[
c=\frac{0.0200}{2.00}
\]

\[
\boxed{c=0.0100\ \mathrm{mol\,L^{-1}}}
\]

That is the concentration you enter into the initial row of your ICE table.

Check your volume units first

Your volume must be in litres, not millilitres.

To convert millilitres into litres, divide by 1000.

A common mistake is to substitute 100 into the formula when the volume is 100 mL. This makes the calculated concentration 1000 times too small.

What if the question gives you mass instead of moles?

You need one extra step.

First, calculate moles using:

\[
n=\frac{m}{M}
\]

where \(m\) is mass in grams and \(M\) is molar mass in g mol^-1.

Then convert those moles into concentration:

\[
c=\frac{m}{MV}
\]

If you need to revise the first calculation, see Moles from Mass and Molar Mass: HSC Chemistry Guide.

02Which volume should you use?

This is where things become slightly more interesting.

The volume you use depends on how the reaction mixture was prepared.

For a gas in a sealed container: Use the volume of the container.

For substances dissolved in water: Use the final volume of the combined solution.

For mixtures prepared from multiple stock solutions: Calculate the moles delivered by each solution, then divide by the final volume of the mixture.

The important point is that you need the volume occupied by the reaction mixture, not automatically the volume written on the original reagent bottle.

A common experimental situation: mixing two solutions

Imagine an equilibrium investigation involving iron(III) ions and thiocyanate ions:

\[
\ce{Fe^3+(aq) + SCN^-(aq) <=> FeSCN^2+(aq)}
\]

The product, \(\mathrm{FeSCN}^{2+}\), is a red-coloured complex ion.

A typical experimental arrangement uses volumetric pipettes to transfer measured amounts of the two stock solutions into a volumetric flask. Water is added until the final volume reaches the calibration mark. The flask is stoppered and mixed.

A sample can then be transferred into a cuvette, where light passes through the solution to a colorimeter. The instrument measures absorbance. With an appropriate calibration, this can be used to determine the equilibrium concentration of the coloured complex.

What should happen if more thiocyanate ions are added while the other conditions remain constant?

You would expect more of the red complex to form as the equilibrium responds to the increased reactant concentration. Within the calibrated range, the colourimeter should register greater absorbance.

However, the colourimeter does not directly tell you the initial concentrations of the reactants. Those come from the amounts of stock solutions transferred and the final volume.

Consider the following illustrative preparation:

• 25.0 mL of 0.0200 mol L^-1 \(\mathrm{Fe}^{3+}\)
• 15.0 mL of 0.0100 mol L^-1 \(\mathrm{SCN}^{-}\)
• Final mixture volume of 100.0 mL

A tempting mistake is to enter 0.0100 mol L^-1 as the initial concentration of \(\mathrm{SCN}^{-}\), because that is the concentration printed on its stock solution.

That is incorrect. Only 15.0 mL of the stock solution was transferred, and it was diluted to a final mixture volume of 100.0 mL.

You need to track the moles that actually entered the reaction mixture.

For any species supplied by a stock solution:

\[
n=c_{\text{stock}}V_{\text{stock}}
\]

Then:

\[
c_{\text{initial}}=
\frac{c_{\text{stock}}V_{\text{stock}}}
{V_{\text{final}}}
\]

This is the essential calculation when preparing an ICE table from mixed solutions.

03Building the ICE table

Once you have converted every initial amount into concentration, you can construct your ICE table.

ICE stands for:

• I – Initial: The concentrations immediately after the reaction mixture has been prepared, before the equilibrium change is accounted for.
• C – Change: The increase or decrease in concentration as the reaction moves towards equilibrium.
• E – Equilibrium: The final concentrations once equilibrium has been established.

The initial row contains concentrations, not moles.

Understanding the change row

Consider the general reaction:

\[
aA+bB\rightleftharpoons cC+dD
\]

If the reaction proceeds in the forward direction, the reactants are consumed and the products are formed.

The change row is therefore:

The coefficients come directly from the balanced chemical equation.

For example:

\[
\ce{N2O4(g) <=> 2NO2(g)}
\]

If the reaction moves forward, one mole of \(\ce{N2O4}\) is consumed for every two moles of \(\ce{NO2}\) produced.

The change row must therefore contain \(-x\) for \(\ce{N2O4}\) and \(+2x\) for \(\ce{NO2}\).

Here, \(x\) represents a concentration change measured in mol L^-1.

How do you know which direction the reaction moves?

If the question gives you \(K_c\), you can compare it with the initial reaction quotient, \(Q_c\).

The expression for \(Q_c\) has the same form as the expression for \(K_c\), but you substitute the initial concentrations.

For a general reaction:

\[
Q_c=\frac{[C]^c[D]^d}{[A]^a[B]^b}
\]

The comparison tells you the direction of the net reaction:

This helps you assign the signs in the change row before solving for \(x\).

If the question explicitly tells you the direction of the shift, use that information instead.

The equilibrium row

The equilibrium row is simply the initial concentration plus the change.

For example:

All entries are in mol L^-1.

Notice that the initial row contains actual numerical concentrations, while the equilibrium row contains algebraic expressions.

These expressions are what you substitute into the \(K_c\) equation.

06Rejecting nonphysical roots

Quadratic equations can produce two mathematical solutions, but that does not mean both solutions describe a possible chemical equilibrium.

A valid solution must satisfy the physical constraints of the reaction.

Use these checks:

• Every equilibrium concentration must be zero or positive.
• You cannot consume more of a reactant than was initially available.
• The changes must obey the stoichiometric coefficients.
• The resulting concentrations must satisfy the given \(K_c\).

There is also an important distinction when interpreting negative values of \(x\).

If you initially assume a forward reaction and obtain a negative \(x\), it may indicate that the reaction actually proceeds backwards. However, if you have already established the direction using \(Q_c\) and defined \(x\) as a positive change in that direction, a negative root contradicts your chosen direction.

Always check the resulting concentrations rather than blindly accepting every algebraic root.

07What happens when the volume changes?

A volume change creates another situation where you must calculate concentrations before constructing an ICE table.

Suppose a gas mixture is already at equilibrium. Its volume is suddenly doubled.

The number of moles of each gas does not immediately change, but every concentration is halved.

Does this always cause the equilibrium to shift?

No. It depends on the stoichiometry of the reaction.

For a general reaction:

\[
aA+bB\rightleftharpoons cC+dD
\]

define:

\[
\Delta\nu=(c+d)-(a+b)
\]

This is the difference between the total stoichiometric coefficients of products and reactants in the equilibrium expression.

If every concentration is multiplied by the same dilution factor \(f\), then:

\[
Q’_c=f^{\Delta\nu}Q_c
\]

This relationship applies when all species in the equilibrium expression are diluted equally.

Three cases are worth recognising:

This explains why you cannot automatically assume that increasing volume shifts every equilibrium towards the side with more particles.

It also reinforces an important rule: whenever the volume changes, recalculate the concentrations before writing a new initial row.

08Questions and solutions

The following questions progress from straightforward concentration conversions to more demanding equilibrium problems. All numerical data are illustrative.

Question 1

A student places 0.0180 mol of gas A and 0.00600 mol of gas B into a sealed 3.00 L container.

The reaction is:

\[
\ce{A(g) + B(g) <=> C(g)}
\]

Initially, no C is present.

Calculate the initial concentration of each species and construct the ICE table, assuming the reaction initially proceeds forwards.

Solution 1

The initial concentrations are 0.00600 mol L^-1 for A, 0.00200 mol L^-1 for B, and zero for C.

First, convert the moles into concentrations.

For A:

\[
[A]_0=\frac{0.0180}{3.00}
=0.00600\ \mathrm{mol\,L^{-1}}
\]

For B:

\[
[B]_0=\frac{0.00600}{3.00}
=0.00200\ \mathrm{mol\,L^{-1}}
\]

C is initially absent:

\[
[C]_0=0
\]

The balanced equation has a 1:1:1 ratio. Therefore, when the reaction proceeds forwards, A and B decrease by \(x\), while C increases by \(x\).

The completed table is:

All concentrations are in mol L^-1.

Notice that you cannot put 0.0180 and 0.00600 directly into the initial row. Those numbers represent moles, not concentrations.

Question 2

A student prepares a mixture for the following equilibrium:

\[
\ce{2A(aq) + B(aq) <=> C(aq)}
\]

The student combines:

• 20.0 mL of 0.0300 mol L^-1 A
• 30.0 mL of 0.0100 mol L^-1 B
• 10.0 mL of 0.0200 mol L^-1 C

The final mixture volume is 100.0 mL.

At the experimental temperature:

\[
K_c=1.00\times10^4\ \mathrm{L^2\,mol^{-2}}
\]

Determine:

a) The initial concentration of each species.

b) Whether the reaction initially proceeds forwards or backwards.

c) The correct change row.

d) The maximum possible value of \(x\) if it proceeds in the reverse direction.

Solution 2

The initial concentrations are 0.00600 mol L^-1 for A, 0.00300 mol L^-1 for B, and 0.00200 mol L^-1 for C. The reaction initially proceeds in reverse, and its change row is \(+2x,+x,-x\).

Part a: Calculate the initial concentrations.

First, calculate the moles delivered by each stock solution.

For A:

\[
n_A=0.0300\times0.0200
\]

\[
n_A=0.000600\ \mathrm{mol}
\]

For B:

\[
n_B=0.0100\times0.0300
\]

\[
n_B=0.000300\ \mathrm{mol}
\]

For C:

\[
n_C=0.0200\times0.0100
\]

\[
n_C=0.000200\ \mathrm{mol}
\]

The final volume is:

\[
V=0.1000\ \mathrm{L}
\]

Therefore:

\[
[A]_0=\frac{0.000600}{0.1000}
=0.00600\ \mathrm{mol\,L^{-1}}
\]

\[
[B]_0=\frac{0.000300}{0.1000}
=0.00300\ \mathrm{mol\,L^{-1}}
\]

\[
[C]_0=\frac{0.000200}{0.1000}
=0.00200\ \mathrm{mol\,L^{-1}}
\]

Part b: Determine the reaction direction.

The equilibrium expression is:

\[
K_c=\frac{[C]}{[A]^2[B]}
\]

Substitute the initial concentrations to calculate \(Q_c\):

\[
Q_c=
\frac{0.00200}{(0.00600)^2(0.00300)}
\]

\[
Q_c=1.85\times10^4
\]

Compare this with the given equilibrium constant:

\[
Q_c>K_c
\]

Therefore, the reaction initially proceeds backwards, towards A and B.

Part c: Construct the change row.

The forward reaction consumes two units of A and one unit of B to produce one unit of C.

The reverse reaction does the opposite.

Therefore:

The full table is:

Part d: Determine the maximum possible value of \(x\).

C is the only species being consumed in the reverse direction.

Its equilibrium concentration cannot become negative:

\[
0.00200-x\geq0
\]

Therefore:

\[
\boxed{x_{\max}=0.00200\ \mathrm{mol\,L^{-1}}}
\]

This is a useful check before solving the equilibrium expression. Any proposed value greater than this would consume more C than the mixture initially contains.

Question 3

A 500.0 mL reaction vessel initially contains:

• 0.00500 mol of A
• 0.0100 mol of B
• 0.00250 mol of C

The reaction is:

\[
\ce{A(aq) + B(aq) <=> C(aq)}
\]

At constant temperature:

\[
K_c=40.0\ \mathrm{L\,mol^{-1}}
\]

Calculate the equilibrium concentration of each species.

Your solution must show the initial concentration conversion, the ICE table, the quadratic equation, and the rejection of any nonphysical root.

Solution 3

The equilibrium concentrations are:

\[
\boxed{[A]_{\text{eq}}=0.00860\ \mathrm{mol\,L^{-1}}}
\]

\[
\boxed{[B]_{\text{eq}}=0.0186\ \mathrm{mol\,L^{-1}}}
\]

\[
\boxed{[C]_{\text{eq}}=0.00640\ \mathrm{mol\,L^{-1}}}
\]

The valid quadratic root is \(x=0.00140\ \mathrm{mol\,L^{-1}}\).

Step 1: Convert the initial moles into concentrations.

The volume is:

\[
500.0\ \mathrm{mL}=0.5000\ \mathrm{L}
\]

For A:

\[
[A]_0=\frac{0.00500}{0.5000}
=0.0100\ \mathrm{mol\,L^{-1}}
\]

For B:

\[
[B]_0=\frac{0.0100}{0.5000}
=0.0200\ \mathrm{mol\,L^{-1}}
\]

For C:

\[
[C]_0=\frac{0.00250}{0.5000}
=0.00500\ \mathrm{mol\,L^{-1}}
\]

Step 2: Determine the reaction direction.

\[
Q_c=\frac{[C]}{[A][B]}
\]

\[
Q_c=\frac{0.00500}{0.0100\times0.0200}
\]

\[
Q_c=25.0\ \mathrm{L\,mol^{-1}}
\]

Since:

\[
Q_c<K_c
\]

the reaction proceeds forwards.

Step 3: Construct the ICE table.

The reaction has a 1:1:1 stoichiometric ratio.

Step 4: Substitute into the equilibrium expression.

\[
40.0=
\frac{0.00500+x}
{(0.0100-x)(0.0200-x)}
\]

Multiply both sides:

\[
40.0(0.0100-x)(0.0200-x)
=0.00500+x
\]

Expand:

\[
40.0(0.000200-0.0300x+x^2)
=0.00500+x
\]

Collect terms:

\[
40.0x^2-2.20x+0.00300=0
\]

Step 5: Solve the quadratic and select the physical root.

The two solutions are:

\[
x=0.00140
\]

or

\[
x=0.0536
\]

The second root is impossible because it exceeds the initial concentration of A:

\[
0.0536>0.0100
\]

It would produce a negative equilibrium concentration for A.

Therefore:

\[
x=0.00140\ \mathrm{mol\,L^{-1}}
\]

Step 6: Calculate the equilibrium concentrations.

For A:

\[
[A]_{\text{eq}}=0.0100-0.00140
\]

\[
=0.00860\ \mathrm{mol\,L^{-1}}
\]

For B:

\[
[B]_{\text{eq}}=0.0200-0.00140
\]

\[
=0.0186\ \mathrm{mol\,L^{-1}}
\]

For C:

\[
[C]_{\text{eq}}=0.00500+0.00140
\]

\[
=0.00640\ \mathrm{mol\,L^{-1}}
\]

The concentrations are all positive and satisfy the equilibrium expression.

Question 4

A gas reaction is already at equilibrium inside a 2.00 L container.

\[
\ce{P(g) <=> 2Q(g)}
\]

The container initially contains:

• 0.0400 mol of P
• 0.0800 mol of Q

The volume is suddenly doubled to 4.00 L at constant temperature.

Determine:

a) The original equilibrium concentrations.

b) The value of \(K_c\).

c) The immediate concentrations after expansion.

d) The direction of the subsequent equilibrium shift.

e) The change row for the new ICE table.

Solution 4

The reaction shifts forwards after expansion. The immediate concentrations become 0.0100 mol L^-1 for P and 0.0200 mol L^-1 for Q, giving \(Q_c=0.0400\), which is less than \(K_c=0.0800\).

Part a: Calculate the original equilibrium concentrations.

For P:

\[
[P]=\frac{0.0400}{2.00}
\]

\[
[P]=0.0200\ \mathrm{mol\,L^{-1}}
\]

For Q:

\[
[Q]=\frac{0.0800}{2.00}
\]

\[
[Q]=0.0400\ \mathrm{mol\,L^{-1}}
\]

Part b: Calculate \(K_c\).

Because the mixture is initially at equilibrium:

\[
K_c=\frac{[Q]^2}{[P]}
\]

\[
K_c=\frac{(0.0400)^2}{0.0200}
\]

\[
\boxed{K_c=0.0800}
\]

Part c: Calculate the concentrations immediately after expansion.

The volume doubles, but the number of moles of each gas remains unchanged immediately after expansion.

For P:

\[
[P]’=\frac{0.0400}{4.00}
\]

\[
[P]’=0.0100\ \mathrm{mol\,L^{-1}}
\]

For Q:

\[
[Q]’=\frac{0.0800}{4.00}
\]

\[
[Q]’=0.0200\ \mathrm{mol\,L^{-1}}
\]

Both concentrations have halved.

Part d: Determine the direction of the shift.

Calculate the new reaction quotient:

\[
Q_c=\frac{(0.0200)^2}{0.0100}
\]

\[
Q_c=0.0400
\]

Since:

\[
Q_c<K_c
\]

the reaction proceeds forwards.

More Q must form to restore equilibrium.

Part e: Construct the new change row.

For:

\[
\ce{P(g) <=> 2Q(g)}
\]

the forward reaction consumes one unit of P and produces two units of Q.

The new ICE table is:

The important detail is that the initial row for the second equilibrium calculation must use the concentrations after expansion, not the concentrations from the original equilibrium.

Question 5

A gas reaction occurs inside a piston-controlled vessel:

\[
\ce{A(g) + B(g) <=> 2C(g)}
\]

The mixture is initially at equilibrium in a 1.00 L vessel and contains:

• 0.0800 mol of A
• 0.120 mol of B
• 0.160 mol of C

The following changes occur in sequence:

Stage 1: The piston is pulled outwards, increasing the volume from 1.00 L to 2.00 L at constant temperature. The piston is then locked.

Stage 2: A side valve injects an additional 0.0400 mol of C into the vessel. The volume remains fixed at 2.00 L.

Calculate the final equilibrium concentration of all three gases after Stage 2.

Your solution must:

a) Calculate the original value of \(K_c\).

b) Explain whether Stage 1 causes an equilibrium shift.

c) Calculate the concentrations immediately after Stage 2.

d) Determine the direction of the subsequent shift.

e) Construct and solve the new ICE table, rejecting any nonphysical root.

Solution 5

The final equilibrium concentrations are:

\[
\boxed{[A]_{\text{eq}}=0.0455\ \mathrm{mol\,L^{-1}}}
\]

\[
\boxed{[B]_{\text{eq}}=0.0655\ \mathrm{mol\,L^{-1}}}
\]

\[
\boxed{[C]_{\text{eq}}=0.0891\ \mathrm{mol\,L^{-1}}}
\]

The crucial observation is that expanding the vessel in Stage 1 does not shift the equilibrium because the total stoichiometric coefficients are equal on both sides. Adding C in Stage 2 does disturb the equilibrium, causing a reverse shift.

Part a: Calculate the original equilibrium constant.

The original vessel volume is 1.00 L.

Therefore:

\[
[A]_0=\frac{0.0800}{1.00}
=0.0800\ \mathrm{mol\,L^{-1}}
\]

\[
[B]_0=\frac{0.120}{1.00}
=0.120\ \mathrm{mol\,L^{-1}}
\]

\[
[C]_0=\frac{0.160}{1.00}
=0.160\ \mathrm{mol\,L^{-1}}
\]

The equilibrium expression is:

\[
K_c=\frac{[C]^2}{[A][B]}
\]

Substitute:

\[
K_c=\frac{(0.160)^2}{0.0800\times0.120}
\]

\[
\boxed{K_c=2.667}
\]

The stoichiometric coefficients add to two on each side, so the concentration expression has no net concentration units.

Part b: Analyse Stage 1.

The volume increases from 1.00 L to 2.00 L.

Immediately after expansion, the number of moles of each gas is unchanged, but the concentrations halve.

Therefore:

\[
[A]’=0.0400\ \mathrm{mol\,L^{-1}}
\]

\[
[B]’=0.0600\ \mathrm{mol\,L^{-1}}
\]

\[
[C]’=0.0800\ \mathrm{mol\,L^{-1}}
\]

Now calculate the reaction quotient:

\[
Q_c=\frac{(0.0800)^2}{0.0400\times0.0600}
\]

\[
Q_c=\frac{0.00640}{0.00240}
\]

\[
Q_c=2.667
\]

Therefore:

\[
Q_c=K_c
\]

The reaction is still at equilibrium.

Why?

The forward reaction has one mole of A and one mole of B forming two moles of C. There are two stoichiometric units of gas on each side.

When all concentrations are halved, the numerator and denominator of the equilibrium expression change by the same overall factor.

This means the reaction quotient remains unchanged.

A student who automatically predicts a shift towards the side with more gas particles would get this question wrong. There is no side with more gas particles in this reaction.

Part c: Calculate the concentrations immediately after Stage 2.

An additional 0.0400 mol of C is injected.

The vessel volume remains 2.00 L.

The increase in concentration is:

\[
\Delta[C]=\frac{0.0400}{2.00}
\]

\[
\Delta[C]=0.0200\ \mathrm{mol\,L^{-1}}
\]

Therefore:

\[
[C]’=0.0800+0.0200
\]

\[
[C]’=0.1000\ \mathrm{mol\,L^{-1}}
\]

A and B are unchanged during the injection.

The concentrations immediately after Stage 2 are:

Part d: Determine the new direction of the reaction.

Calculate the reaction quotient:

\[
Q_c=\frac{(0.1000)^2}{0.0400\times0.0600}
\]

\[
Q_c=\frac{0.0100}{0.00240}
\]

\[
Q_c=4.167
\]

Since:

\[
Q_c>K_c
\]

the reaction proceeds backwards.

It must consume some of the additional C and regenerate A and B.

Part e: Construct the new ICE table.

The reaction proceeds in reverse:

\[
\ce{2C(g) -> A(g) + B(g)}
\]

Therefore, the change row is:

• A increases by \(x\).
• B increases by \(x\).
• C decreases by \(2x\).

The table becomes:

Substitute into the equilibrium expression:

\[
2.667=
\frac{(0.1000-2x)^2}
{(0.0400+x)(0.0600+x)}
\]

Using the unrounded value \(K_c=8/3\):

\[
\frac{8}{3}=
\frac{(0.1000-2x)^2}
{(0.0400+x)(0.0600+x)}
\]

Cross-multiply:

\[
3(0.1000-2x)^2
=
8(0.0400+x)(0.0600+x)
\]

Expand:

\[
0.0300-1.20x+12x^2
=
0.0192+0.800x+8x^2
\]

Collect terms:

\[
4x^2-2.00x+0.0108=0
\]

Solving gives:

\[
x=0.00546
\]

or

\[
x=0.4945
\]

Reject the nonphysical root.

The reverse reaction consumes two units of C for every unit of \(x\).

Since the initial concentration of C is 0.1000 mol L^-1:

\[
0.1000-2x\geq0
\]

Therefore:

\[
x\leq0.0500
\]

The root \(x=0.4945\) exceeds this limit and would give a negative concentration of C.

It is physically impossible.

The valid root is:

\[
\boxed{x=0.00546\ \mathrm{mol\,L^{-1}}}
\]

Calculate the final equilibrium concentrations.

For A:

\[
[A]_{\text{eq}}=0.0400+0.00546
\]

\[
\boxed{[A]_{\text{eq}}=0.0455\ \mathrm{mol\,L^{-1}}}
\]

For B:

\[
[B]_{\text{eq}}=0.0600+0.00546
\]

\[
\boxed{[B]_{\text{eq}}=0.0655\ \mathrm{mol\,L^{-1}}}
\]

For C:

\[
[C]_{\text{eq}}=0.1000-2(0.00546)
\]

\[
\boxed{[C]_{\text{eq}}=0.0891\ \mathrm{mol\,L^{-1}}}
\]

The final concentrations are consistent with a reverse shift: A and B increase, while C decreases.

The interesting part is the sequence. Expanding the vessel did not disturb the first equilibrium, but adding C did. You had to recalculate the concentrations after each change before constructing the next ICE table.

09The method to remember

Before starting any \(K_c\) ICE table, follow this sequence:

1. Identify the initial amount of every species.
2. Convert masses into moles if necessary.
3. Calculate the moles delivered by any stock solutions.
4. Identify the actual final reaction volume.
5. Convert every initial amount into mol L^-1.
6. Write the initial row using those concentrations.
7. Determine the direction of the reaction, if \(K_c\) is supplied.
8. Construct the change row using the balanced equation.
9. Add the initial and change rows to obtain the equilibrium expressions.
10. Solve for \(x\) and reject any root that produces an impossible concentration.

Remember that the initial row is not a record of the original stock solutions. It represents the concentrations in the prepared reaction mixture before its equilibrium composition changes.

Once this conversion is reliable, the next step is using the completed ICE table to calculate equilibrium concentrations and \(K_c\). Continue with ICE Tables for Equilibrium: Calculate Concentrations and Use Kc.

For gas equilibria, also practise recognising when a change in container volume requires you to recalculate the entire initial concentration row, as covered in How to Calculate Gas Equilibrium at Constant Volume.