How to Use Stoichiometric Coefficients in an ICE Table

A sealed flask contains nitrogen dioxide and dinitrogen tetroxide. The reaction is moving towards equilibrium:

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

Suppose the concentration of \(\ce{N2O4}\) increases by \(0.10\ \mathrm{M}\). How much \(\ce{NO2}\) must disappear?

A common mistake is to say \(0.10\ \mathrm{M}\). But the balanced equation tells us that two molecules of \(\ce{NO2}\) combine to form one molecule of \(\ce{N2O4}\). Therefore, \(0.20\ \mathrm{M}\) of \(\ce{NO2}\) must be consumed.

That factor of two is exactly what the stoichiometric coefficients tell you to put in the change row of an ICE table.

The important skill is knowing how to translate the balanced equation into concentration changes without accidentally changing the initial concentrations or producing an impossible equilibrium.

All numerical data in this resource are illustrative. Unless stated otherwise, assume constant temperature and volume, and express concentrations in \(\mathrm{mol\,L^{-1}}\), or M.

01Why stoichiometric coefficients matter in an ICE table

Consider the reversible reaction:

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

The coefficients give us the ratio:

\[ 2\text{ mol }\ce{NO2}:1\text{ mol }\ce{N2O4} \]

This means that whenever the reaction proceeds in the forward direction:

• Two units of \(\ce{NO2}\) are consumed.
• One unit of \(\ce{N2O4}\) is produced.

The same ratio applies to changes in concentration when the reaction occurs at constant volume.

What the three ICE rows represent

ICE stands for Initial, Change, and Equilibrium. Each row has a different job.

The change row is where the coefficients become important.

For our nitrogen dioxide reaction, suppose the initial concentrations are:

\[ [\ce{NO2}]_0=0.60\ \mathrm{M} \]

\[ [\ce{N2O4}]_0=0.10\ \mathrm{M} \]

Let \(x\) represent the reaction progress in the forward direction, measured in M.

The ICE table becomes:

Notice that the change row does not contain \(-x\) and \(+x\).

The coefficient of 2 belongs to \(\ce{NO2}\), so its concentration decreases by \(2x\). The coefficient of 1 belongs to \(\ce{N2O4}\), so its concentration increases by \(x\).

This is the central rule:

Multiply the reaction progress, \(x\), by the stoichiometric coefficient of each substance.

2. The general method for any balanced equation

The same approach works for any reversible reaction.

Consider:

\[ \ce{aA + bB <=> cC + dD} \]

Here, \(a\), \(b\), \(c\), and \(d\) are the stoichiometric coefficients.

If the reaction proceeds forwards by an amount \(x\), the change row is:

There are three rules to remember.

Rule 1: Reactants decrease in the forward direction.

Their changes are negative.

Rule 2: Products increase in the forward direction.

Their changes are positive.

Rule 3: Multiply every change by its own coefficient.

For example, consider:

\[ \ce{2SO2(g) + O2(g) <=> 2SO3(g)} \]

The coefficients are \(2:1:2\).

Therefore, the change row is:

If \(x=0.020\ \mathrm{M}\), the actual concentration changes are:

• \(\ce{SO2}\): decreases by \(0.040\ \mathrm{M}\).
• \(\ce{O2}\): decreases by \(0.020\ \mathrm{M}\).
• \(\ce{SO3}\): increases by \(0.040\ \mathrm{M}\).

The coefficients control the ratios of the changes, not the numerical value of \(x\).

3. What if the reaction proceeds backwards?

This is where students sometimes become confused about the signs.

The easiest approach is to choose one direction and stick with it.

In this resource, we define \(x\) as positive in the forward direction, as written in the balanced equation.

For:

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

the forward change row is always:

\[ -2x,\quad +x \]

If the reaction actually shifts backwards, \(x\) will be negative.

For example, if \(x=-0.05\ \mathrm{M}\):

\[ \Delta[\ce{NO2}]=-2(-0.05)=+0.10\ \mathrm{M} \]

\[ \Delta[\ce{N2O4}]=+(-0.05)=-0.05\ \mathrm{M} \]

So \(\ce{NO2}\) increases, while \(\ce{N2O4}\) decreases.

That is exactly what we expect from the reverse reaction.

You can also define a new variable \(y\) that is positive in the reverse direction. In that case, the change row becomes \(+2y\) for \(\ce{NO2}\) and \(-y\) for \(\ce{N2O4}\).

Both methods work. The important thing is not to mix the two sign conventions.

If you need to determine the direction using the reaction quotient:

• \(Q_c<K_c\): the reaction proceeds forwards.
• \(Q_c>K_c\): the reaction proceeds backwards.
• \(Q_c=K_c\): the system is already at equilibrium.

4. A common misconception: confusing coefficients with Kc powers

There is another reason students sometimes get the change row wrong.

Stoichiometric coefficients also appear as powers in the equilibrium constant expression. However, these are two separate applications of the same balanced equation.

Compare the following:

For the second reaction:

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

the equilibrium expression is:

\[ K_c=\frac{[\ce{N2O4}]}{[\ce{NO2}]^2} \]

Notice the two different uses of the coefficient 2:

1. In the change row, it multiplies \(x\), giving \(-2x\).
2. In the equilibrium expression, it becomes the power of the concentration, giving \([\ce{NO2}]^2\).

Do not put the square into the change row, and do not use \(2x\) as the exponent in the equilibrium expression.

5. How to recognise and reject nonphysical roots

Once you substitute the equilibrium row into the \(K_c\) expression, you may need to solve a quadratic equation.

Sometimes, the quadratic produces two mathematical solutions.

However, not every mathematical solution represents a physically possible chemical system.

The most important condition is that concentrations cannot be negative.

Return to:

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

Suppose the initial concentrations are:

\[ [\ce{NO2}]_0=0.60\ \mathrm{M} \]

\[ [\ce{N2O4}]_0=0.10\ \mathrm{M} \]

The equilibrium row is:

\[ [\ce{NO2}]_{\mathrm{eq}}=0.60-2x \]

\[ [\ce{N2O4}]_{\mathrm{eq}}=0.10+x \]

Both concentrations must be non-negative.

For nitrogen dioxide:

\[ 0.60-2x\geq0 \]

\[ x\leq0.30 \]

For dinitrogen tetroxide:

\[ 0.10+x\geq0 \]

\[ x\geq-0.10 \]

Combining these conditions gives:

\[ \boxed{-0.10\leq x\leq0.30} \]

This is the physically possible range of \(x\).

Any mathematical root outside this interval must be rejected because it would produce a negative concentration.

There is an important distinction here: a negative value of \(x\) is not automatically wrong. It can simply mean the reaction proceeds backwards. What matters is whether every resulting concentration is physically possible.

06Worked examples

03Questions and solutions

The following questions progress from straightforward change-row calculations to more demanding equilibrium problems. Use the forward direction as the positive direction for \(x\), unless the question specifies otherwise.

Question 1

Hydrogen gas reacts with bromine gas according to:

\[ \ce{H2(g) + Br2(g) <=> 2HBr(g)} \]

A vessel initially contains:

• \([\ce{H2}]_0=0.40\ \mathrm{M}\)
• \([\ce{Br2}]_0=0.35\ \mathrm{M}\)
• \([\ce{HBr}]_0=0.10\ \mathrm{M}\)

The reaction proceeds forwards with \(x=0.080\ \mathrm{M}\).

Construct the complete ICE table and calculate the equilibrium concentration of each substance.

Solution 1

The equilibrium concentrations are \(0.32\ \mathrm{M}\) hydrogen, \(0.27\ \mathrm{M}\) bromine, and \(0.26\ \mathrm{M}\) hydrogen bromide.

The balanced equation has coefficients \(1:1:2\).

Therefore, the change row is:

\[ -x,\quad -x,\quad +2x \]

Substitute \(x=0.080\ \mathrm{M}\):

\[ \begin{aligned} \Delta[\ce{H2}]&=-0.080\ \mathrm{M}\\ \Delta[\ce{Br2}]&=-0.080\ \mathrm{M}\\ \Delta[\ce{HBr}]&=+0.160\ \mathrm{M} \end{aligned} \]

Now add each change to its initial concentration.

\[ [\ce{H2}]_{\mathrm{eq}}=0.40-0.080=0.32\ \mathrm{M} \]

\[ [\ce{Br2}]_{\mathrm{eq}}=0.35-0.080=0.27\ \mathrm{M} \]

\[ [\ce{HBr}]_{\mathrm{eq}}=0.10+0.160=0.26\ \mathrm{M} \]

The completed table is:

The important detail is that hydrogen bromide increases by twice the amount that either reactant decreases.

Question 2

Sulfur dioxide reacts with oxygen to produce sulfur trioxide:

\[ \ce{2SO2(g) + O2(g) <=> 2SO3(g)} \]

A reaction vessel initially contains:

\[ [\ce{SO2}]_0=0.80\ \mathrm{M} \]

\[ [\ce{O2}]_0=0.60\ \mathrm{M} \]

\[ [\ce{SO3}]_0=0.10\ \mathrm{M} \]

At equilibrium, the oxygen concentration is measured to be \(0.35\ \mathrm{M}\).

Calculate:

(a) The reaction progress, \(x\).

(b) The equilibrium concentrations of sulfur dioxide and sulfur trioxide.

(c) Whether the resulting concentrations are physically possible.

Solution 2

The reaction progress is \(x=0.25\ \mathrm{M}\), giving equilibrium concentrations of \(0.30\ \mathrm{M}\) sulfur dioxide and \(0.60\ \mathrm{M}\) sulfur trioxide.

Start by constructing the change row:

(a) Find \(x\).

We are given the equilibrium oxygen concentration:

\[ [\ce{O2}]_{\mathrm{eq}}=0.35\ \mathrm{M} \]

Using the equilibrium row:

\[ 0.60-x=0.35 \]

Rearrange:

\[ x=0.60-0.35 \]

\[ \boxed{x=0.25\ \mathrm{M}} \]

(b) Calculate the other equilibrium concentrations.

For sulfur dioxide:

\[ [\ce{SO2}]_{\mathrm{eq}}=0.80-2x \]

\[ =0.80-2(0.25) \]

\[ \boxed{[\ce{SO2}]_{\mathrm{eq}}=0.30\ \mathrm{M}} \]

For sulfur trioxide:

\[ [\ce{SO3}]_{\mathrm{eq}}=0.10+2x \]

\[ =0.10+2(0.25) \]

\[ \boxed{[\ce{SO3}]_{\mathrm{eq}}=0.60\ \mathrm{M}} \]

(c) Check physical possibility.

All three equilibrium concentrations are positive:

• \(\ce{SO2}=0.30\ \mathrm{M}\)
• \(\ce{O2}=0.35\ \mathrm{M}\)
• \(\ce{SO3}=0.60\ \mathrm{M}\)

The result is physically possible.

A common error would be to use \(-x\) for sulfur dioxide. However, its coefficient is 2, so it must decrease by \(2x=0.50\ \mathrm{M}\), not \(0.25\ \mathrm{M}\).

Question 3

Carbon monoxide reacts with oxygen to produce carbon dioxide:

\[ \ce{2CO(g) + O2(g) <=> 2CO2(g)} \]

A vessel initially contains:

\[ [\ce{CO}]_0=0.50\ \mathrm{M} \]

\[ [\ce{O2}]_0=0.40\ \mathrm{M} \]

\[ [\ce{CO2}]_0=0.10\ \mathrm{M} \]

At equilibrium, the carbon dioxide concentration is \(0.50\ \mathrm{M}\).

Calculate:

(a) The value of \(x\).

(b) The equilibrium concentrations of carbon monoxide and oxygen.

(c) The equilibrium constant, \(K_c\).

Explain why using \(x=0.40\ \mathrm{M}\) in the change row would be incorrect.

Solution 3

The reaction progress is \(x=0.20\ \mathrm{M}\), the equilibrium concentrations are \(0.10\ \mathrm{M}\) carbon monoxide and \(0.20\ \mathrm{M}\) oxygen, and \(K_c=125\ \mathrm{M^{-1}}\).

(a) Find \(x\).

The balanced equation has coefficients \(2:1:2\).

The ICE table is:

We are told that:

\[ [\ce{CO2}]_{\mathrm{eq}}=0.50\ \mathrm{M} \]

Therefore:

\[ 0.10+2x=0.50 \]

\[ 2x=0.40 \]

\[ \boxed{x=0.20\ \mathrm{M}} \]

(b) Find the remaining equilibrium concentrations.

For carbon monoxide:

\[ [\ce{CO}]_{\mathrm{eq}} =0.50-2(0.20) \]

\[ \boxed{[\ce{CO}]_{\mathrm{eq}}=0.10\ \mathrm{M}} \]

For oxygen:

\[ [\ce{O2}]_{\mathrm{eq}} =0.40-0.20 \]

\[ \boxed{[\ce{O2}]_{\mathrm{eq}}=0.20\ \mathrm{M}} \]

(c) Calculate \(K_c\).

The equilibrium expression is:

\[ K_c=\frac{[\ce{CO2}]^2} {[\ce{CO}]^2[\ce{O2}]} \]

Substitute the equilibrium concentrations:

\[ K_c=\frac{(0.50)^2} {(0.10)^2(0.20)} \]

\[ K_c=\frac{0.25}{0.002} \]

\[ \boxed{K_c=125\ \mathrm{M^{-1}}} \]

Why is \(x=0.40\ \mathrm{M}\) incorrect?

The carbon dioxide concentration increases by \(0.40\ \mathrm{M}\), but its coefficient is 2.

Therefore:

\[ \Delta[\ce{CO2}]=+2x \]

The value of \(x\) must be half the actual concentration increase.

Confusing the total concentration change with the reaction progress would make every other entry in the ICE table incorrect.

Question 4

Consider the reversible reaction:

\[ \ce{2CO(g) + O2(g) <=> 2CO2(g)} \]

A reaction vessel initially contains:

\[ [\ce{CO}]_0=0.50\ \mathrm{M} \]

\[ [\ce{O2}]_0=0.40\ \mathrm{M} \]

\[ [\ce{CO2}]_0=0.60\ \mathrm{M} \]

At equilibrium, the measured carbon dioxide concentration is \(0.40\ \mathrm{M}\).

Using \(x\) as the reaction progress in the forward direction:

(a) Construct the ICE table.

(b) Calculate \(x\) and the equilibrium concentrations of the other two gases.

(c) Calculate \(K_c\).

(d) Explain why a negative value of \(x\) is not a physically impossible result.

Solution 4

The reaction progress is \(x=-0.10\ \mathrm{M}\), giving equilibrium concentrations of \(0.70\ \mathrm{M}\) carbon monoxide, \(0.50\ \mathrm{M}\) oxygen, and \(0.40\ \mathrm{M}\) carbon dioxide.

(a) Construct the ICE table.

The coefficients are \(2:1:2\).

Using the forward direction as positive:

(b) Calculate \(x\) and the equilibrium concentrations.

We are given:

\[ [\ce{CO2}]_{\mathrm{eq}}=0.40\ \mathrm{M} \]

Therefore:

\[ 0.60+2x=0.40 \]

\[ 2x=-0.20 \]

\[ \boxed{x=-0.10\ \mathrm{M}} \]

Now calculate the other concentrations.

For carbon monoxide:

\[ [\ce{CO}]_{\mathrm{eq}} =0.50-2(-0.10) \]

\[ \boxed{[\ce{CO}]_{\mathrm{eq}}=0.70\ \mathrm{M}} \]

For oxygen:

\[ [\ce{O2}]_{\mathrm{eq}} =0.40-(-0.10) \]

\[ \boxed{[\ce{O2}]_{\mathrm{eq}}=0.50\ \mathrm{M}} \]

(c) Calculate \(K_c\).

The equilibrium expression is:

\[ K_c=\frac{[\ce{CO2}]^2} {[\ce{CO}]^2[\ce{O2}]} \]

Substitute:

\[ K_c=\frac{(0.40)^2} {(0.70)^2(0.50)} \]

\[ K_c=\frac{0.16}{0.245} \]

\[ \boxed{K_c=0.653\ \mathrm{M^{-1}}} \]

(d) Explain the negative value of \(x\).

The negative value does not mean that a concentration is negative.

It means the reaction has proceeded in the reverse direction.

Initially, the reaction quotient is:

\[ Q_c=\frac{(0.60)^2}{(0.50)^2(0.40)} \]

\[ Q_c=3.60\ \mathrm{M^{-1}} \]

Since:

\[ Q_c>K_c \]

the mixture contains relatively too much product compared with the equilibrium ratio.

The reaction therefore shifts backwards, consuming carbon dioxide and producing carbon monoxide and oxygen.

Our negative value of \(x\) correctly represents this reverse reaction.

All three equilibrium concentrations are positive, so the result is physically valid.

Question 5

A sealed vessel contains nitrogen dioxide and dinitrogen tetroxide at equilibrium:

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

At a constant temperature, the equilibrium concentrations are:

\[ [\ce{NO2}]=0.50\ \mathrm{M} \]

\[ [\ce{N2O4}]=0.50\ \mathrm{M} \]

The equilibrium constant is therefore:

\[ K_c=2.0\ \mathrm{M^{-1}} \]

The vessel is suddenly compressed to half its original volume. No gas is added or removed, and the temperature remains constant.

Determine:

(a) The concentrations of both gases immediately after compression, before any further reaction occurs.

(b) Whether the reaction subsequently shifts forwards or backwards. Justify your answer using \(Q_c\) and \(K_c\).

(c) The new equilibrium concentrations of both gases.

(d) Why the immediate concentration changes caused by compression must not be entered as the stoichiometric change row of the ICE table.

Solution 5

Immediately after compression, both concentrations double to \(1.00\ \mathrm{M}\). The reaction then shifts forwards, and the new equilibrium concentrations are \(0.75\ \mathrm{M}\) nitrogen dioxide and \(1.125\ \mathrm{M}\) dinitrogen tetroxide.

This question tests an important distinction: a change in concentration caused by changing the volume is not the same thing as a change caused by a chemical reaction.

(a) Calculate the concentrations immediately after compression.

The volume is halved, while the number of moles of each gas remains unchanged at the instant of compression.

Using:

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

halving the volume doubles the concentration of every gas.

Therefore:

\[ [\ce{NO2}]_{\mathrm{new}}=2(0.50) \]

\[ \boxed{[\ce{NO2}]_{\mathrm{new}}=1.00\ \mathrm{M}} \]

Similarly:

\[ [\ce{N2O4}]_{\mathrm{new}}=2(0.50) \]

\[ \boxed{[\ce{N2O4}]_{\mathrm{new}}=1.00\ \mathrm{M}} \]

These are the concentrations immediately after compression, before the reaction has had time to respond.

(b) Determine the direction of the reaction.

The equilibrium constant does not change because the temperature remains constant.

Therefore:

\[ K_c=2.0\ \mathrm{M^{-1}} \]

Calculate the new reaction quotient using the concentrations immediately after compression:

\[ Q_c=\frac{[\ce{N2O4}]}{[\ce{NO2}]^2} \]

Substitute:

\[ Q_c=\frac{1.00}{(1.00)^2} \]

\[ Q_c=1.0\ \mathrm{M^{-1}} \]

Compare the two values:

\[ Q_c=1.0<K_c=2.0 \]

The reaction must proceed forwards to increase the relative amount of dinitrogen tetroxide.

This also agrees with the fact that the forward reaction reduces the number of gas molecules from two to one.

(c) Calculate the new equilibrium concentrations.

This is where we construct a new ICE table.

The important point is that the initial row now contains the concentrations immediately after compression, not the concentrations before compression.

For:

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

the table becomes:

Substitute the equilibrium row into the \(K_c\) expression:

\[ 2.0=\frac{1.00+x}{(1.00-2x)^2} \]

Rearrange:

\[ 2.0(1.00-2x)^2=1.00+x \]

Expand:

\[ 2.0-8x+8x^2=1.00+x \]

Collect terms:

\[ 8x^2-9x+1=0 \]

Factorise:

\[ (8x-1)(x-1)=0 \]

Therefore:

\[ x=0.125\ \mathrm{M} \]

or:

\[ x=1.00\ \mathrm{M} \]

We must check both roots.

The nitrogen dioxide concentration must remain non-negative:

\[ 1.00-2x\geq0 \]

\[ x\leq0.50 \]

Therefore, \(x=1.00\ \mathrm{M}\) is impossible.

It would produce:

\[ [\ce{NO2}]=1.00-2(1.00) =-1.00\ \mathrm{M} \]

Reject this root.

The valid solution is:

\[ x=0.125\ \mathrm{M} \]

Calculate the new equilibrium concentrations.

For nitrogen dioxide:

\[ [\ce{NO2}]_{\mathrm{eq}} =1.00-2(0.125) \]

\[ \boxed{[\ce{NO2}]_{\mathrm{eq}}=0.75\ \mathrm{M}} \]

For dinitrogen tetroxide:

\[ [\ce{N2O4}]_{\mathrm{eq}} =1.00+0.125 \]

\[ \boxed{[\ce{N2O4}]_{\mathrm{eq}}=1.125\ \mathrm{M}} \]

Check the equilibrium constant:

\[ K_c=\frac{1.125}{(0.75)^2} \]

\[ K_c=\frac{1.125}{0.5625} \]

\[ K_c=2.0\ \mathrm{M^{-1}} \]

The result matches the original equilibrium constant, as expected at constant temperature.

(d) Explain why compression is not part of the change row.

This is the most important conceptual point in the question.

The immediate doubling of both concentrations happens because the same number of gas molecules now occupies half the volume.

It is not caused by a chemical reaction.

The stoichiometric change row only describes the subsequent chemical reaction. For this reaction, that means:

\[ \ce{2NO2 -> N2O4} \]

and therefore:

\[ -2x,\quad +x \]

If you incorrectly treated compression as part of this row, you would assume that the concentration changes caused by the volume reduction must follow the chemical ratio of \(2:1\).

They do not. Both gas concentrations initially double by the same factor.

The correct method is to calculate the new concentrations after compression, use them as the initial row of a fresh ICE table, and then apply the stoichiometric coefficients to the chemical changes.

This distinction becomes particularly useful in questions involving changes in gas pressure and volume.

04What to remember for your next equilibrium calculation

When constructing the change row of an ICE table, keep these four ideas in mind:

1. Start with the correctly balanced chemical equation.
2. Give reactants negative changes and products positive changes when using the forward direction.
3. Multiply \(x\) by each substance’s own stoichiometric coefficient.
4. Check that every equilibrium concentration is physically possible before accepting a mathematical root.

Remember that the coefficients control the ratio of concentration changes, while the initial concentrations tell you where the system starts.

Once your change row is correct, the equilibrium row follows by adding the initial concentrations and changes. You can then substitute those equilibrium concentrations into the \(K_c\) expression.

For a broader explanation of the complete calculation, read ICE Tables for Equilibrium: Calculate Concentrations and Use Kc.

If you want to strengthen your understanding of why balanced-equation coefficients determine mole ratios in the first place, see Mass Changes from Stoichiometric Equations: HSC Chemistry.