How to Find Initial Concentration from Equilibrium Data

A sealed flask starts with an unknown concentration of colourless nitrogen tetroxide, \(\ce{N2O4(g)}\). After equilibrium is established, a measurement shows that the concentration of brown nitrogen dioxide, \(\ce{NO2(g)}\), is \(0.100\ \mathrm{mol\,L^{-1}}\).

Can you work backwards from that measurement to find the original concentration of \(\ce{N2O4}\)?

Yes. You need to use the equilibrium measurement to reconstruct the equilibrium concentrations, then use the balanced equation to calculate how much of each substance must have reacted. From there, you can recover the initial concentrations.

The key relationship is:

\[ \boxed{\text{Initial}=\text{Equilibrium}-\text{Change}} \]

This is the reverse of the usual ICE-table calculation. It is particularly useful when a question gives you an experimentally measured equilibrium concentration and asks you to determine the starting concentration.

01Why an equilibrium measurement can reveal the starting concentration

Imagine investigating the reversible reaction:

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

Nitrogen tetroxide is colourless, while nitrogen dioxide is brown. This makes the reaction useful for illustrating how an equilibrium composition can be measured.

How would you measure the equilibrium concentration?

Consider an illustrative experimental arrangement.

A known volume of nitrogen tetroxide is placed inside a sealed, transparent optical cell. The cell sits inside a temperature-controlled bath to keep the temperature constant.

A light source shines through the cell. A detector on the opposite side measures how much light passes through the gas. As nitrogen dioxide forms, it absorbs some of the light.

Using a calibration between absorbance and concentration, the detector’s reading can be converted into an equilibrium concentration of \(\ce{NO2}\).

What would you expect the detector to record as time passes?

Initially, the cell contains no nitrogen dioxide, so the absorbance associated with \(\ce{NO2}\) is zero. As nitrogen tetroxide dissociates, nitrogen dioxide forms and the absorbance increases.

Eventually, the reading becomes constant.

This plateau indicates that the macroscopic concentration has stopped changing. The forward and reverse reactions are still occurring, but at equal rates. The system has reached dynamic equilibrium.

There is an important experimental detail here.

Removing a gas sample for separate analysis could disturb the equilibrium mixture. A sealed optical cell allows the composition to be monitored without withdrawing gas. The rigid vessel also prevents volume changes during the measurement.

This is an illustrative experimental design, rather than a description of a particular historical experiment.

The important idea is that the detector measures the final composition, not the starting composition. Chemistry and stoichiometry let us reconstruct the earlier state.

02The backwards ICE-table method

ICE stands for Initial, Change, and Equilibrium.

Each row has a different job:

The equilibrium row is calculated using:

\[ E=I+C \]

Rearranging gives:

\[ \boxed{I=E-C} \]

Here, \(C\) represents the signed concentration change. A decrease is negative, while an increase is positive.

This relationship is the foundation of finding an initial concentration from an equilibrium measurement.

Step 1: Use the balanced equation to establish the changes

Consider:

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

The balanced equation tells us that one mole of nitrogen tetroxide produces two moles of nitrogen dioxide.

If the forward reaction consumes \(x\ \mathrm{mol\,L^{-1}}\) of \(\ce{N2O4}\), it produces \(2x\ \mathrm{mol\,L^{-1}}\) of \(\ce{NO2}\).

The change row is therefore:

Notice that the product changes by \(2x\), not \(x\).

This is one of the easiest places to lose marks. The coefficients in the balanced equation determine the relative concentration changes.

Step 2: Construct the three rows

Suppose the initial concentration of nitrogen tetroxide is unknown, so we call it \(c_0\).

Assume there is initially no nitrogen dioxide.

The ICE table becomes:

All concentrations are in \(\mathrm{mol\,L^{-1}}\).

The equilibrium row is obtained by adding the initial and change rows.

For example:

\[ [\ce{NO2}]_{\mathrm{eq}}=0+2x \]

Therefore:

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

This is particularly useful when the experiment measures the product concentration. You can immediately determine \(x\).

Step 3: Use \(K_c\) to find missing equilibrium concentrations

The equilibrium constant connects the concentrations of the substances once equilibrium has been established.

For the nitrogen tetroxide reaction:

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

Notice that the concentrations in this expression are equilibrium concentrations.

If you know \(K_c\) and the equilibrium concentration of nitrogen dioxide, you can rearrange the expression to find the equilibrium concentration of nitrogen tetroxide:

\[ \boxed{ [\ce{N2O4}]_{\mathrm{eq}} = \frac{[\ce{NO2}]_{\mathrm{eq}}^2}{K_c} } \]

You now have two pieces of information:

• How much nitrogen tetroxide remains at equilibrium.
• How much nitrogen tetroxide was consumed during the reaction.

Add these together to recover the initial concentration.

A useful decision guide

There is one important limitation: a single equilibrium measurement does not automatically reveal every initial concentration. You need enough additional information, such as \(K_c\), known starting concentrations, and the balanced equation, to determine the missing values.

If the reaction direction is uncertain and all initial concentrations are known, you can compare \(Q_c\) with \(K_c\). If an initial concentration is itself unknown, keep it as an unknown rather than assuming a reaction direction.

03Worked examples

04When this method becomes more difficult

There are three situations worth watching for in HSC Chemistry questions.

1. The reaction shifts backwards.

If a reactant has a higher equilibrium concentration than its initial concentration, the reaction may have proceeded in the reverse direction.

The change row must reflect that.

For example, for:

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

a reverse shift gives:

You must work backwards using these signs, not automatically subtract \(x\) from every reactant.

2. More than one mathematical root is possible.

A quadratic equation can produce two algebraic solutions, but that does not mean both describe possible chemical mixtures.

Check every initial and equilibrium concentration.

All must be non-negative. Also check that the extent of reaction does not exceed the amount of any substance available in the relevant direction.

3. The volume changes during the experiment.

This is particularly important for gas equilibria.

Concentration is the amount of substance divided by volume:

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

If the volume changes, the concentration can change immediately, even before any chemical reaction occurs.

You must distinguish a concentration change caused by dilution or expansion from a concentration change caused by the reaction itself.

The final practice question explores this distinction.

05Questions and solutions

The following questions progress from a straightforward back-calculation to problems involving reverse reactions, inconsistent measurements, and changing volumes.

Question 1

Consider the equilibrium:

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

At a particular temperature:

\[ K_c=2.00 \]

A vessel initially contains pure A. At equilibrium, a measurement gives:

\[ [B]_{\mathrm{eq}}=0.300\ \mathrm{mol\,L^{-1}} \]

Calculate the initial concentration of A.

Solution 1

The initial concentration of A is \(0.450\ \mathrm{mol\,L^{-1}}\).

First, use the equilibrium expression:

\[ K_c=\frac{[B]_{\mathrm{eq}}}{[A]_{\mathrm{eq}}} \]

Rearrange:

\[ [A]_{\mathrm{eq}} = \frac{[B]_{\mathrm{eq}}}{K_c} \]

Substitute:

\[ [A]_{\mathrm{eq}} = \frac{0.300}{2.00} \]

\[ =0.150\ \mathrm{mol\,L^{-1}} \]

Now consider the ICE table.

Since the initial B concentration was zero:

\[ x=0.300\ \mathrm{mol\,L^{-1}} \]

Therefore, the initial concentration of A was:

\[ [A]_0=[A]_{\mathrm{eq}}+x \]

\[ =0.150+0.300 \]

\[ \boxed{[A]_0=0.450\ \mathrm{mol\,L^{-1}}} \]

The tempting mistake is to use the measured equilibrium concentration of B as the amount of A initially present. They are not the same quantity. The equilibrium constant tells us how much A remains, while stoichiometry tells us how much A reacted.

Question 2

Nitrogen tetroxide dissociates according to:

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

A sealed vessel initially contains:

\[ [\ce{N2O4}]_0=0.250\ \mathrm{M} \]

The initial concentration of nitrogen dioxide is unknown.

At equilibrium, an optical measurement gives:

\[ [\ce{NO2}]_{\mathrm{eq}}=0.300\ \mathrm{M} \]

At this temperature:

\[ K_c=0.450\ \mathrm{mol\,L^{-1}} \]

Calculate the initial concentration of nitrogen dioxide.

Solution 2

The initial nitrogen dioxide concentration is \(0.200\ \mathrm{mol\,L^{-1}}\).

Start by using the equilibrium expression:

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

Rearrange:

\[ [\ce{N2O4}]_{\mathrm{eq}} = \frac{(0.300)^2}{0.450} \]

\[ =0.200\ \mathrm{M} \]

Now calculate the change in nitrogen tetroxide:

\[ \Delta[\ce{N2O4}] =0.200-0.250 \]

\[ =-0.050\ \mathrm{M} \]

The forward reaction has consumed \(0.050\ \mathrm{M}\) of nitrogen tetroxide.

According to the balanced equation, twice this concentration of nitrogen dioxide must have formed:

\[ \Delta[\ce{NO2}] =2(0.050) \]

\[ =+0.100\ \mathrm{M} \]

Therefore:

\[ [\ce{NO2}]_0 = [\ce{NO2}]_{\mathrm{eq}} -\Delta[\ce{NO2}] \]

\[ =0.300-0.100 \]

\[ \boxed{ [\ce{NO2}]_0=0.200\ \mathrm{mol\,L^{-1}} } \]

The important detail is that nitrogen dioxide was already present initially. The measured equilibrium concentration of \(0.300\ \mathrm{M}\) includes both the original nitrogen dioxide and the additional nitrogen dioxide formed during dissociation.

Question 3

Hydrogen and iodine react according to:

\[ \ce{H2(g) + I2(g) <=> 2HI(g)} \]

A vessel initially contains:

\[ [\ce{H2}]_0=0.100\ \mathrm{M} \]

\[ [\ce{I2}]_0=0.200\ \mathrm{M} \]

The initial concentration of hydrogen iodide is unknown.

At equilibrium, measurements give:

\[ [\ce{H2}]_{\mathrm{eq}}=0.250\ \mathrm{M} \]

\[ [\ce{I2}]_{\mathrm{eq}}=0.350\ \mathrm{M} \]

The equilibrium constant is \(K_c=4.00\).

Calculate the initial concentration of hydrogen iodide.

Explain which direction the reaction proceeded.

Solution 3

The initial hydrogen iodide concentration was approximately \(0.892\ \mathrm{mol\,L^{-1}}\), and the reaction proceeded in the reverse direction.

First, notice that both hydrogen and iodine increased in concentration:

\[ 0.250-0.100=0.150\ \mathrm{M} \]

\[ 0.350-0.200=0.150\ \mathrm{M} \]

Therefore, the reaction proceeded backwards, producing hydrogen and iodine while consuming hydrogen iodide.

The change in hydrogen iodide must be twice the change in either reactant:

\[ \Delta[\ce{HI}]=-2(0.150) \]

\[ =-0.300\ \mathrm{M} \]

Next, calculate the equilibrium concentration of hydrogen iodide using \(K_c\):

\[ K_c= \frac{[\ce{HI}]_{\mathrm{eq}}^2} {[\ce{H2}]_{\mathrm{eq}}[\ce{I2}]_{\mathrm{eq}}} \]

Rearrange:

\[ [\ce{HI}]_{\mathrm{eq}}^2 = K_c[\ce{H2}]_{\mathrm{eq}}[\ce{I2}]_{\mathrm{eq}} \]

Substitute:

\[ [\ce{HI}]_{\mathrm{eq}}^2 = 4.00(0.250)(0.350) \]

\[ =0.350 \]

Take the positive square root because a concentration cannot be negative:

\[ [\ce{HI}]_{\mathrm{eq}} =\sqrt{0.350} \]

\[ =0.5916\ \mathrm{M} \]

Now work backwards.

Hydrogen iodide decreased by \(0.300\ \mathrm{M}\), so its initial concentration must have been higher than its equilibrium concentration:

\[ [\ce{HI}]_0 = [\ce{HI}]_{\mathrm{eq}}+0.300 \]

\[ =0.5916+0.300 \]

\[ \boxed{ [\ce{HI}]_0=0.892\ \mathrm{mol\,L^{-1}} } \]

The trap is to assume that every equilibrium calculation involves a forward reaction.

Here, the measured reactant concentrations are higher than their initial values. This is direct evidence that the net reaction proceeded towards the reactants.

Question 4

Consider the reaction:

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

A solution initially contains:

\[ [A]_0=0.600\ \mathrm{M} \]

\[ [B]_0=0.400\ \mathrm{M} \]

The initial concentration of C is unknown.

At equilibrium, measurements give:

\[ [A]_{\mathrm{eq}}=0.450\ \mathrm{M} \]

\[ [C]_{\mathrm{eq}}=0.300\ \mathrm{M} \]

The reported equilibrium constant is:

\[ K_c=2.00\ \mathrm{L\,mol^{-1}} \]

A student calculates an initial concentration for C using stoichiometry and concludes that the data are consistent.

Determine the student’s calculated initial concentration of C. Then assess whether all the supplied data can describe the same equilibrium mixture.

Solution 4

Stoichiometry gives a candidate initial concentration of \(0.150\ \mathrm{mol\,L^{-1}}\), but the complete data set is inconsistent with the stated equilibrium constant.

First, calculate the change in A:

\[ \Delta[A]=0.450-0.600 \]

\[ =-0.150\ \mathrm{M} \]

Therefore, \(0.150\ \mathrm{M}\) of A has reacted.

The balanced equation has a \(1:1:1\) ratio, so:

• B decreases by \(0.150\ \mathrm{M}\).
• C increases by \(0.150\ \mathrm{M}\).

The equilibrium concentration of B must therefore be:

\[ [B]_{\mathrm{eq}} =0.400-0.150 \]

\[ =0.250\ \mathrm{M} \]

Now calculate the initial concentration of C using the measured equilibrium concentration:

\[ [C]_0=[C]_{\mathrm{eq}}-0.150 \]

\[ =0.300-0.150 \]

\[ \boxed{[C]_0=0.150\ \mathrm{M}} \]

So far, the stoichiometry is satisfied.

However, we must also check the equilibrium constant.

For this reaction:

\[ K_c= \frac{[C]_{\mathrm{eq}}} {[A]_{\mathrm{eq}}[B]_{\mathrm{eq}}} \]

Substitute the calculated equilibrium concentrations:

\[ K_c= \frac{0.300}{(0.450)(0.250)} \]

\[ K_c=2.67\ \mathrm{L\,mol^{-1}} \]

But the question states:

\[ K_c=2.00\ \mathrm{L\,mol^{-1}} \]

These values do not agree.

In fact, the stated equilibrium constant would require:

\[ [B]_{\mathrm{eq}} = \frac{0.300}{(2.00)(0.450)} \]

\[ =0.333\ \mathrm{M} \]

This contradicts the \(0.250\ \mathrm{M}\) required by stoichiometry.

Therefore, no initial concentration of C can make all the supplied information consistent.

This is an important distinction: satisfying the balanced equation does not automatically mean that a calculated mixture is at equilibrium. You must satisfy both the stoichiometric relationships and the equilibrium expression.

Question 5

This final question combines equilibrium calculations with conservation of matter and a changing gas volume.

A sealed vessel initially contains pure nitrogen tetroxide:

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

The original vessel volume is \(1.00\ \mathrm{L}\).

The mixture is allowed to reach equilibrium at a constant temperature. The vessel is then expanded isothermally to \(2.00\ \mathrm{L}\), without allowing any gas to escape.

The mixture is allowed to reach equilibrium again.

At the final equilibrium, an optical measurement gives:

\[ [\ce{NO2}]_{\mathrm{eq}}=0.100\ \mathrm{M} \]

At this temperature:

\[ K_c=0.200\ \mathrm{mol\,L^{-1}} \]

Calculate the original initial concentration of nitrogen tetroxide in the \(1.00\ \mathrm{L}\) vessel.

Explain why treating the final equilibrium concentrations as though the volume had remained constant gives an incorrect answer.

Solution 5

The original initial concentration of nitrogen tetroxide was:

\[ \boxed{ [\ce{N2O4}]_0=0.200\ \mathrm{mol\,L^{-1}} } \]

The crucial idea is that the number of moles of nitrogen-containing particles is conserved, even though the vessel volume changes.

A simple ICE table using the original \(1.00\ \mathrm{L}\) volume would give the wrong answer because the final equilibrium measurements refer to a \(2.00\ \mathrm{L}\) vessel.

Step 1: Calculate the final equilibrium concentration of nitrogen tetroxide.

Use the equilibrium expression:

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

Rearrange:

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

Substitute:

\[ [\ce{N2O4}]_{\mathrm{eq}} = \frac{(0.100)^2}{0.200} \]

\[ =0.0500\ \mathrm{M} \]

So the final equilibrium mixture contains:

\[ [\ce{N2O4}]_{\mathrm{eq}}=0.0500\ \mathrm{M} \]

\[ [\ce{NO2}]_{\mathrm{eq}}=0.100\ \mathrm{M} \]

Step 2: Recognise what the balanced equation conserves.

The reaction is:

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

One molecule of nitrogen tetroxide contains two nitrogen atoms.

Those two nitrogen atoms remain in the products when nitrogen tetroxide dissociates into two nitrogen dioxide molecules.

Therefore, we can count the total amount of nitrogen tetroxide units using:

\[ n_{\mathrm{units}} = n_{\ce{N2O4}}+ \frac{1}{2}n_{\ce{NO2}} \]

This quantity remains constant because no nitrogen enters or leaves the vessel.

Step 3: Calculate the total concentration of nitrogen tetroxide units at the final equilibrium.

Express the conserved quantity as a concentration:

\[ c_{\mathrm{units}} = [\ce{N2O4}]_{\mathrm{eq}} + \frac{1}{2}[\ce{NO2}]_{\mathrm{eq}} \]

Substitute:

\[ c_{\mathrm{units}} = 0.0500+\frac{1}{2}(0.100) \]

\[ =0.100\ \mathrm{mol\,L^{-1}} \]

This is the concentration of conserved nitrogen tetroxide units in the final \(2.00\ \mathrm{L}\) vessel.

Step 4: Convert this concentration into the total number of moles.

Use:

\[ n=cV \]

The final volume is \(2.00\ \mathrm{L}\):

\[ n_{\mathrm{units}} = 0.100\times2.00 \]

\[ =0.200\ \mathrm{mol} \]

The vessel initially contained pure nitrogen tetroxide, and no gas escaped.

Therefore, the original amount of nitrogen tetroxide was also:

\[ n_{\ce{N2O4},0}=0.200\ \mathrm{mol} \]

Step 5: Calculate the original initial concentration.

The original volume was \(1.00\ \mathrm{L}\).

Therefore:

\[ c_0=\frac{n_0}{V_0} \]

\[ c_0=\frac{0.200}{1.00} \]

\[ \boxed{ [\ce{N2O4}]_0=0.200\ \mathrm{mol\,L^{-1}} } \]

Why is the tempting alternative incorrect?

You might initially try to treat the final measured concentration as though the original volume had remained unchanged.

That would give:

\[ 2x=0.100 \]

\[ x=0.0500\ \mathrm{M} \]

You could then add the final nitrogen tetroxide concentration to \(x\):

\[ 0.0500+0.0500=0.100\ \mathrm{M} \]

But this calculation incorrectly assumes that the final concentrations refer to the original \(1.00\ \mathrm{L}\) volume.

They actually refer to the expanded \(2.00\ \mathrm{L}\) vessel.

The correct calculation must account for the volume change and conserve the total number of moles.

This is the deeper principle: chemical reactions redistribute atoms between substances, but they do not destroy the atoms themselves. When volume changes, conserve moles first and calculate concentrations using the appropriate volume.

06Where to go next

Once you can reconstruct initial concentrations from equilibrium measurements, you can tackle the reverse problem of calculating equilibrium concentrations from known starting amounts and \(K_c\).

For a full foundation in constructing ICE tables and using equilibrium expressions, see ICE Tables for Equilibrium: Calculate Concentrations and Use Kc.

If a question gives you starting amounts in moles rather than concentrations, revise How to Convert Moles into Concentration for an ICE Table.

The decision to remember is simple: use \(K_c\) to reconstruct the equilibrium mixture, stoichiometry to reconstruct the reaction changes, and \(I=E-C\) to recover the starting concentrations. Always check that the resulting mixture is physically possible.