How to Explain Equilibrium Colour Changes Using Actual Species

A red equilibrium mixture becomes a deeper red after a few drops of iron(III) nitrate are added. Why?

The strongest HSC Chemistry explanation doesn’t stop at saying that equilibrium shifts to the right. It identifies the actual chemical species responsible for the colour, explains how the disturbance changes the forward and reverse reaction rates, and connects the resulting change in concentration to what you observe.

For example, if the red species is \(\mathrm{FeSCN}^{2+}\), a deeper red colour means that its concentration has increased. Your job is to explain why that happened.

There is a reliable way to build this explanation. We’ll use an iron(III) thiocyanate equilibrium to develop it, then test it against dilution, experimental measurements, and some less obvious situations.

01Identify the species responsible for the colour

Consider the following equilibrium reaction:

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

The reaction involves three important species.

The red colour comes from the iron(III) thiocyanate complex, \(\mathrm{FeSCN}^{2+}\).

This gives us our first important relationship:

• Increasing \([\mathrm{FeSCN}^{2+}]\) makes the red colour more intense.
• Decreasing \([\mathrm{FeSCN}^{2+}]\) makes the red colour less intense.

Here, square brackets indicate concentration, usually measured in \(\mathrm{mol\,L^{-1}}\).

Notice that iron(III) ions are pale yellow rather than colourless. This matters when interpreting an actual experiment because adding an iron(III) solution can affect the overall appearance independently of the red complex.

For this explanation, assume a dilute acidic aqueous solution in which the simplified equilibrium above is the dominant reaction. The acidic conditions help limit iron(III) hydrolysis, and nitrate ions introduced with iron(III) nitrate are treated as spectators.

At sufficiently high thiocyanate concentrations, additional iron(III)-thiocyanate complexes may become significant. The single-complex model is most useful under the stated dilute conditions.

02Establish the system boundary

Before explaining an equilibrium disturbance, identify exactly what belongs to the chemical system.

Imagine an equilibrium mixture inside a transparent cuvette. A pipette containing concentrated iron(III) nitrate is positioned above it.

The system is the solution inside the cuvette. It contains:

• Iron(III) ions, \(\mathrm{Fe}^{3+}\).
• Thiocyanate ions, \(\mathrm{SCN}^{-}\).
• Iron(III) thiocyanate complexes, \(\mathrm{FeSCN}^{2+}\).
• Water and the other dissolved ions.

The pipette, stock solution, bench, and surrounding air are outside the system.

When iron(III) nitrate is added, the added solution crosses the system boundary. Once mixed, its dissolved contents become part of the system.

This is important because you need to distinguish between:

1. The external change made to the system.
2. The resulting changes in the concentrations of its chemical species.
3. The subsequent reaction that restores equilibrium.

For our example, assume that a small volume of concentrated iron(III) nitrate is added, so the change in total solution volume is negligible.

This means we can treat the immediate disturbance as an increase in \([\mathrm{Fe}^{3+}]\), rather than having to account for dilution at the same time.

If the added volume is significant, however, dilution must also be considered.

03Explain the colour change using forward and reverse rates

At dynamic equilibrium, the forward and reverse reactions continue occurring.

However, they occur at equal rates.

We can represent this as:

\[
r_{\mathrm{forward}}=r_{\mathrm{reverse}}
\]

Here:

• \(r_{\mathrm{forward}}\) is the rate at which \(\mathrm{Fe}^{3+}\) and \(\mathrm{SCN}^{-}\) form \(\mathrm{FeSCN}^{2+}\).
• \(r_{\mathrm{reverse}}\) is the rate at which \(\mathrm{FeSCN}^{2+}\) dissociates into its ions.

Equal rates mean that the concentrations of the species remain constant. They do not mean that the reactions have stopped.

What happens when iron(III) nitrate is added?

Predict what happens first.

We increase the concentration of one reactant, \(\mathrm{Fe}^{3+}\), while initially leaving the concentrations of \(\mathrm{SCN}^{-}\) and \(\mathrm{FeSCN}^{2+}\) unchanged.

The forward reaction now has more iron(III) ions available to react with thiocyanate ions.

Consequently, the forward reaction rate increases.

The reverse reaction rate is initially unchanged because the concentration of its reactant, \(\mathrm{FeSCN}^{2+}\), has not yet changed.

Therefore:

\[
r_{\mathrm{forward}}>r_{\mathrm{reverse}}
\]

The forward reaction temporarily occurs faster than the reverse reaction.

As a result:

• \([\mathrm{FeSCN}^{2+}]\) increases.
• \([\mathrm{Fe}^{3+}]\) and \([\mathrm{SCN}^{-}]\) are consumed by the forward reaction.
• The red colour becomes more intense.

As more \(\mathrm{FeSCN}^{2+}\) forms, the reverse reaction rate increases. The forward and reverse rates eventually become equal again.

A new dynamic equilibrium is established.

What has actually changed?

The important result is that the new equilibrium contains a greater concentration of the red complex than the original equilibrium.

The reaction has not stopped. Instead, the forward and reverse rates have adjusted until they are equal again.

This is a much stronger explanation than writing:

The equilibrium shifts to the right, so the solution becomes redder.

That statement identifies the direction of change but doesn’t explain the chemistry responsible for it.

A complete explanation connects the added reactant to the reaction rates and then to the concentration of the coloured species.

04How the experiment demonstrates the change

There is a measurement problem worth considering.

Suppose a student adds iron(III) nitrate to a red equilibrium mixture and notices that its appearance changes.

Does this observation alone prove that more \(\mathrm{FeSCN}^{2+}\) has formed?

Not necessarily.

Iron(III) ions are themselves pale yellow. The added solution may also change the total volume, affecting the concentration of the red complex.

A simple visual comparison therefore has limitations.

Improving the experimental design

A more reliable experiment uses matched samples and a colorimeter.

First, prepare an equilibrium mixture containing iron(III) and thiocyanate ions in acidic solution. Divide it into equal portions.

Use three samples:

The apparatus has a straightforward arrangement.

The prepared solution sits inside a transparent cuvette. Light from the colorimeter passes through the solution and reaches a detector on the opposite side.

Choose a wavelength at which the red complex absorbs strongly. The colorimeter measures how much light is absorbed.

The prediction is that the test sample will develop a greater concentration of \(\mathrm{FeSCN}^{2+}\), producing an increase in its corrected absorbance.

The control should remain approximately unchanged if the added solvent has no other effect.

The reagent blank helps distinguish absorption caused by the added iron(III) ions from absorption caused by the red complex.

Interpreting illustrative experimental data

Suppose the baseline-corrected absorbance of the test sample is recorded over time.

The following values are illustrative data, not measurements from a reported experiment.

Assume the calibration relationship is:

\[
A=kc
\]

where:

• \(A\) is absorbance, which is dimensionless;
• \(k=2.00\times10^3\ \mathrm{L\,mol^{-1}}\) is the calibration slope at the chosen wavelength and fixed path length;
• \(c\) is the concentration of \(\mathrm{FeSCN}^{2+}\), in \(\mathrm{mol\,L^{-1}}\).

The absorbance increases and eventually reaches a constant value.

This supports the interpretation that the concentration of the red complex has increased and then stabilised.

The plateau is particularly important.

It indicates that the concentration of the coloured species has become constant. It does not indicate that the forward and reverse reactions have stopped.

They continue at equal rates.

05A contrasting case: Why dilution makes the red colour fade

Now consider a different disturbance.

An equilibrium mixture contains the same three species, but instead of adding iron(III) nitrate, a student adds water.

Predict what happens to the red colour.

It is tempting to say that dilution simply slows the reaction. However, that doesn’t explain which direction the equilibrium moves.

We need to compare the forward and reverse rates again.

The immediate effect of adding water

Adding water increases the total volume of the solution.

The concentrations of all three dissolved species decrease.

If the volume doubles, each concentration is initially halved.

This includes the red complex, \(\mathrm{FeSCN}^{2+}\).

Therefore, its colour intensity decreases immediately because there is now less red complex per unit volume.

But something else happens.

Dilution also changes the relative rates of the forward and reverse reactions.

Calculate the direction of the equilibrium change

For the reaction:

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

the equilibrium constant expression is:

\[
K_c=
\frac{[\mathrm{FeSCN}^{2+}]}
{[\mathrm{Fe}^{3+}][\mathrm{SCN}^{-}]}
\]

The reaction quotient, \(Q_c\), uses the same expression but the concentrations at the particular moment being considered.

At equilibrium:

\[
Q_c=K_c
\]

Suppose the original equilibrium concentrations are:

\[
[\mathrm{Fe}^{3+}]=1.00\times10^{-3}\ \mathrm{M}
\]

\[
[\mathrm{SCN}^{-}]=2.00\times10^{-4}\ \mathrm{M}
\]

\[
[\mathrm{FeSCN}^{2+}]=2.00\times10^{-4}\ \mathrm{M}
\]

Here, \(\mathrm{M}\) means \(\mathrm{mol\,L^{-1}}\).

The equilibrium constant is therefore:

\[
K_c=
\frac{2.00\times10^{-4}}
{(1.00\times10^{-3})(2.00\times10^{-4})}
\]

\[
K_c=1.00\times10^3\ \mathrm{L\,mol^{-1}}
\]

Now double the volume by adding water.

The new concentrations immediately after dilution are:

Calculate the new reaction quotient:

\[
Q_c=
\frac{1.00\times10^{-4}}
{(5.00\times10^{-4})(1.00\times10^{-4})}
\]

\[
Q_c=2.00\times10^3\ \mathrm{L\,mol^{-1}}
\]

Compare the values:

\[
Q_c>K_c
\]

This means the system has a greater product-to-reactant concentration ratio than it should at equilibrium.

The reverse reaction therefore occurs faster than the forward reaction overall.

More \(\mathrm{FeSCN}^{2+}\) dissociates to form \(\mathrm{Fe}^{3+}\) and \(\mathrm{SCN}^{-}\).

The concentration of the red complex decreases further, making the solution even paler.

Why do the rates become unequal?

In a simplified one-step mass-action model, the forward rate depends on encounters between \(\mathrm{Fe}^{3+}\) and \(\mathrm{SCN}^{-}\), while the reverse rate depends on the concentration of \(\mathrm{FeSCN}^{2+}\).

If all three concentrations are halved:

• The forward rate falls to one-quarter of its original value in this model.
• The reverse rate falls to one-half of its original value.

The reverse rate is therefore greater than the forward rate.

The precise microscopic rate law depends on the reaction mechanism, so these numerical rate factors are a model rather than a universal rule. The comparison of \(Q_c\) with \(K_c\), however, establishes the net direction without requiring that simplified rate law.

This example reveals an important distinction:

Dilution first decreases the concentration of the coloured species directly. The subsequent reverse reaction decreases it further.

Both effects contribute to the fading colour.

06A reliable method for explaining any equilibrium colour change

You can now build an explanation using five steps.

A useful way to check your answer is to ask:

1. Have I named the actual coloured chemical species?
2. Have I identified what was added, removed, or changed?
3. Have I compared the forward and reverse reaction rates?
4. Have I explained which species is produced or consumed?
5. Have I connected that concentration change to the observed colour?
6. Have I explained why the reaction rates eventually become equal again?

If the question provides equilibrium concentrations and \(K_c\), you can also use \(Q_c\) as a mathematical check:

Remember that \(K_c\) remains constant when concentration changes at constant temperature. It is the reaction quotient, \(Q_c\), that changes immediately after the disturbance.

For a broader explanation of how concentration affects reaction rates, see How to Explain Reaction-Rate Changes in HSC Chemistry. The important extra step here is comparing both directions of a reversible reaction.

07Questions and solutions

The following questions move from a standard HSC explanation towards situations where colour, concentration, and equilibrium don’t give you the same obvious answer.

Question 1

An acidic equilibrium mixture contains \(\mathrm{Fe}^{3+}\), \(\mathrm{SCN}^{-}\), and \(\mathrm{FeSCN}^{2+}\).

A student divides the mixture into two equal samples. They add a small amount of concentrated iron(III) nitrate to Sample A and the same volume of acidified water to Sample B.

Assume that the volume changes are negligible and both samples remain at constant temperature.

Explain which sample develops the more intense red colour. Include the changes to the forward and reverse reaction rates and identify the species responsible for the colour.

Solution 1

Sample A develops the more intense red colour because adding iron(III) nitrate increases the concentration of \(\mathrm{Fe}^{3+}\), causing the forward reaction rate to become greater than the reverse reaction rate.

The equilibrium reaction is:

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

Step 1: Identify the coloured species.

The red colour is produced by \(\mathrm{FeSCN}^{2+}\), not by the iron(III) ions themselves.

Step 2: Analyse Sample A.

Adding iron(III) nitrate increases \([\mathrm{Fe}^{3+}]\).

This increases the forward reaction rate, while the reverse reaction rate initially remains unchanged because \([\mathrm{FeSCN}^{2+}]\) has not yet changed.

Therefore:

\[
r_{\mathrm{forward}}>r_{\mathrm{reverse}}
\]

The forward reaction produces additional \(\mathrm{FeSCN}^{2+}\), increasing its concentration and intensifying the red colour.

Step 3: Analyse Sample B.

Adding water dilutes the equilibrium mixture.

The concentration of the red complex decreases immediately. The dilution also changes the relative forward and reverse rates, causing a subsequent net reverse reaction.

Therefore, Sample B becomes paler rather than developing a deeper red colour.

The two samples respond differently because they experience different disturbances.

Question 2

An equilibrium mixture has the following concentrations at a constant temperature:

\[
[\mathrm{Fe}^{3+}]=1.50\times10^{-3}\ \mathrm{M}
\]

\[
[\mathrm{SCN}^{-}]=4.00\times10^{-4}\ \mathrm{M}
\]

\[
[\mathrm{FeSCN}^{2+}]=4.80\times10^{-4}\ \mathrm{M}
\]

The equilibrium constant is:

\[
K_c=8.00\times10^2\ \mathrm{L\,mol^{-1}}
\]

A student doubles the volume of the solution by adding water.

a) Calculate the new concentrations immediately after dilution.

b) Calculate the new reaction quotient, \(Q_c\).

c) Determine whether the forward or reverse reaction occurs at a greater rate.

d) Explain the immediate colour change and the subsequent change as equilibrium is re-established.

Solution 2

The solution becomes paler immediately after dilution and then fades further because the reverse reaction becomes faster than the forward reaction.

Step 1: Calculate the new concentrations.

The volume doubles, so the concentration of each dissolved species is halved.

For iron(III) ions:

\[
[\mathrm{Fe}^{3+}]_{\mathrm{new}}
=\frac{1.50\times10^{-3}}{2}
\]

\[
=7.50\times10^{-4}\ \mathrm{M}
\]

For thiocyanate ions:

\[
[\mathrm{SCN}^{-}]_{\mathrm{new}}
=\frac{4.00\times10^{-4}}{2}
\]

\[
=2.00\times10^{-4}\ \mathrm{M}
\]

For the red complex:

\[
[\mathrm{FeSCN}^{2+}]_{\mathrm{new}}
=\frac{4.80\times10^{-4}}{2}
\]

\[
=2.40\times10^{-4}\ \mathrm{M}
\]

Step 2: Calculate the new reaction quotient.

Use:

\[
Q_c=
\frac{[\mathrm{FeSCN}^{2+}]}
{[\mathrm{Fe}^{3+}][\mathrm{SCN}^{-}]}
\]

Substitute the new concentrations:

\[
Q_c=
\frac{2.40\times10^{-4}}
{(7.50\times10^{-4})(2.00\times10^{-4})}
\]

\[
Q_c=
\frac{2.40\times10^{-4}}
{1.50\times10^{-7}}
\]

\[
Q_c=1.60\times10^3\ \mathrm{L\,mol^{-1}}
\]

Step 3: Compare the reaction quotient with the equilibrium constant.

\[
Q_c=1.60\times10^3
\]

\[
K_c=8.00\times10^2
\]

Therefore:

\[
Q_c>K_c
\]

The reverse reaction occurs faster than the forward reaction overall.

Step 4: Explain the colour change.

Immediately after dilution, the concentration of \(\mathrm{FeSCN}^{2+}\) is halved. This directly reduces the intensity of the red colour.

The subsequent net reverse reaction consumes more \(\mathrm{FeSCN}^{2+}\), producing \(\mathrm{Fe}^{3+}\) and \(\mathrm{SCN}^{-}\).

The concentration of the red complex decreases further until a new equilibrium is established.

The key distinction is that dilution causes an immediate concentration change, followed by an additional chemical change as the system re-establishes equilibrium.

Question 3

A student investigates the effect of adding iron(III) nitrate to an equilibrium mixture using a colorimeter.

The initial baseline-corrected absorbance of the mixture is 0.240.

After adding iron(III) nitrate, the test sample reaches a stable raw absorbance of 0.310.

A matched iron(III)-only reagent blank contributes an absorbance of 0.020 at the same wavelength.

A control sample receiving the same volume of acidified solvent remains at an absorbance of 0.240.

The calibration relationship for the red complex is:

\[
A_{\mathrm{complex}}
=(2.00\times10^3)c
\]

where \(c\) is measured in \(\mathrm{mol\,L^{-1}}\).

a) Calculate the increase in the concentration of \(\mathrm{FeSCN}^{2+}\), accounting for the reagent blank.

b) Explain why the control and reagent blank are both useful.

c) Explain what the stable final absorbance indicates about the forward and reverse reaction rates.

Solution 3

The concentration of \(\mathrm{FeSCN}^{2+}\) increases by \(2.50\times10^{-5}\ \mathrm{mol\,L^{-1}}\).

The control and reagent blank allow the student to distinguish an actual increase in the red complex from other changes in measured absorbance.

Step 1: Correct the final absorbance for the reagent blank.

The final raw absorbance is 0.310.

The iron(III)-only blank contributes 0.020.

Therefore:

\[
A_{\mathrm{final,corrected}}
=0.310-0.020
\]

\[
=0.290
\]

Step 2: Calculate the increase in absorbance attributable to the red complex.

The initial baseline-corrected absorbance is 0.240.

Therefore:

\[
\Delta A=0.290-0.240
\]

\[
\Delta A=0.050
\]

Step 3: Convert the absorbance increase into a concentration increase.

Using:

\[
A=kc
\]

Rearrange:

\[
\Delta c=\frac{\Delta A}{k}
\]

Substitute:

\[
\Delta c=
\frac{0.050}{2.00\times10^3}
\]

\[
\Delta c=2.50\times10^{-5}\ \mathrm{mol\,L^{-1}}
\]

The concentration of the red complex has increased by this amount.

Step 4: Explain the purpose of the controls.

The solvent control receives the same volume of liquid as the test sample. It helps identify changes caused by dilution, handling, or measurement over time.

The iron(III)-only blank accounts for background absorption from the added iron(III) solution.

Without the blank, the student could incorrectly attribute all the increase in absorbance to the formation of \(\mathrm{FeSCN}^{2+}\).

Step 5: Interpret the stable final absorbance.

A stable absorbance indicates that the concentration of the absorbing red complex is no longer changing.

The forward and reverse reactions have reached equal rates, so the system is at dynamic equilibrium.

The reactions continue, but there is no net change in the concentration of \(\mathrm{FeSCN}^{2+}\).

Question 4

An equilibrium mixture initially contains:

\[
[\mathrm{Fe}^{3+}]=1.00\times10^{-3}\ \mathrm{M}
\]

\[
[\mathrm{SCN}^{-}]=2.00\times10^{-4}\ \mathrm{M}
\]

\[
[\mathrm{FeSCN}^{2+}]=2.00\times10^{-4}\ \mathrm{M}
\]

At the experimental temperature:

\[
K_c=1.00\times10^3\ \mathrm{L\,mol^{-1}}
\]

A student first doubles the volume by adding water. They then add concentrated iron(III) nitrate.

Immediately after both additions, the concentrations are:

\[
[\mathrm{Fe}^{3+}]=1.50\times10^{-3}\ \mathrm{M}
\]

\[
[\mathrm{SCN}^{-}]=1.00\times10^{-4}\ \mathrm{M}
\]

\[
[\mathrm{FeSCN}^{2+}]=1.00\times10^{-4}\ \mathrm{M}
\]

Assume the iron(III) nitrate addition has a negligible volume effect and the temperature remains constant.

a) Explain the immediate colour change caused by the first addition of water.

b) Calculate \(Q_c\) after both additions.

c) Determine the net direction of the reaction after the second addition.

d) Explain what happens to the red colour as the system approaches its new equilibrium.

Solution 4

The red colour first becomes paler because of dilution. After iron(III) nitrate is added, the forward reaction becomes favoured, increasing the concentration of the red complex.

Step 1: Explain the effect of the first addition.

Adding water doubles the volume.

The concentration of every dissolved species is therefore halved immediately.

In particular:

\[
[\mathrm{FeSCN}^{2+}]
=1.00\times10^{-4}\ \mathrm{M}
\]

The lower concentration of the red complex makes the solution paler.

Step 2: Identify the effect of the second addition.

Adding iron(III) nitrate increases the concentration of \(\mathrm{Fe}^{3+}\).

The final concentrations provided in the question account for both disturbances.

We must use these final concentrations rather than the concentrations immediately after dilution alone.

Step 3: Calculate the reaction quotient.

\[
Q_c=
\frac{[\mathrm{FeSCN}^{2+}]}
{[\mathrm{Fe}^{3+}][\mathrm{SCN}^{-}]}
\]

Substitute:

\[
Q_c=
\frac{1.00\times10^{-4}}
{(1.50\times10^{-3})(1.00\times10^{-4})}
\]

\[
Q_c=
\frac{1.00\times10^{-4}}
{1.50\times10^{-7}}
\]

\[
Q_c=6.67\times10^2\ \mathrm{L\,mol^{-1}}
\]

Step 4: Determine the direction of the reaction.

Compare:

\[
Q_c=6.67\times10^2
\]

\[
K_c=1.00\times10^3
\]

Therefore:

\[
Q_c<K_c
\]

The forward reaction occurs faster than the reverse reaction overall.

The system produces more \(\mathrm{FeSCN}^{2+}\) as it approaches its new equilibrium.

Step 5: Explain the final colour change.

The second addition increases \([\mathrm{Fe}^{3+}]\), causing net forward reaction.

This consumes \(\mathrm{Fe}^{3+}\) and \(\mathrm{SCN}^{-}\) and produces additional \(\mathrm{FeSCN}^{2+}\).

Consequently, the red colour becomes more intense relative to its diluted state.

This question demonstrates why you cannot automatically assume that dilution determines the final direction of an equilibrium reaction. A second disturbance can change the result.

Question 5

Two separate cuvettes, A and B, contain the iron(III)-thiocyanate equilibrium system at the same temperature.

Both have the same path length and the same measured, baseline-corrected absorbance attributable to the red complex.

Cuvette A is at equilibrium and contains:

\[
[\mathrm{Fe}^{3+}]=1.00\times10^{-3}\ \mathrm{M}
\]

\[
[\mathrm{SCN}^{-}]=2.00\times10^{-4}\ \mathrm{M}
\]

\[
[\mathrm{FeSCN}^{2+}]=2.00\times10^{-4}\ \mathrm{M}
\]

Cuvette B initially contains the same concentrations of \(\mathrm{SCN}^{-}\) and \(\mathrm{FeSCN}^{2+}\), but its iron(III) concentration is:

\[
[\mathrm{Fe}^{3+}]=2.00\times10^{-3}\ \mathrm{M}
\]

The equilibrium constant at this temperature is:

\[
K_c=1.00\times10^3\ \mathrm{L\,mol^{-1}}
\]

A student argues:

Both cuvettes have the same red absorbance, so both must be at equilibrium.

Evaluate this claim.

Your answer must:

a) Determine whether each cuvette is initially at equilibrium.

b) Calculate the reaction quotient for each cuvette.

c) Explain how the forward and reverse rates compare in cuvette B.

d) Predict which cuvette has the greater red-complex concentration once both systems have reached equilibrium.

e) Explain why measuring the concentration of only the coloured species is insufficient to establish whether an equilibrium system is at equilibrium.

Solution 5

Only cuvette A is initially at equilibrium. Cuvette B has the same concentration of red complex but a greater concentration of free iron(III) ions, so its reaction quotient is below the equilibrium constant and its forward reaction occurs faster than its reverse reaction.

Step 1: Calculate the reaction quotient for cuvette A.

Use:

\[
Q_c=
\frac{[\mathrm{FeSCN}^{2+}]}
{[\mathrm{Fe}^{3+}][\mathrm{SCN}^{-}]}
\]

Substitute the concentrations in A:

\[
Q_{c,A}=
\frac{2.00\times10^{-4}}
{(1.00\times10^{-3})(2.00\times10^{-4})}
\]

\[
Q_{c,A}=1.00\times10^3\ \mathrm{L\,mol^{-1}}
\]

Since:

\[
Q_{c,A}=K_c
\]

cuvette A is at equilibrium.

Its forward and reverse reaction rates are equal.

Step 2: Calculate the reaction quotient for cuvette B.

The concentration of \(\mathrm{FeSCN}^{2+}\) is unchanged, but the concentration of \(\mathrm{Fe}^{3+}\) has doubled.

Therefore:

\[
Q_{c,B}=
\frac{2.00\times10^{-4}}
{(2.00\times10^{-3})(2.00\times10^{-4})}
\]

\[
Q_{c,B}=
\frac{2.00\times10^{-4}}
{4.00\times10^{-7}}
\]

\[
Q_{c,B}=5.00\times10^2\ \mathrm{L\,mol^{-1}}
\]

Compare this with the equilibrium constant:

\[
Q_{c,B}<K_c
\]

Cuvette B is not at equilibrium.

Step 3: Explain the reaction rates in cuvette B.

Because \(Q_c<K_c\), the forward reaction occurs faster than the reverse reaction overall.

The reaction:

\[
\ce{Fe^3+ + SCN^- -> FeSCN^2+}
\]

proceeds with a net forward rate.

This consumes free \(\mathrm{Fe}^{3+}\) and \(\mathrm{SCN}^{-}\), producing additional \(\mathrm{FeSCN}^{2+}\).

The reverse reaction becomes increasingly significant as more red complex forms, until the two rates become equal.

Step 4: Predict the final red-complex concentrations.

Cuvette A is already at equilibrium, so its concentration of \(\mathrm{FeSCN}^{2+}\) remains:

\[
2.00\times10^{-4}\ \mathrm{M}
\]

Cuvette B starts with exactly the same concentration, but its forward reaction proceeds faster than its reverse reaction.

Therefore, its concentration of \(\mathrm{FeSCN}^{2+}\) must increase before equilibrium is reached.

Its final red-complex concentration will be greater than that of cuvette A.

This conclusion follows directly from the fact that B starts with \(Q_c<K_c\).

Step 5: Explain why the student’s reasoning is incorrect.

The student’s mistake is assuming that equal colour means equal equilibrium conditions.

However, the measured red-complex absorbance tells us about the concentration of \(\mathrm{FeSCN}^{2+}\), not the concentrations of every species in the equilibrium system.

Cuvettes A and B have identical initial concentrations of the coloured complex, but different concentrations of free \(\mathrm{Fe}^{3+}\).

That difference changes the reaction quotient and therefore the relative forward and reverse reaction rates.

The two cuvettes can look equally red while only one is at equilibrium.

This is the deeper principle: a colour measurement can tell you about a coloured species, but it cannot establish dynamic equilibrium without considering the other species in the system.

08Where to take this next

Once you can explain a colour change through actual species and reaction rates, the next useful step is understanding how equilibrium composition relates to the amount of product obtained from a reversible reaction.

A reaction reaching equilibrium does not necessarily mean that all reactants have been converted into products. This distinction becomes important when interpreting incomplete reactions and calculating theoretical and actual yields.

See Theoretical vs Actual Yield: Calculate Yield and Explain Losses for the related stoichiometric calculations.

For colour-change questions, though, keep returning to the same reasoning: identify the coloured species, establish the system boundary, compare forward and reverse rates, and explain how the resulting concentration change produces the observation.