How to Solve Equilibrium After Removing a Reactant

An equilibrium mixture contains phosphorus pentachloride, \(\ce{PCl5}\), phosphorus trichloride, \(\ce{PCl3}\), and chlorine gas, \(\ce{Cl2}\). You selectively remove some of the \(\ce{PCl5}\) while keeping the temperature and volume constant.

What happens next? The reaction shifts backwards, producing some \(\ce{PCl5}\) from \(\ce{PCl3}\) and \(\ce{Cl2}\). However, it does not replace everything that was removed. It moves towards a new equilibrium.

To calculate that new equilibrium, you need to treat the concentrations immediately after removal as a new starting point, calculate the reaction quotient, and construct an initial-change-equilibrium (ICE) table.

The method is straightforward once you understand which concentrations belong in each row.

01Why removing a reactant makes the reaction shift backwards

Consider the general equilibrium:

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

The equilibrium constant expression is:

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

At equilibrium, the reaction quotient, \(Q_c\), equals the equilibrium constant, \(K_c\).

The reaction quotient has exactly the same mathematical expression as \(K_c\), but uses the concentrations at the current moment rather than necessarily at equilibrium.

This gives us a useful way to predict what happens after a disturbance.

Now consider what happens when you remove reactant \(A\).

Because \(A\) appears in the denominator of the equilibrium expression, decreasing its concentration makes the denominator smaller.

The value of \(Q_c\) therefore increases.

Immediately after removing \(A\):

\[ Q_c>K_c \]

The reaction shifts backwards, consuming products and producing more reactants until:

\[ Q_c=K_c \]

The system has reached its new equilibrium.

Why doesn’t the equilibrium constant change?

This is an important distinction.

Removing a reactant changes the reaction quotient, \(Q_c\), but does not change the equilibrium constant, \(K_c\), provided the temperature stays constant.

The equilibrium constant is determined by temperature. Changing concentration changes the reaction’s immediate position relative to equilibrium, not the value of \(K_c\).

You can remember the difference like this:

• Concentration disturbance: changes \(Q_c\).
• Temperature disturbance: changes \(K_c\).

02A colour experiment: removing a reactant from equilibrium

A useful way to understand this process is through the cobalt(II) chloride equilibrium.

The reaction is:

\[ \ce{[Co(H2O)6]^2+(aq) + 4Cl^-(aq)} \]

\[ \ce{<=> [CoCl4]^2-(aq) + 6H2O(l)} \]

The two cobalt complexes have different colours:

• \(\ce{[Co(H2O)6]^2+}\) is pink.
• \(\ce{[CoCl4]^2-}\) is blue.

The chloride ions are reactants in the forward reaction.

How could we test what happens when chloride is removed?

Start with an equilibrium mixture containing both cobalt complexes.

Divide equal portions into two clear test tubes.

Add silver nitrate solution, \(\ce{AgNO3}\), to the first tube. Add an equal volume of a suitable nitrate control solution to the second.

Silver ions react with chloride ions:

\[ \ce{Ag+(aq) + Cl^-(aq) -> AgCl(s)} \]

Silver chloride forms a white precipitate. This removes free chloride ions from the equilibrium mixture.

What should happen?

The cobalt equilibrium has lost some of its chloride reactant.

Its equilibrium expression is:

\[ K_c= \frac{[\mathrm{CoCl_{4}}^{2-}]} {[\ce{Co(H2O)6^2+}][\mathrm{Cl}^{-}]^4} \]

Liquid water is omitted from this expression.

When chloride concentration decreases, the denominator decreases. The immediate reaction quotient becomes greater than the equilibrium constant.

Therefore, the reaction shifts backwards:

\[ \ce{[CoCl4]^2- -> [Co(H2O)6]^2+ + 4Cl^-} \]

The characteristic result is an increase in the pink colour as more of the pink cobalt complex forms.

The white precipitate is evidence that chloride has been removed. The pink colour change is evidence of the subsequent equilibrium shift.

Why does the control matter?

A simple comparison between a tube receiving silver nitrate and an untouched tube has a weakness.

Adding a solution also changes the volume, potentially diluting the equilibrium mixture. A colour change might therefore be partly caused by dilution rather than selective chloride removal.

A better design uses a control receiving the same volume of a nitrate solution that does not precipitate chloride. This controls for the added volume and nitrate ions, although it does not perfectly reproduce every ionic-strength effect.

A spectrophotometer could provide a more quantitative comparison by measuring changes in the light absorbed by the cobalt complexes.

This is a qualitative laboratory demonstration to analyse, not a home experiment. Cobalt salts and silver nitrate require appropriate laboratory supervision.

A small calculation: how much can Qc change?

Suppose the initial chloride concentration is:

\[ [\mathrm{Cl}^{-}]=0.80\ \mathrm{mol\,L^{-1}} \]

After selective removal, it becomes:

\[ [\mathrm{Cl}^{-}]=0.60\ \mathrm{mol\,L^{-1}} \]

Assume the other concentrations have not yet changed.

The initial mixture was at equilibrium, so \(Q_c=K_c\).

The new reaction quotient relative to the old equilibrium constant is:

\[ \frac{Q_{c,\mathrm{new}}}{K_c} = \left(\frac{0.80}{0.60}\right)^4 \]

\[ =3.16 \]

Therefore:

\[ Q_{c,\mathrm{new}}=3.16K_c \]

The reaction quotient is now substantially greater than the equilibrium constant.

Notice the fourth power. The balanced equation contains four chloride ions, so the chloride concentration appears to the fourth power in the denominator.

This is why the balanced chemical equation matters when predicting equilibrium changes.

The values above are illustrative, not experimental measurements.

03How to construct an ICE table after removing a reactant

An ICE table organises your calculation into three rows:

• Initial (I): concentrations immediately after the disturbance.
• Change (C): concentration changes as the reaction moves towards equilibrium.
• Equilibrium (E): final concentrations after the shift.

The most important detail is that the initial row must represent the mixture after the reactant has been removed.

Do not reuse the original equilibrium concentrations as your new initial concentrations.

Step 1: Calculate the original equilibrium constant

Use the original equilibrium concentrations to calculate \(K_c\), if it has not already been supplied.

You need this value to determine the new equilibrium.

Step 2: Construct the new initial row

Apply the removal first.

If you remove \(0.20\ \mathrm{mol\,L^{-1}}\) of reactant \(A\), subtract that amount from its original concentration.

Leave the other concentrations unchanged immediately after selective removal, provided the volume is constant and no other species are affected.

Step 3: Calculate the new reaction quotient

Substitute the post-removal concentrations into the equilibrium expression.

Compare \(Q_c\) with \(K_c\).

If a reactant has been selectively removed, you will generally find:

\[ Q_c>K_c \]

The reaction therefore shifts backwards.

Step 4: Fill in the change row

For the general reaction:

\[ A+B\rightleftharpoons C \]

a backwards reaction:

• Increases \(A\).
• Increases \(B\).
• Decreases \(C\).

Let \(x\) represent the amount of reaction progress in the backwards direction, measured in \(\mathrm{mol\,L^{-1}}\).

Your ICE table becomes:

The coefficients in the balanced equation determine the multipliers of \(x\).

For example, if the reaction is:

\[ 2A+B\rightleftharpoons C \]

the backwards changes become:

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

Always follow the balanced equation.

Step 5: Substitute into Kc and solve

Use the equilibrium row to construct the equilibrium expression.

For the general reaction:

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

Substitute the expressions containing \(x\), then solve for \(x\).

Depending on the reaction, this may produce a linear or quadratic equation.

Step 6: Reject non-physical roots

This is where you need to check your algebra against chemistry.

Every equilibrium concentration must be zero or greater.

For example, if:

\[ [C]_{\mathrm{eq}}=0.20-x \]

then:

\[ x\leq 0.20 \]

A mathematical solution of \(x=0.35\) would give:

\[ [C]_{\mathrm{eq}}=0.20-0.35 \]

\[ =-0.15\ \mathrm{mol\,L^{-1}} \]

A negative concentration is impossible, so reject that root.

Do not accept a quadratic solution just because your calculator produced it. Check every root against the equilibrium table and the amount of each substance initially available.

A negative \(x\) is also inconsistent with a direction you have already defined as positive. If you choose the direction correctly using \(Q_c\) and \(K_c\), the physically relevant extent should agree with that direction.

For a fuller treatment of ICE tables and equilibrium calculations in rigid gas vessels, see How to Calculate Gas Equilibrium at Constant Volume.

04Worked examples

05Important distinction: selective removal versus removing the whole mixture

The method above assumes that you selectively remove one reacting species.

Not every type of removal does this.

Case 1: Selectively removing one reactant

Suppose you remove some \(A\) from:

\[ A+B\rightleftharpoons C \]

while keeping the volume constant.

Only \([A]\) decreases immediately.

The denominator of \(Q_c\) becomes smaller, so:

\[ Q_c>K_c \]

The reaction shifts backwards.

Case 2: Withdrawing a sample of a homogeneous solution

Now imagine taking \(25\%\) of a well-mixed equilibrium solution out of a beaker.

You have removed \(25\%\) of every dissolved species, but you have also removed \(25\%\) of the solution volume.

The number of moles decreases, but the concentrations remain unchanged.

For example:

Because the concentrations have not changed:

\[ Q_c=K_c \]

There is no equilibrium shift.

This is a very different situation from selectively removing a reactant.

The same distinction between moles and concentration appears in pH After Dilution and Acid-Base Mixing | HSC Chemistry.

Case 3: Removing part of a pure solid reactant

There is another important exception.

Pure solids are omitted from equilibrium expressions.

For example, consider:

\[ \ce{CaCO3(s) <=> CaO(s) + CO2(g)} \]

The equilibrium expression is:

\[ K_c=[\ce{CO2}] \]

Removing some solid \(\ce{CaCO3}\) does not change its concentration in the equilibrium expression, provided some solid remains.

Therefore, simply removing part of the solid does not cause an equilibrium shift.

The key question is not just whether a reactant has been removed. It is whether its concentration appears in the equilibrium expression and has actually changed.

06Questions and solutions

Question 1

Consider the equilibrium:

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

At \(25^\circ\mathrm{C}\), the system reaches equilibrium with:

\[ [A]=0.400\ \mathrm{mol\,L^{-1}} \]

\[ [B]=0.500\ \mathrm{mol\,L^{-1}} \]

\[ [C]=0.600\ \mathrm{mol\,L^{-1}} \]

A selective process removes \(0.200\ \mathrm{mol\,L^{-1}}\) of \(B\), without changing the volume or temperature.

Calculate:

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

b) The new reaction quotient, \(Q_c\).

c) The direction in which the reaction shifts.

Solution 1

The reaction shifts backwards because the new reaction quotient is greater than the equilibrium constant.

Step 1: Calculate the original equilibrium constant.

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

\[ K_c=\frac{0.600}{(0.400)(0.500)} \]

\[ \boxed{K_c=3.00} \]

Step 2: Calculate the concentration of \(B\) immediately after removal.

\[ [B]_{\mathrm{initial}}=0.500-0.200 \]

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

The other concentrations remain unchanged.

Step 3: Calculate the new reaction quotient.

\[ Q_c=\frac{0.600}{(0.400)(0.300)} \]

\[ \boxed{Q_c=5.00} \]

Step 4: Compare \(Q_c\) with \(K_c\).

\[ 5.00>3.00 \]

Therefore:

\[ Q_c>K_c \]

The reaction shifts backwards.

In the ICE table, the change row would be:

The reverse reaction produces more \(A\) and \(B\), while consuming \(C\), until equilibrium is restored.

Question 2

The following equilibrium is established in an aqueous cobalt chloride mixture:

\[ \ce{[Co(H2O)6]^2+ + 4Cl^- <=> [CoCl4]^2- + 6H2O} \]

A student divides the mixture between two test tubes.

• Test tube X receives silver nitrate solution.
• Test tube Y receives an equal volume of sodium nitrate solution.

Test tube X develops a white precipitate and becomes noticeably pinker. Test tube Y does not show the same colour change.

The student concludes:

“The equilibrium constant must have decreased because chloride ions were removed.”

Evaluate this statement using the equilibrium expression, the reaction quotient, and the experimental observations.

Solution 2

The student’s conclusion is incorrect. The equilibrium constant has not decreased; instead, removing chloride increases the reaction quotient, causing the equilibrium to shift backwards.

Step 1: Identify the role of chloride.

Chloride ions are reactants in the forward reaction.

Silver ions remove them through precipitation:

\[ \ce{Ag+ + Cl^- -> AgCl(s)} \]

The formation of solid silver chloride reduces the concentration of free chloride ions.

Step 2: Explain the effect on the reaction quotient.

The equilibrium expression is:

\[ K_c= \frac{[\mathrm{CoCl_{4}}^{2-}]} {[\ce{Co(H2O)6^2+}][\mathrm{Cl}^{-}]^4} \]

When chloride concentration decreases, the denominator decreases.

Consequently, the new reaction quotient increases:

\[ Q_c>K_c \]

Step 3: Predict the equilibrium shift.

The reaction shifts backwards:

\[ \ce{[CoCl4]^2- -> [Co(H2O)6]^2+ + 4Cl^-} \]

This produces more pink \(\ce{[Co(H2O)6]^2+}\), explaining the observed colour change.

Step 4: Explain why the equilibrium constant remains unchanged.

The temperature has not changed.

Therefore, \(K_c\) remains constant. Removing chloride changes the current concentrations and hence \(Q_c\), not the equilibrium constant.

Step 5: Explain the purpose of the control.

The sodium nitrate solution controls for the addition of extra liquid and nitrate ions. It helps distinguish the effect of chloride removal from a simple change in volume.

However, the control does not perfectly reproduce the ionic-strength changes in the silver nitrate tube.

The observations are consistent with the predicted equilibrium shift, rather than a decrease in \(K_c\).

Question 3

A simplified equilibrium is represented by:

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

At \(25^\circ\mathrm{C}\), the original equilibrium concentrations are:

\[ [A]=0.600\ \mathrm{mol\,L^{-1}} \]

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

Some of reactant \(A\) is selectively removed. The system then reaches a new equilibrium, where measurements show:

\[ [A]=0.400\ \mathrm{mol\,L^{-1}} \]

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

Assuming constant temperature and volume, calculate the concentration of \(A\) that was removed.

Solution 3

The concentration of \(A\) removed was:

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

The important step is recognising that the final concentration of \(A\) is not the concentration immediately after removal.

Step 1: Calculate the original equilibrium constant.

\[ K_c=\frac{[B]}{[A]} \]

\[ K_c=\frac{0.300}{0.600} \]

\[ K_c=0.500 \]

Step 2: Work out the change in \(B\).

Only \(A\) was removed.

Therefore, the concentration of \(B\) immediately after removal must still have been:

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

At the new equilibrium:

\[ [B]_{\mathrm{eq}}=0.200\ \mathrm{mol\,L^{-1}} \]

Therefore, \(B\) decreased by:

\[ x=0.300-0.200 \]

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

Step 3: Determine the backwards reaction change.

Because \(A\) and \(B\) have a 1:1 mole ratio, the backwards reaction increases \(A\) by the same amount that it decreases \(B\).

Therefore:

\[ [A]_{\mathrm{eq}}=[A]_{\mathrm{initial}}+0.100 \]

Rearrange:

\[ 0.400=[A]_{\mathrm{initial}}+0.100 \]

\[ [A]_{\mathrm{initial}}=0.300 \]

Step 4: Calculate the amount of \(A\) removed.

The original concentration was \(0.600\ \mathrm{mol\,L^{-1}}\).

The concentration immediately after removal was \(0.300\ \mathrm{mol\,L^{-1}}\).

Therefore:

\[ \text{Amount removed} =0.600-0.300 \]

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

A tempting mistake is to subtract the final concentration from the original concentration:

\[ 0.600-0.400=0.200 \]

That gives the net decrease in \(A\), not the amount initially removed.

The backwards reaction regenerated \(0.100\ \mathrm{mol\,L^{-1}}\) of \(A\). You must account for that restoration when working backwards from the final equilibrium.

Question 4

Consider:

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

The system is initially at equilibrium with:

\[ [A]=0.400\ \mathrm{mol\,L^{-1}} \]

\[ [B]=0.600\ \mathrm{mol\,L^{-1}} \]

\[ [C]=0.480\ \mathrm{mol\,L^{-1}} \]

Two changes occur simultaneously at constant temperature and volume:

• \(0.200\ \mathrm{mol\,L^{-1}}\) of \(A\) is selectively removed.
• \(0.600\ \mathrm{mol\,L^{-1}}\) of \(B\) is added.

Determine whether the equilibrium shifts forwards, shifts backwards, or remains at equilibrium.

Explain why considering only the removal of \(A\) would produce an incorrect prediction.

Solution 4

There is no net equilibrium shift. The simultaneous changes leave the reaction quotient equal to the equilibrium constant.

Step 1: Calculate the original equilibrium constant.

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

\[ K_c=\frac{0.480}{(0.400)(0.600)} \]

\[ \boxed{K_c=2.00} \]

Step 2: Calculate the new concentrations immediately after both changes.

For \(A\):

\[ [A]=0.400-0.200=0.200 \]

For \(B\):

\[ [B]=0.600+0.600=1.200 \]

For \(C\):

\[ [C]=0.480 \]

Step 3: Calculate the new reaction quotient.

\[ Q_c=\frac{0.480}{(0.200)(1.200)} \]

\[ \boxed{Q_c=2.00} \]

Step 4: Compare the reaction quotient with the equilibrium constant.

\[ Q_c=K_c \]

Therefore, the system remains at equilibrium.

There is no net reaction in either direction.

Why is this result surprising?

If you considered only the removal of \(A\), you would predict a backwards shift because the denominator of \(Q_c\) decreases.

However, the simultaneous addition of \(B\) increases the other term in the denominator.

In fact, the product of the two reactant concentrations remains unchanged:

Before the disturbance:

\[ [A][B]=(0.400)(0.600)=0.240 \]

After the disturbance:

\[ [A][B]=(0.200)(1.200)=0.240 \]

The two changes exactly compensate in the equilibrium expression.

This is why you must calculate \(Q_c\) after applying all simultaneous disturbances, rather than predicting the effect of each one independently.

Question 5

A homogeneous aqueous equilibrium follows:

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

At \(25^\circ\mathrm{C}\), a \(1.00\ \mathrm{L}\) solution reaches equilibrium with:

\[ [A]=0.600\ \mathrm{mol\,L^{-1}} \]

\[ [B]=0.400\ \mathrm{mol\,L^{-1}} \]

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

The following events then occur:

1. A \(0.250\ \mathrm{L}\) sample of the homogeneous equilibrium solution is withdrawn.
2. From the remaining solution, a selective process removes \(0.150\ \mathrm{mol}\) of \(A\).
3. The selective removal does not change the solution volume or affect \(B\) or \(C\).
4. The system is allowed to re-establish equilibrium at the original temperature.

Calculate the final equilibrium concentrations of \(A\), \(B\), and \(C\).

Explain why the first removal does not cause an equilibrium shift, even though the amount of every dissolved species decreases.

Solution 5

The final equilibrium concentrations are:

\[ \boxed{[A]=0.449\ \mathrm{mol\,L^{-1}}} \]

\[ \boxed{[B]=0.449\ \mathrm{mol\,L^{-1}}} \]

\[ \boxed{[C]=0.251\ \mathrm{mol\,L^{-1}}} \]

The key is to distinguish withdrawing part of a homogeneous solution from selectively removing one dissolved reactant.

Step 1: Calculate the original equilibrium constant.

The expression is:

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

Substitute the original concentrations:

\[ K_c=\frac{0.300}{(0.600)(0.400)} \]

\[ \boxed{K_c=1.25} \]

Step 2: Analyse the first removal.

A \(0.250\ \mathrm{L}\) sample is withdrawn from the original \(1.00\ \mathrm{L}\) solution.

This removes \(25\%\) of the solution.

However, it also removes \(25\%\) of every dissolved species.

The remaining volume is:

\[ V=1.00-0.250 \]

\[ =0.750\ \mathrm{L} \]

The remaining amounts are:

For \(A\):

\[ n(A)=0.600(0.750)=0.450\ \mathrm{mol} \]

For \(B\):

\[ n(B)=0.400(0.750)=0.300\ \mathrm{mol} \]

For \(C\):

\[ n(C)=0.300(0.750)=0.225\ \mathrm{mol} \]

Divide each amount by the remaining volume:

\[ [A]=\frac{0.450}{0.750}=0.600\ \mathrm{mol\,L^{-1}} \]

\[ [B]=\frac{0.300}{0.750}=0.400\ \mathrm{mol\,L^{-1}} \]

\[ [C]=\frac{0.225}{0.750}=0.300\ \mathrm{mol\,L^{-1}} \]

The concentrations are unchanged.

Therefore:

\[ Q_c=K_c \]

The first removal does not disturb the equilibrium.

Step 3: Apply the selective removal of \(A\).

Now \(0.150\ \mathrm{mol}\) of \(A\) is selectively removed from the remaining \(0.750\ \mathrm{L}\) solution.

The decrease in concentration is:

\[ \Delta[A]=\frac{n}{V} \]

\[ \Delta[A]=\frac{0.150}{0.750} \]

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

The new initial concentration is:

\[ [A]_{\mathrm{initial}}=0.600-0.200 \]

\[ =0.400\ \mathrm{mol\,L^{-1}} \]

The other concentrations remain unchanged.

Therefore, the new initial row is:

\[ [A]=0.400,\quad [B]=0.400,\quad [C]=0.300 \]

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

Step 4: Calculate the new reaction quotient.

\[ Q_c=\frac{0.300}{(0.400)(0.400)} \]

\[ \boxed{Q_c=1.875} \]

Compare this with the original equilibrium constant:

\[ Q_c=1.875>K_c=1.25 \]

The reaction shifts backwards.

Step 5: Construct the ICE table.

For the backwards reaction:

• \(A\) increases.
• \(B\) increases.
• \(C\) decreases.

Let \(x\) represent the backwards reaction progress.

Step 6: Substitute into the equilibrium expression.

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

Substitute the equilibrium expressions:

\[ 1.25= \frac{0.300-x}{(0.400+x)^2} \]

Cross-multiply:

\[ 1.25(0.400+x)^2=0.300-x \]

Expand:

\[ 1.25(0.160+0.800x+x^2)=0.300-x \]

\[ 0.200+1.000x+1.25x^2=0.300-x \]

Rearrange:

\[ 1.25x^2+2.00x-0.100=0 \]

Step 7: Solve the quadratic equation.

Using the quadratic formula:

\[ x=\frac{-2.00\pm \sqrt{(2.00)^2-4(1.25)(-0.100)}}{2(1.25)} \]

This gives:

\[ x=0.0485 \]

or:

\[ x=-1.649 \]

Reject the negative root.

It would produce negative equilibrium concentrations for \(A\) and \(B\), which are physically impossible.

Therefore:

\[ x=0.0485 \]

Step 8: Calculate the final equilibrium concentrations.

For \(A\):

\[ [A]_{\mathrm{eq}}=0.400+0.0485 \]

\[ \boxed{[A]_{\mathrm{eq}}=0.449\ \mathrm{mol\,L^{-1}}} \]

For \(B\):

\[ [B]_{\mathrm{eq}}=0.400+0.0485 \]

\[ \boxed{[B]_{\mathrm{eq}}=0.449\ \mathrm{mol\,L^{-1}}} \]

For \(C\):

\[ [C]_{\mathrm{eq}}=0.300-0.0485 \]

\[ \boxed{[C]_{\mathrm{eq}}=0.251\ \mathrm{mol\,L^{-1}}} \]

The first removal changed the amount of solution but not its concentrations. The second removal selectively decreased the concentration of \(A\), causing the reaction to shift backwards.

This is the important distinction: equilibrium responds to changes in the concentrations of reacting species, not simply to a reduction in the total amount of material in the container.

07Where to go next

Once you can solve selective reactant removal problems, the next useful step is combining concentration changes with volume or pressure changes.

In those questions, the same principle applies: calculate every immediate concentration change first, determine the new \(Q_c\), and only then construct your ICE table.

The difficult part is usually not the algebra. It is identifying exactly what changed before the reaction had time to respond.