How to Check an ICE-Table Answer by Substituting into Keq

Your ICE-table calculation gives an equilibrium concentration of \(0.0030\ \mathrm{mol\,L^{-1}}\) for a red iron-thiocyanate complex. The question gives an equilibrium constant of \(K_c=25.0\). Your answer looks reasonable, but how can you tell whether it actually satisfies the equilibrium equation?

Substitute your calculated equilibrium concentrations back into the original \(K_c\) expression. If the ratio reproduces the given equilibrium constant, within reasonable rounding, your calculation passes the numerical check. You must also confirm that no concentration is negative and that the changes obey the balanced equation.

The important detail is that you substitute the entire equilibrium row, not just the value of \(x\).

01Which ICE-table values belong in the equilibrium expression?

Consider the reversible reaction:

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

Iron(III) ions and thiocyanate ions combine to form a red complex ion, \(\mathrm{FeSCN}^{2+}\).

Each complex formed uses one \(\mathrm{Fe}^{3+}\) ion and one \(\mathrm{SCN}^{-}\) ion. The reverse reaction can separate the complex again.

Suppose the initial concentrations are:

• \([\mathrm{Fe}^{3+}]_0=0.0200\ \mathrm{mol\,L^{-1}}\)
• \([\mathrm{SCN}^{-}]_0=0.0100\ \mathrm{mol\,L^{-1}}\)
• \([\mathrm{FeSCN}^{2+}]_0=0\)

These are illustrative values, not measurements from a reported experiment.

Understanding the three ICE rows

An ICE table separates three different stages.

Let \(x\) represent the concentration of complex formed through the net forward reaction.

All concentrations and changes are measured in \(\mathrm{mol\,L^{-1}}\), abbreviated M.

The sign convention is:

• Positive \(x\): net forward reaction.
• Negative \(x\): net reverse reaction.
• Zero \(x\): no net change from the initial mixture.

For this particular mixture, \(x\) cannot be negative because no complex is initially present to undergo the reverse net change. It also cannot exceed \(0.0100\ \mathrm{M}\), because that would consume more thiocyanate than was initially available.

Therefore, the physically allowed range is:

\[ 0\leq x\leq0.0100\ \mathrm{M} \]

If setting up the change row is still unfamiliar, the full method is covered in ICE Tables for Equilibrium: Calculate Concentrations and Use Kc. Here, the focus is on testing the answer once the table has been solved.

03What would an actual experimental check add?

Back-substitution checks whether your calculated concentrations are consistent with the equilibrium equation. It does not independently measure those concentrations.

For the iron-thiocyanate reaction, the red complex offers a way to test the model experimentally.

Imagine preparing a mixture from known initial amounts and placing it in a transparent cuvette. A colourimeter directs light of a selected wavelength through the liquid. Some light is absorbed by the red complex, and the detector measures the light that passes through.

Predict what would happen as more red complex forms.

At a suitable wavelength, the solution would generally absorb more light. But simply seeing a darker red solution does not tell us the exact equilibrium concentration, or whether the mixture satisfies \(K_c=25.0\).

Improving the measurement

A first design that judges colour by eye cannot reliably separate concentration changes from differences in lighting, sample thickness, or background absorption.

A more useful design would:

1. Prepare calibration standards with known complex concentrations and measure their absorbances.
2. Use a suitable blank and the same optical path length.
3. Measure the equilibrium mixture at a controlled temperature.
4. Use its calibrated absorbance to estimate \([\mathrm{FeSCN}^{2+}]_{\mathrm{eq}}\).
5. Calculate the remaining free-ion concentrations using the initial amounts and the reaction’s 1:1:1 stoichiometry.

Suppose, purely as illustrative data, that this procedure estimated the equilibrium complex concentration as \(0.0030\ \mathrm{M}\).

The equilibrium ratio calculated earlier would be approximately 25.2. That would support the supplied value of 25.0 within the rounding of the example.

A substantially different measured result would call for investigation. The assumed reaction model, calibration, initial concentrations, temperature, or measurement uncertainty might need checking.

This is the key experimental distinction: a calculation reproducing its own input \(K_c\) establishes mathematical consistency, while independently measured concentrations provide evidence about the chemical model.

Real iron-thiocyanate solutions can also involve additional equilibria and ion interactions. Our numerical model assumes the stated reaction adequately describes the mixture.

05Why matching \(K_c\) is necessary but not sufficient

There are several ways an answer can reproduce the given equilibrium constant and still be wrong.

The second and third tests catch mistakes that the first cannot.

For instance, consider:

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

Its equilibrium expression is:

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

If an invalid solution gives a negative concentration for A, squaring that concentration can still produce a positive numerical \(K_c\).

A calculator has no way of knowing that negative amounts of gas are impossible.

There is another limitation. An incorrect set of entirely positive concentrations might produce exactly the right equilibrium ratio but not be reachable from the specified initial mixture.

To catch that, you need the ICE table’s stoichiometric constraints, not just the numerical equilibrium constant.

06Questions and solutions

All numerical data in the following questions are original, illustrative values. Assume constant temperature, ideal or sufficiently dilute behaviour for concentration-based calculations, and no unlisted reactions. Concentrations are in \(\mathrm{mol\,L^{-1}}\).

Question 1

For the reversible reaction:

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

the initial concentrations are \([A]_0=0.300\ \mathrm{M}\) and \([B]_0=0.100\ \mathrm{M}\).

At this temperature, \(K_c=1.00\).

A student calculates that \(0.100\ \mathrm{M}\) of A is converted into B.

Use back-substitution to determine whether the answer is correct.

Solution 1

The student’s answer is correct. It gives positive equilibrium concentrations that reproduce \(K_c=1.00\).

For a forward change of \(x=0.100\ \mathrm{M}\):

\[ \begin{aligned} [A]_{\mathrm{eq}} &=0.300-0.100\\ &=0.200\ \mathrm{M}\\ [B]_{\mathrm{eq}} &=0.100+0.100\\ &=0.200\ \mathrm{M} \end{aligned} \]

The equilibrium expression is:

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

Substituting the equilibrium values:

\[ K_{\mathrm{back}} =\frac{0.200}{0.200} =1.00 \]

This matches the given value exactly.

The initial concentrations were different, but the equilibrium concentrations are equal. That is consistent with \(K_c=1.00\) for this one-to-one reaction.

The mistake to avoid is substituting the initial values, which would give \(0.100/0.300=0.333\). That is the initial reaction quotient, not the final equilibrium constant.

Question 2

A hypothetical gas-phase dimerisation follows:

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

Initially:

\[ [A]_0=0.400\ \mathrm{M}, \qquad [B]_0=0.0500\ \mathrm{M} \]

The numerical equilibrium constant is \(K_c=1.00\) when concentrations are expressed in M.

Define \(x\) as the signed forward change, so that:

\[ [A]_{\mathrm{eq}}=0.400-2x \]

\[ [B]_{\mathrm{eq}}=0.0500+x \]

Solve the equilibrium equation, test both mathematical roots by back-substitution, and determine which root is physically possible. Explain why the other root can reproduce \(K_c\) despite being wrong.

Solution 2

The only physical root is \(x=0.04549\ \mathrm{M}\). The second root also reproduces the equilibrium ratio, but gives a negative concentration of A.

The equilibrium expression is:

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

Substituting the ICE expressions, using numerical concentrations in M:

\[ 1.00=\frac{0.0500+x}{(0.400-2x)^2} \]

Rearranging:

\[ \begin{aligned} (0.400-2x)^2&=0.0500+x\\ 4x^2-2.60x+0.110&=0 \end{aligned} \]

Solving the quadratic gives:

\[ x=0.04549\ \mathrm{M} \]

or

\[ x=0.60451\ \mathrm{M} \]

For the first root:

\[ \begin{aligned} [A]_{\mathrm{eq}} &=0.400-2(0.04549)\\ &=0.30902\ \mathrm{M}\\ [B]_{\mathrm{eq}} &=0.0500+0.04549\\ &=0.09549\ \mathrm{M} \end{aligned} \]

Back-substitution gives:

\[ K_{\mathrm{back}} =\frac{0.09549}{(0.30902)^2} \approx1.00 \]

Both concentrations are positive. The result also obeys the two-to-one stoichiometric relationship.

For the second root:

\[ \begin{aligned} [A]_{\mathrm{eq}} &=0.400-2(0.60451)\\ &=-0.80902\ \mathrm{M}\\ [B]_{\mathrm{eq}} &=0.0500+0.60451\\ &=0.65451\ \mathrm{M} \end{aligned} \]

Surprisingly, its substitution also gives:

\[ K_{\mathrm{back}} =\frac{0.65451}{(-0.80902)^2} \approx1.00 \]

The square removes the negative sign, hiding the impossible concentration.

An independent constraint comes from the initial amount of A:

\[ 0.400-2x\geq0 \]

Therefore:

\[ x\leq0.200\ \mathrm{M} \]

The larger root exceeds this limit.

The initial quotient also supports the smaller positive root:

\[ Q_{c,0} =\frac{0.0500}{(0.400)^2} =0.3125 \]

Since \(Q_{c,0}<K_c\), the net reaction proceeds forwards.

Only \(x=0.04549\ \mathrm{M}\) satisfies all the conditions.

Question 3

A sealed, constant-volume vessel contains the reaction:

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

Its initial concentrations are:

At this temperature, the concentration-based equilibrium constant is \(K_c=\frac{4}{3}\), using concentrations in M.

An analyst reports the following equilibrium concentrations:

The reported concentrations are all positive and reproduce the supplied \(K_c\) exactly.

A separate, reliable measurement confirms that \([B]_{\mathrm{eq}}=0.300\ \mathrm{M}\).

Can the analyst’s complete set of concentrations be correct? Determine the equilibrium concentrations required by the given initial mixture and explain why matching \(K_c\) is not enough.

Solution 3

The reported set is impossible for the specified closed system, despite satisfying \(K_c\). The equilibrium concentrations must be \([A]=0.500\ \mathrm{M}\), \([B]=0.300\ \mathrm{M}\), and \([C]=0.200\ \mathrm{M}\).

First, verify the analyst’s numerical check:

\[ \begin{aligned} K_{\mathrm{back}} &=\frac{[C]}{[A][B]}\\ &=\frac{0.160}{(0.400)(0.300)}\\ &=\frac{0.160}{0.120}\\ &=\frac{4}{3} \end{aligned} \]

That ratio is correct, but it only tests one relationship.

The balanced reaction requires every mole of B consumed to correspond to one mole of A consumed and one mole of C formed.

Since the reliable B measurement shows a decrease from \(0.400\) to \(0.300\ \mathrm{M}\):

\[ x=0.400-0.300=0.100\ \mathrm{M} \]

The corresponding equilibrium concentrations must be:

\[ \begin{aligned} [A]_{\mathrm{eq}}&=0.600-0.100\\ &=0.500\ \mathrm{M}\\ [B]_{\mathrm{eq}}&=0.300\ \mathrm{M}\\ [C]_{\mathrm{eq}}&=0.100+0.100\\ &=0.200\ \mathrm{M} \end{aligned} \]

Substituting these values:

\[ K_{\mathrm{back}} =\frac{0.200}{(0.500)(0.300)} =\frac{4}{3} \]

So the corrected concentrations satisfy both the equilibrium equation and the stoichiometry.

We can make the contradiction even clearer.

Because A and C change in opposite directions by equal amounts, their sum must remain constant in this closed, constant-volume system:

\[ [A]+[C]=0.700\ \mathrm{M} \]

Initially, this is \(0.600+0.100=0.700\ \mathrm{M}\).

The analyst’s concentrations give only:

\[ 0.400+0.160=0.560\ \mathrm{M} \]

Likewise, \([B]+[C]\) must remain \(0.500\ \mathrm{M}\), but the analyst reports only \(0.460\ \mathrm{M}\).

Both conserved quantities have been violated.

The analyst’s values describe a concentration ratio consistent with the given \(K_c\), but not a state reachable from this initial mixture under the stated assumptions.

This is why a correct equilibrium ratio is necessary but cannot replace the material-balance check.

Question 4

A closed gas system undergoes:

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

It is initially at equilibrium in a vessel of volume \(V\), with:

\[ [A]=0.400\ \mathrm{M} \]

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

\[ [C]=0.160\ \mathrm{M} \]

The equilibrium constant is \(K_c=2.00\), using concentrations in M.

A piston rapidly doubles the vessel’s volume to \(2V\), without changing its temperature or allowing gas to escape.

A student substitutes the original equilibrium concentrations back into \(K_c\), gets 2.00, and concludes that nothing changes.

Explain why this check is insufficient. Calculate the concentrations immediately after expansion and after the gas mixture reaches its new equilibrium. Verify the final result by back-substitution.

Solution 4

The student is incorrect. Doubling the volume initially halves every concentration, producing a mixture that is no longer at equilibrium. It then undergoes a net reverse reaction until the quotient returns to \(K_c=2.00\).

The original equilibrium expression is:

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

The original state satisfies:

\[ \frac{0.160}{(0.400)(0.200)}=2.00 \]

However, those concentrations are no longer the current values once the volume doubles.

Immediately after expansion, the amounts of each gas have not yet changed chemically, but the same amounts occupy twice the volume.

Therefore:

\[ \begin{aligned} [A]_{\mathrm{new}}&=0.200\ \mathrm{M}\\ [B]_{\mathrm{new}}&=0.100\ \mathrm{M}\\ [C]_{\mathrm{new}}&=0.0800\ \mathrm{M} \end{aligned} \]

These are the new starting concentrations for the reaction’s subsequent shift.

The reaction quotient immediately after expansion is:

\[ \begin{aligned} Q_c &=\frac{0.0800}{(0.200)(0.100)}\\ &=4.00 \end{aligned} \]

Since \(Q_c>K_c\), the reaction shifts in reverse.

Let \(y\) be the positive concentration change associated with the reverse reaction.

Substitute into the given equilibrium expression:

\[ 2.00= \frac{0.0800-y} {(0.200+y)(0.100+y)} \]

Rearranging the numerical equation:

\[ 2y^2+1.60y-0.0400=0 \]

The physical root is:

\[ y\approx0.024264\ \mathrm{M} \]

The other quadratic root would give negative reactant concentrations and must be rejected.

Therefore:

\[ \begin{aligned} [A]_{\mathrm{eq}}&=0.224264\ \mathrm{M}\\ [B]_{\mathrm{eq}}&=0.124264\ \mathrm{M}\\ [C]_{\mathrm{eq}}&=0.055736\ \mathrm{M} \end{aligned} \]

Now perform the required back-substitution:

\[ \begin{aligned} K_{\mathrm{back}} &=\frac{0.055736} {(0.224264)(0.124264)}\\ &\approx2.00 \end{aligned} \]

The final concentrations reproduce the original equilibrium constant, which has not changed because temperature was held constant.

The important distinction is between the instant after expansion and the subsequent establishment of equilibrium.

The original mixture satisfied \(K_c\), but the rapid physical change altered its concentrations. Only the new equilibrium concentrations should be used for the final check.

Question 5

Consider a different sealed gas equilibrium:

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

The vessel is initially at equilibrium at constant temperature, with volume \(V\) and concentrations:

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

\[ [B]=0.500\ \mathrm{M} \]

\[ [C]=0.400\ \mathrm{M} \]

The piston then doubles the volume to \(2V\). No gas escapes, the temperature remains constant, and only the stated reaction occurs.

After the volume change, two concentration analysers report different results:

Both sets are positive. Both also reproduce the original \(K_c\) exactly.

One student argues that either set could represent the equilibrium state because both pass back-substitution.

Another argues that doubling the volume must cause some chemical shift, even if the equilibrium quotient does not change.

Determine which analyser’s result is physically possible. Decide whether any net reaction shift occurs, and explain why the equilibrium-constant check cannot distinguish these two datasets.

Your explanation should use the balanced equation, the volume change, and the mathematical structure of \(K_c\), rather than relying only on a calculated result.

Solution 5

Only Analyser 1 is consistent with the stated closed system and volume change. The reaction undergoes no net shift after the expansion, and both analysers pass the \(K_c\) check because this particular equilibrium expression is unchanged when all concentrations are multiplied by the same factor.

The first step is to calculate the original equilibrium constant:

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

\[ \begin{aligned} K_c &=\frac{(0.400)^2}{(0.300)(0.500)}\\ &=\frac{0.160}{0.150}\\ &=\frac{16}{15} \end{aligned} \]

Now check Analyser 1:

\[ \begin{aligned} K_{\mathrm{back,1}} &=\frac{(0.200)^2}{(0.150)(0.250)}\\ &=\frac{0.0400}{0.0375}\\ &=\frac{16}{15} \end{aligned} \]

Analyser 2 also passes:

\[ \begin{aligned} K_{\mathrm{back,2}} &=\frac{(0.240)^2}{(0.180)(0.300)}\\ &=\frac{0.0576}{0.0540}\\ &=\frac{16}{15} \end{aligned} \]

So the numerical equilibrium check cannot separate the two answers.

Why?

Notice that every concentration reported by Analyser 2 is \(1.20\) times the corresponding Analyser 1 value.

If all concentrations in this reaction are multiplied by a common positive factor \(s\), the new reaction quotient becomes:

\[ Q_c'= \frac{(s[C])^2}{(s[A])(s[B])} \]

Simplifying:

\[ Q_c'= \frac{s^2[C]^2}{s^2[A][B]} =Q_c \]

The factor \(s^2\) cancels.

That happens because the total stoichiometric coefficient on the product side is 2, and the total on the reactant side is also 2.

This means a uniform concentration error can leave the calculated equilibrium quotient completely unchanged.

But there is another constraint that the equilibrium expression does not test.

The reaction:

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

converts two moles of reactant gas into two moles of product gas. Therefore, chemical reaction alone does not change the total number of gas moles.

Because the system is sealed, the total amount of gas must remain constant.

Initially, the total molar concentration is:

\[ 0.300+0.500+0.400 =1.200\ \mathrm{M} \]

When the volume doubles, the total concentration must become:

\[ \frac{1.200}{2} =0.600\ \mathrm{M} \]

This remains true regardless of the net reaction extent, because the total number of gas moles is conserved.

Analyser 1 reports:

\[ 0.150+0.250+0.200 =0.600\ \mathrm{M} \]

Analyser 2 reports:

\[ 0.180+0.300+0.240 =0.720\ \mathrm{M} \]

Analyser 2 violates the known volume and mole constraints. Its result cannot represent the stated system.

We can also determine whether a reaction shift is needed.

Immediately after expansion, before any possible chemical change, each concentration is exactly half its original value:

\[ \begin{aligned} [A]&=0.150\ \mathrm{M}\\ [B]&=0.250\ \mathrm{M}\\ [C]&=0.200\ \mathrm{M} \end{aligned} \]

As already calculated, these concentrations give:

\[ Q_c=\frac{16}{15}=K_c \]

The mixture is already at equilibrium immediately after expansion. No net chemical shift is required.

If a net forward change occurred, C would increase while A and B decreased, causing \(Q_c\) to rise above \(K_c\). A reverse change would cause the opposite. Neither would be a new equilibrium state under the stated conditions.

There is one further insight. The second dataset is exactly 60% of the original concentrations. It could describe the same mixture after expansion to \(5V/3\), rather than \(2V\), with no reaction shift. Alternatively, a common calibration error could make an analyser over-report every concentration.

The \(K_c\) calculation alone cannot identify which of those explanations applies.

That is the conceptual limit of back-substitution: an equilibrium ratio does not independently establish that the correct amount of material is present, that the volume is correct, or that a measurement is accurate.

The safest method is to use back-substitution alongside physical constraints and stoichiometric conservation. If you’ve obtained your ICE result using an approximation rather than solving the full equation, the next useful step is to check whether that equilibrium approximation is valid. A small calculated change is only trustworthy when the resulting concentrations still satisfy the equilibrium conditions to the required accuracy.