How to Check Whether an Equilibrium Approximation Is Valid
Learn how to use an ICE table and the 5% rule to test equilibrium approximations. Includes worked examples covering selective approximations, exact calculations, and rejecting nonphysical quadratic roots.
An equilibrium calculation gives \(x=0.0040\ \mathrm{mol\,L^{-1}}\). Is that small enough to ignore?
It depends on what you’re comparing it with. If the initial concentration was \(0.200\ \mathrm{mol\,L^{-1}}\), the change is only 2%. If it was \(0.0100\ \mathrm{mol\,L^{-1}}\), the change is 40%.
The second approximation would be a poor choice, even though \(x\) is exactly the same.
To check whether an equilibrium approximation is valid, calculate the percentage change in every concentration term you intend to approximate. A common rule of thumb is that each ignored change should be no more than 5% of its corresponding initial concentration. If a change exceeds 5%, you should retain that term and solve the full equilibrium expression.
The method starts with an initial-change-equilibrium (ICE) table. You then use the stoichiometric coefficients, test the approximation, and check that your final concentrations are physically possible.
01Understand the three rows of an ICE table
Consider the general reversible reaction:
\[ aA+bB\rightleftharpoons cC+dD \]
An ICE table tracks the concentrations of each substance as the reaction moves towards equilibrium.
The three rows have different jobs:
| Row | Meaning | What you enter |
|---|---|---|
| Initial (I) | Concentrations before the net reaction shift | Starting concentrations |
| Change (C) | How concentrations change during the reaction | Changes determined by stoichiometry |
| Equilibrium (E) | Concentrations after equilibrium is established | Initial + Change |
The relationship connecting the rows is:
\[ \boxed{E=I+C} \]
Let’s build one.
Suppose the reaction is:
\[ \ce{2A(g) + B(g) <=> C(g)} \]
Assume the forward reaction proceeds by an amount \(x\), measured in \(\mathrm{mol\,L^{-1}}\).
The balanced equation tells us that:
- Two units of A are consumed.
- One unit of B is consumed.
- One unit of C is produced.
Therefore, the change row is \(-2x\), \(-x\), and \(+x\), respectively.
Suppose the initial concentrations are:
\[ [A]_0=0.100,\quad [B]_0=0.0500 \]
and no C is initially present.
The ICE table becomes:
| A | B | C | |
|---|---|---|---|
| Initial | 0.100 | 0.0500 | 0 |
| Change | \(-2x\) | \(-x\) | \(+x\) |
| Equilibrium | \(0.100-2x\) | \(0.0500-x\) | \(x\) |
All concentrations and changes are in \(\mathrm{mol\,L^{-1}}\).
Notice that the coefficient of A is 2, so its concentration decreases by \(2x\), not \(x\).
This is one of the most important details when checking an approximation.
What if the reaction moves backwards?
The table above assumes a forward reaction.
If we define \(x\) as a signed forward change, a negative value means the net reaction has proceeded backwards. Alternatively, we can define a new variable that is positive in the reverse direction.
If the direction is unclear, calculate the reaction quotient, \(Q_c\), and compare it with \(K_c\):
- \(Q_c<K_c\): net forward reaction.
- \(Q_c>K_c\): net reverse reaction.
- \(Q_c=K_c\): the mixture is already at equilibrium.
Do not automatically assume that reactants decrease and products increase. The initial composition determines the direction.
Also, make sure your initial row contains concentrations, not unconverted mole amounts. If a question gives moles and volume, convert them first using \(c=n/V\). See How to Convert Moles into Concentration for an ICE Table.
02The 5% test: how to check an approximation
Suppose an equilibrium expression contains:
\[ [A]_{\mathrm{eq}}=[A]_0-x \]
If \(x\) is sufficiently small compared with the initial concentration, we can approximate:
\[ [A]_{\mathrm{eq}}\approx[A]_0 \]
But how small is sufficiently small?
Calculate the percentage change:
\[ \boxed{ \%\text{ change} = \frac{\text{magnitude of concentration change}} {\text{initial concentration}} \times100 } \]
For a species A with a stoichiometric coefficient \(a\):
\[ \boxed{ \%\text{ change in A} = \frac{|ax|}{[A]_0}\times100 } \]
The absolute value is important because the percentage change should be positive, regardless of reaction direction.
Applying the 5% rule
A common classroom rule of thumb is:
| Percentage change | Decision |
|---|---|
| Less than 5% | Approximation is generally acceptable |
| Around 5% | Borderline; consider solving exactly |
| Greater than 5% | Approximation is not justified |
The 5% value is a practical guideline, not a universal mathematical boundary. Follow any specific tolerance given in a question.
There are three details worth remembering.
First: compare the change with the correct initial concentration.
Do not simply say that \(x=0.001\) is small. Compare it with the concentration of the species whose expression you are simplifying.
Second: include the stoichiometric coefficient.
If the equilibrium concentration is \(0.100-2x\), your percentage calculation must use \(2x\), not \(x\).
Third: only test the terms you actually approximate.
Suppose a product starts at zero, so its equilibrium concentration is simply:
\[ [C]_{\mathrm{eq}}=x \]
You cannot replace this with zero using the 5% rule. Its initial concentration is zero, so the percentage-change calculation is undefined.
You must keep the \(x\) term.
Likewise, if you decide to keep a particular concentration expression exactly, a large change in that species does not automatically invalidate an approximation elsewhere.
This last point becomes important in more complicated questions.
03Worked example: Using a pH measurement to check an acid approximation
A laboratory investigation provides a useful way to see where the numbers come from.
A student prepares a \(0.0500\ \mathrm{mol\,L^{-1}}\) solution of a monoprotic weak acid, HA.
The solution is prepared using a volumetric pipette and volumetric flask. An aliquot is transferred to a beaker in a constant-temperature water bath, and a calibrated pH electrode is immersed in the solution.
Once the reading stabilises, the pH meter displays 3.03.
The student wants to calculate \(K_a\) and assumes that the equilibrium acid concentration is approximately equal to its initial concentration.
Determine whether this approximation is justified.
The measurement below is illustrative, not a record of an actual experiment.
Step 1
The volumetric glassware establishes the initial acid concentration.
The water bath keeps the temperature constant, while the pH electrode measures the solution’s pH.
A stable pH reading indicates that the measured composition is no longer changing macroscopically.
However, a stable reading alone does not prove that the acid concentration has changed by only a small amount. We need to calculate that change.
Step 2
If the acid ionises only slightly, the hydronium concentration should be much smaller than the initial acid concentration.
For a dilute solution, using the usual HSC concentration approximation:
\[ [H_3O^+]\approx10^{-\mathrm{pH}} \]
Substituting the measured pH:
\[ [H_3O^+]=10^{-3.03} \]
\[ [H_3O^+]=9.33\times10^{-4} \ \mathrm{mol\,L^{-1}} \]
Assuming initially negligible hydronium and conjugate-base concentrations, the amount of acid ionised per litre is:
\[ x=9.33\times10^{-4}\ \mathrm{mol\,L^{-1}} \]
Step 3
The acid ionisation reaction is:
\[ \ce{HA(aq) + H2O(l) <=> H3O+(aq) + A^-(aq)} \]
Water is omitted from the equilibrium expression.
| HA | \(\ce{H3O+}\) | \(\mathrm{A}^{-}\) | |
|---|---|---|---|
| Initial | 0.0500 | 0 | 0 |
| Change | \(-x\) | \(+x\) | \(+x\) |
| Equilibrium | \(0.0500-x\) | \(x\) | \(x\) |
The acid dissociation constant is:
\[ K_a=\frac{[H_3O^+][A^-]}{[HA]} \]
Substituting the ICE expressions:
\[ K_a=\frac{x^2}{0.0500-x} \]
Step 4
The acid concentration decreases by \(x\).
Therefore:
\[ \%\text{ change} = \frac{9.33\times10^{-4}}{0.0500} \times100 \]
\[ \boxed{\%\text{ change}=1.87\%} \]
This is below 5%.
We can therefore approximate:
\[ [HA]_{\mathrm{eq}}\approx0.0500 \ \mathrm{mol\,L^{-1}} \]
Step 5
Using the approximation:
\[ K_a\approx\frac{x^2}{0.0500} \]
\[ K_a\approx \frac{(9.33\times10^{-4})^2}{0.0500} \]
\[ \boxed{K_a\approx1.74\times10^{-5}} \]
Without the approximation:
\[ K_a= \frac{(9.33\times10^{-4})^2} {0.0500-9.33\times10^{-4}} \]
\[ \boxed{K_a\approx1.78\times10^{-5}} \]
The two results differ by approximately 1.87%.
The approximation is justified for this illustrative solution.
There is an experimental limitation worth noting: a real pH electrode measures hydrogen-ion activity rather than concentration directly. We have used the usual dilute-solution approximation that activity is approximately equal to concentration. We have also assumed negligible background ionisation from water and no other source of hydronium ions.
A different solution concentration or a different acid could produce a much larger percentage change. You must check each new situation rather than assuming every weak acid permits the same approximation.
04Worked example: When one reactant changes too much
Now consider a different equilibrium:
\[ \ce{A(g) + B(g) <=> C(g)} \]
An equilibrium mixture is prepared with:
- Initial \([A]=0.100\ \mathrm{mol\,L^{-1}}\)
- Initial \([B]=0.0100\ \mathrm{mol\,L^{-1}}\)
- Initial \([C]=0\)
- \(K_c=4.0\ \mathrm{L\,mol^{-1}}\)
Determine whether you can approximate both reactant concentrations using their initial values.
If not, solve the exact equilibrium expression and identify the physical root.
Step 1
The forward reaction is favoured initially because no product is present.
| A | B | C | |
|---|---|---|---|
| Initial | 0.100 | 0.0100 | 0 |
| Change | \(-x\) | \(-x\) | \(+x\) |
| Equilibrium | \(0.100-x\) | \(0.0100-x\) | \(x\) |
Step 2
The equilibrium expression is:
\[ K_c=\frac{[C]}{[A][B]} \]
Substituting the ICE expressions:
\[ 4.0= \frac{x}{(0.100-x)(0.0100-x)} \]
Suppose we initially assume that both reactant changes are negligible.
Then:
\[ 4.0\approx \frac{x}{0.100\times0.0100} \]
Rearranging:
\[ x\approx4.0\times0.00100 \]
\[ \boxed{x_{\mathrm{approx}}=0.00400\ \mathrm{mol\,L^{-1}}} \]
Step 3
For A:
\[ \%\text{ change in A} = \frac{0.00400}{0.100}\times100 \]
\[ \boxed{4.00\%} \]
This approximation passes the usual 5% test.
Now test B:
\[ \%\text{ change in B} = \frac{0.00400}{0.0100}\times100 \]
\[ \boxed{40.0\%} \]
This approximation fails.
The important detail is that the same \(x\) is being compared with two different initial concentrations.
You cannot approximate both reactants by their initial values.
Step 4
Return to the original expression:
\[ 4.0= \frac{x}{(0.100-x)(0.0100-x)} \]
Multiplying through:
\[ 4.0(0.100-x)(0.0100-x)=x \]
Expanding and rearranging, using concentrations numerically in \(\mathrm{mol\,L^{-1}}\):
\[ \boxed{4x^2-1.44x+0.00400=0} \]
Using the quadratic formula:
\[ x=\frac{-b\pm\sqrt{b^2-4ac}}{2a} \]
The two mathematical roots are:
\[ x_1=0.0027995\ \mathrm{mol\,L^{-1}} \]
\[ x_2=0.3572005\ \mathrm{mol\,L^{-1}} \]
Step 5
The equilibrium concentrations must all be non-negative.
In particular:
\[ [A]_{\mathrm{eq}}=0.100-x\geq0 \]
\[ [B]_{\mathrm{eq}}=0.0100-x\geq0 \]
Since B is the limiting reactant, the physically allowed interval is:
\[ 0\leq x\leq0.0100 \]
The larger root, \(0.3572005\), is outside this interval. It would produce a negative equilibrium concentration of B.
Therefore, the physical root is:
\[ \boxed{x=0.00280\ \mathrm{mol\,L^{-1}}} \]
The equilibrium concentrations are approximately:
\[ [A]_{\mathrm{eq}}=0.0972\ \mathrm{mol\,L^{-1}} \]
\[ [B]_{\mathrm{eq}}=0.00720\ \mathrm{mol\,L^{-1}} \]
\[ [C]_{\mathrm{eq}}=0.00280\ \mathrm{mol\,L^{-1}} \]
This example reveals two separate checks:
- The approximation failed because one reactant changed by 40%.
- The exact quadratic produced two mathematical roots, but only one was physically possible.
A mathematically correct root is not automatically a chemically possible result.
For a more detailed treatment of root selection, see How to Choose the Physical Root of an Equilibrium Quadratic.
05A reliable decision process
When you encounter an equilibrium approximation question, use this sequence.
1. Write the balanced equation.
Check that every stoichiometric coefficient is correct.
2. Construct the ICE table.
Enter the initial concentrations, calculate changes using the mole ratios, and write the equilibrium expressions.
3. Determine the reaction direction.
If necessary, compare \(Q_c\) with \(K_c\).
4. Identify exactly which terms you want to approximate.
For example:
\[ [A]_0-2x\approx[A]_0 \]
This means you are neglecting a change of \(2x\), not \(x\).
5. Solve the approximate expression.
Calculate the approximate value of \(x\).
6. Apply the percentage-change test.
For every term you replaced with its initial concentration, calculate:
\[ \frac{|\text{concentration change}|} {\text{initial concentration}} \times100 \]
7. Check physical feasibility.
Every equilibrium concentration must be non-negative. If you solve a quadratic, check both roots against the physically allowed range.
If an approximation fails, return to the full equilibrium expression. Do not simply keep an answer because the algebra was easy.
06Questions and solutions
Question 1
A sealed vessel initially contains \(0.200\ \mathrm{mol\,L^{-1}}\) of gas A and no gas B.
The reaction is:
\[ \ce{A(g) <=> B(g)} \]
At a constant temperature, \(K_c=2.0\times10^{-3}\).
Assuming that \([A]_{\mathrm{eq}}\approx[A]_0\), calculate the approximate value of \(x\) and determine whether the approximation is valid using the 5% rule.
Solution 1
The approximation is valid because the calculated concentration change is only 0.20% of the initial concentration.
Step 1: Construct the ICE table.
| A | B | |
|---|---|---|
| Initial | 0.200 | 0 |
| Change | \(-x\) | \(+x\) |
| Equilibrium | \(0.200-x\) | \(x\) |
Step 2: Write the equilibrium expression.
\[ K_c=\frac{[B]}{[A]} \]
Using the approximation:
\[ 2.0\times10^{-3} \approx\frac{x}{0.200} \]
Rearranging:
\[ x=(2.0\times10^{-3})(0.200) \]
\[ \boxed{x=4.0\times10^{-4}\ \mathrm{mol\,L^{-1}}} \]
Step 3: Calculate the percentage change.
\[ \%\text{ change} = \frac{4.0\times10^{-4}}{0.200} \times100 \]
\[ \boxed{0.20\%} \]
This is comfortably below 5%.
The approximation is valid, giving:
\[ [A]_{\mathrm{eq}}\approx0.200\ \mathrm{mol\,L^{-1}} \]
\[ [B]_{\mathrm{eq}}\approx4.0\times10^{-4}\ \mathrm{mol\,L^{-1}} \]
Notice that we kept the product concentration equal to \(x\). Its initial concentration was zero, so it could not be approximated as unchanged.
Question 2
Consider the equilibrium:
\[ \ce{2A(g) + B(g) <=> C(g)} \]
The initial concentrations are:
\[ [A]_0=0.100\ \mathrm{mol\,L^{-1}} \]
\[ [B]_0=0.0200\ \mathrm{mol\,L^{-1}} \]
\[ [C]_0=0 \]
At constant temperature:
\[ K_c=4.0\ \mathrm{L^2\,mol^{-2}} \]
Assume the forward reaction proceeds by \(x\).
Calculate the approximate \(x\), then determine whether both reactant approximations are valid.
Solution 2
Both approximations are valid because the percentage changes are 1.60% for A and 4.00% for B.
Step 1: Construct the ICE table.
| A | B | C | |
|---|---|---|---|
| Initial | 0.100 | 0.0200 | 0 |
| Change | \(-2x\) | \(-x\) | \(+x\) |
| Equilibrium | \(0.100-2x\) | \(0.0200-x\) | \(x\) |
Step 2: Write the equilibrium expression.
\[ K_c=\frac{[C]}{[A]^2[B]} \]
Assuming both reactant changes are small:
\[ 4.0\approx \frac{x}{(0.100)^2(0.0200)} \]
Rearranging:
\[ x=4.0(0.100)^2(0.0200) \]
\[ \boxed{x=8.00\times10^{-4}\ \mathrm{mol\,L^{-1}}} \]
Step 3: Test the approximation for A.
Remember that A changes by \(2x\).
\[ \%\text{ change in A} = \frac{2(0.000800)}{0.100}\times100 \]
\[ \boxed{1.60\%} \]
A passes.
Step 4: Test the approximation for B.
\[ \%\text{ change in B} = \frac{0.000800}{0.0200}\times100 \]
\[ \boxed{4.00\%} \]
B also passes.
Both changes are below 5%, so the approximations are justified.
The approximate equilibrium concentrations are:
\[ [A]_{\mathrm{eq}}\approx0.0984 \]
\[ [B]_{\mathrm{eq}}\approx0.0192 \]
\[ [C]_{\mathrm{eq}}\approx0.000800 \]
All concentrations are in \(\mathrm{mol\,L^{-1}}\).
The key detail is that the coefficient of A doubles its concentration change. Ignoring that coefficient would give the wrong percentage test.
Question 3
Consider the equilibrium:
\[ \ce{A(aq) <=> B(aq)} \]
The initial concentrations are:
\[ [A]_0=0.100\ \mathrm{mol\,L^{-1}} \]
\[ [B]_0=0.00200\ \mathrm{mol\,L^{-1}} \]
The equilibrium constant is:
\[ K_c=0.0500 \]
A student proposes the approximation:
\[ [A]_{\mathrm{eq}}\approx0.100 \]
but keeps the product expression exact:
\[ [B]_{\mathrm{eq}}=0.00200+x \]
Another student argues that the approximation must be invalid because the product concentration changes by more than 100% of its initial value.
Determine which student is correct. Calculate the approximate \(x\) and check the approximation.
Solution 3
The first student is correct: the selective approximation is valid under the usual 5% test.
The second student is testing a concentration change that has not actually been approximated.
Step 1: Determine the reaction direction.
Initially:
\[ Q_c=\frac{0.00200}{0.100} \]
\[ Q_c=0.0200 \]
Since:
\[ Q_c<K_c \]
the reaction proceeds forwards.
Step 2: Construct the ICE table.
| A | B | |
|---|---|---|
| Initial | 0.100 | 0.00200 |
| Change | \(-x\) | \(+x\) |
| Equilibrium | \(0.100-x\) | \(0.00200+x\) |
Step 3: Apply the proposed approximation.
The equilibrium expression is:
\[ 0.0500=\frac{0.00200+x}{0.100-x} \]
Approximating only the A concentration:
\[ 0.0500\approx \frac{0.00200+x}{0.100} \]
Rearranging:
\[ 0.00500=0.00200+x \]
\[ \boxed{x=0.00300\ \mathrm{mol\,L^{-1}}} \]
Step 4: Test the concentration that was approximated.
Only the A concentration was replaced with its initial value.
Therefore:
\[ \%\text{ change in A} = \frac{0.00300}{0.100}\times100 \]
\[ \boxed{3.00\%} \]
The approximation passes.
What about B?
Its percentage increase is:
\[ \frac{0.00300}{0.00200}\times100 =150\% \]
That looks large, but we have not approximated B.
We retained the complete expression \(0.00200+x\), so the large product change does not invalidate our approximation.
Step 5: Check against the exact solution.
Starting with:
\[ 0.0500= \frac{0.00200+x}{0.100-x} \]
Rearranging:
\[ 0.00500-0.0500x=0.00200+x \]
\[ 1.05x=0.00300 \]
\[ x=0.002857\ \mathrm{mol\,L^{-1}} \]
The approximate value was \(0.00300\ \mathrm{mol\,L^{-1}}\), which differs from the exact value by 5.00%.
This is close to the usual tolerance, but the approximation passes the concentration-change test.
The important lesson is that you test the terms you actually discard, not every concentration in the ICE table.
Question 4
A reversible gas reaction is:
\[ \ce{A(g) + 2B(g) <=> C(g)} \]
Initially:
\[ [A]_0=0.100\ \mathrm{mol\,L^{-1}} \]
\[ [B]_0=0.0100\ \mathrm{mol\,L^{-1}} \]
\[ [C]_0=0 \]
The equilibrium constant is:
\[ K_c=300\ \mathrm{L^2\,mol^{-2}} \]
A student assumes both reactants remain approximately at their initial concentrations and calculates \(x\).
Their percentage-change calculation for A is below 5%.
Should their approximation be accepted? Explain why checking A alone is insufficient.
Solution 4
The approximation is invalid because B changes by 60%, even though A changes by only 3%.
Step 1: Construct the ICE table.
| A | B | C | |
|---|---|---|---|
| Initial | 0.100 | 0.0100 | 0 |
| Change | \(-x\) | \(-2x\) | \(+x\) |
| Equilibrium | \(0.100-x\) | \(0.0100-2x\) | \(x\) |
Step 2: Calculate the approximate \(x\).
The equilibrium expression is:
\[ K_c=\frac{[C]}{[A][B]^2} \]
Applying the proposed approximation:
\[ 300\approx \frac{x}{(0.100)(0.0100)^2} \]
Therefore:
\[ x\approx300(0.100)(0.0100)^2 \]
\[ \boxed{x=0.00300\ \mathrm{mol\,L^{-1}}} \]
Step 3: Test A.
\[ \%\text{ change in A} = \frac{0.00300}{0.100}\times100 \]
\[ \boxed{3.00\%} \]
A passes.
Step 4: Test B.
The balanced equation shows that B changes by \(2x\).
\[ \%\text{ change in B} = \frac{2(0.00300)}{0.0100}\times100 \]
\[ \boxed{60.0\%} \]
B fails.
The approximation must therefore be rejected.
There is an additional useful check. B is the limiting reactant, so:
\[ 0.0100-2x\geq0 \]
\[ x\leq0.00500 \]
The approximate \(x=0.00300\) is physically possible because it is within this range.
However, being physically possible does not make an approximation accurate.
The full equilibrium expression is:
\[ 300= \frac{x} {(0.100-x)(0.0100-2x)^2} \]
The small-\(x\) approximation cannot be used to justify the answer. You must retain the full expressions if an exact equilibrium calculation is required.
Question 5
This final problem tests whether you can distinguish between a large concentration change and a large change in a term that has actually been approximated.
Consider:
\[ \ce{A(g) + B(g) <=> C(g)} \]
At a constant temperature, a mixture initially contains:
\[ [A]_0=1.00\ \mathrm{mol\,L^{-1}} \]
\[ [B]_0=0.0100\ \mathrm{mol\,L^{-1}} \]
\[ [C]_0=0.0200\ \mathrm{mol\,L^{-1}} \]
The equilibrium constant is:
\[ K_c=1.00\ \mathrm{L\,mol^{-1}} \]
Two students disagree.
- Student 1 argues that the reaction proceeds backwards and that an approximation may be possible because A is present in a much greater concentration than the other substances.
- Student 2 argues that any approximation must be invalid if the change in B is greater than 5% of its initial concentration.
Investigate the two arguments.
Determine whether it is possible to make a valid approximation by replacing only the equilibrium concentration of A with its initial concentration, while retaining the changes in B and C.
Verify your conclusion using the exact equilibrium equation.
Solution 5
Student 1 is correct, provided the approximation is applied selectively. Student 2’s argument incorrectly assumes that every concentration change must be less than 5%, even when some changes are retained exactly.
This problem requires three linked decisions: determine the direction, choose which term to approximate, and test that particular approximation.
Step 1: Determine the direction of the reaction.
Calculate the initial reaction quotient:
\[ Q_c=\frac{[C]_0}{[A]_0[B]_0} \]
Substituting:
\[ Q_c=\frac{0.0200}{1.00\times0.0100} \]
\[ Q_c=2.00\ \mathrm{L\,mol^{-1}} \]
Since:
\[ Q_c>K_c \]
the reaction proceeds in the reverse direction.
Step 2: Define a reverse reaction variable.
Let \(y\) represent the concentration change in the reverse direction, measured in \(\mathrm{mol\,L^{-1}}\).
The changes are:
- A increases by \(y\).
- B increases by \(y\).
- C decreases by \(y\).
The ICE table is:
| A | B | C | |
|---|---|---|---|
| Initial | 1.00 | 0.0100 | 0.0200 |
| Change | \(+y\) | \(+y\) | \(-y\) |
| Equilibrium | \(1.00+y\) | \(0.0100+y\) | \(0.0200-y\) |
Step 3: Apply the proposed selective approximation.
We want to investigate whether:
\[ [A]_{\mathrm{eq}}=1.00+y \]
can be approximated as:
\[ [A]_{\mathrm{eq}}\approx1.00 \]
However, we will keep the other two concentrations exact:
\[ [B]_{\mathrm{eq}}=0.0100+y \]
\[ [C]_{\mathrm{eq}}=0.0200-y \]
The equilibrium expression becomes:
\[ 1.00\approx \frac{0.0200-y} {1.00(0.0100+y)} \]
Step 4: Calculate the approximate reverse extent.
Rearranging:
\[ 0.0100+y=0.0200-y \]
\[ 2y=0.0100 \]
\[ \boxed{y=0.00500\ \mathrm{mol\,L^{-1}}} \]
Step 5: Test the term that was approximated.
Only A was approximated.
Therefore:
\[ \%\text{ change in A} = \frac{0.00500}{1.00}\times100 \]
\[ \boxed{0.500\%} \]
This is comfortably below 5%.
The approximation is justified.
Now consider the other two species.
For B:
\[ \%\text{ change in B} = \frac{0.00500}{0.0100}\times100 \]
\[ =50.0\% \]
For C:
\[ \%\text{ change in C} = \frac{0.00500}{0.0200}\times100 \]
\[ =25.0\% \]
Both changes are substantial.
But neither concentration was approximated. Their complete expressions remained in the equilibrium equation.
This is why Student 2’s argument is incorrect.
Step 6: Verify the approximation using the exact equation.
The exact equilibrium expression is:
\[ 1.00= \frac{0.0200-y} {(1.00+y)(0.0100+y)} \]
Rearranging:
\[ (1.00+y)(0.0100+y)=0.0200-y \]
Expanding:
\[ 0.0100+1.01y+y^2=0.0200-y \]
Therefore:
\[ y^2+2.01y-0.0100=0 \]
Using the quadratic formula gives:
\[ y_1=0.00496287\ \mathrm{mol\,L^{-1}} \]
\[ y_2=-2.01496287\ \mathrm{mol\,L^{-1}} \]
The reverse extent must satisfy:
\[ 0\leq y\leq0.0200 \]
The upper limit comes from the initial concentration of C. The reverse reaction cannot consume more product than is available.
Therefore, the negative root is nonphysical, and the valid root is:
\[ \boxed{y=0.004963\ \mathrm{mol\,L^{-1}}} \]
The exact equilibrium concentrations are approximately:
\[ [A]_{\mathrm{eq}}=1.00496 \]
\[ [B]_{\mathrm{eq}}=0.01496 \]
\[ [C]_{\mathrm{eq}}=0.01504 \]
All concentrations are in \(\mathrm{mol\,L^{-1}}\).
Compare the two results:
| Quantity | Approximate | Exact |
|---|---|---|
| Reverse extent \(y\) | 0.00500 | 0.004963 |
| A concentration | 1.000 | 1.00496 |
| B concentration | 0.0150 | 0.01496 |
| C concentration | 0.0150 | 0.01504 |
The approximate reverse extent differs from the exact result by less than 1%.
The approximation works because the only concentration change we ignored was small relative to the initial concentration of A.
The final distinction to remember: a concentration can change substantially without invalidating an approximation, provided you keep its change in the equilibrium expression. What matters is the size of the changes you actually neglect.
07Where to go next
Once you have established that an approximation is valid, use the resulting equilibrium concentrations to calculate \(K_c\), \(K_a\), or the quantity requested in the question.
If the approximation fails, return to the complete ICE expressions and solve the exact equilibrium equation. Then check every mathematical root against the physically possible concentration range.
That is the difference between obtaining a convenient answer and obtaining an answer that actually describes the equilibrium mixture.