How to Calculate Gas Equilibrium at Constant Volume

At constant volume, calculate a gas equilibrium by using an initial-change-equilibrium (ICE) table. First, convert the initial amounts of gas into concentrations using \(c=n/V\). Next, use the balanced chemical equation to express each concentration change in terms of a single variable, \(x\). Finally, substitute the equilibrium concentrations into the equilibrium constant expression, solve for \(x\), and reject any solution that produces a negative concentration.

The important detail is that constant volume does not mean constant concentration. As the reaction proceeds, the amounts of reactants and products change, so their concentrations change too. The fixed volume simply means you can calculate those changes directly from the reaction’s mole ratios.

01Why constant volume makes equilibrium calculations easier

Imagine placing a mixture of gases in a sealed, rigid container. The container cannot expand or contract, and its volume stays fixed throughout the reaction.

As the reaction proceeds, some gas molecules are consumed and others are produced. The total number of gas molecules might increase or decrease, but the volume remains unchanged.

This is useful because concentration is defined as:

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

where:

• \(c\) is concentration in mol L\(^{-1}\).
• \(n\) is the amount of substance in moles.
• \(V\) is volume in litres.

When volume is constant, a change in the number of moles produces a proportional change in concentration:

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

For example, if a reaction consumes 0.20 mol of a gas in a 2.00 L container, its concentration decreases by:

\[ \Delta c=\frac{-0.20}{2.00} =-0.10\ \mathrm{mol\,L^{-1}} \]

The negative sign means the concentration decreases.

You do not need to recalculate the volume at each stage. You only need to track how much of each substance reacts.

02Build an ICE table

Consider the general gas reaction:

\[ aA(g)+bB(g)\rightleftharpoons cC(g)+dD(g) \]

An ICE table organises the calculation into three rows:

The change row is where most of the important reasoning happens.

Assume the forward reaction proceeds by an amount represented by \(x\), measured in mol L\(^{-1}\). The balanced equation determines how much each concentration changes.

For example, consider:

\[ A(g)+2B(g)\rightleftharpoons 3C(g) \]

If the forward reaction proceeds, the changes are:

• \(A\) decreases by \(x\).
• \(B\) decreases by \(2x\).
• \(C\) increases by \(3x\).

The resulting ICE table is:

All concentrations are in mol L\(^{-1}\). The values are illustrative.

Notice that the coefficients in the balanced equation become the multipliers of \(x\). This is not a guess. It follows directly from the mole ratios.

A useful sign convention: In this table, positive \(x\) represents a shift in the forward direction. If you initially choose the forward direction but the reaction actually shifts backwards, the algebra may produce a negative \(x\). Alternatively, you can first compare the reaction quotient, \(Q_c\), with \(K_c\), then define \(x\) as positive in the direction the reaction will actually proceed.

For a physically possible equilibrium, every concentration in the equilibrium row must be zero or greater.

03See how the method works in an actual gas equilibrium

Knowing how to build an ICE table is useful. But where do the equilibrium concentrations come from in the first place?

Consider the reversible reaction between nitrogen tetroxide and nitrogen dioxide:

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

Nitrogen tetroxide, \(\ce{N2O4}\), is colourless. Nitrogen dioxide, \(\ce{NO2}\), is brown. This makes the reaction useful for illustrating how a gas equilibrium can be investigated.

How would the experiment work?

A useful experimental design follows these steps:

1. Place a known amount of pure \(\ce{N2O4}\) into a rigid, sealed vessel of known volume.
2. Keep the vessel at a constant temperature using a temperature-controlled bath.
3. Allow the nitrogen tetroxide to dissociate, producing brown nitrogen dioxide.
4. Pass light through the vessel and measure the amount absorbed by the gas mixture.
5. Use a calibration between absorbance and nitrogen dioxide concentration to determine its concentration at equilibrium.

What should happen to the colour?

Initially, the gas is colourless. As \(\ce{N2O4}\) dissociates, the concentration of brown \(\ce{NO2}\) increases. The brown colour becomes more intense, then its intensity stops changing.

This stable reading indicates that the composition has stopped changing overall. The system has reached dynamic equilibrium: the forward and reverse reactions continue, but occur at equal rates.

There is an important experimental design choice here. A flexible container or gas syringe would allow the volume to change as the number of gas molecules changes. That would make it unsuitable for an investigation requiring constant volume. The rigid optical cell avoids this problem.

The temperature also needs to remain constant because the equilibrium constant depends on temperature.

This is an illustrative experimental design, rather than a claim about a particular historical experiment.

04Calculate equilibrium concentrations when \(K_c\) is given

When the equilibrium constant is supplied, you need to combine the ICE table with the equilibrium expression.

There is one decision worth making before you start substituting values: which direction will the reaction shift?

This is particularly important when both reactants and products are already present.

Use the reaction quotient to choose the direction

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

For:

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

the reaction quotient is:

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

Compare \(Q_c\) with \(K_c\):

Think of \(Q_c\) as a snapshot of the current mixture. \(K_c\) is the ratio that the mixture must reach at equilibrium at that temperature.

This comparison prevents a common mistake: automatically writing negative changes for reactants and positive changes for products.

Those signs depend on the direction of the shift, not just on which side of the equation a substance appears.

05What changes when the volume is not constant?

So far, the container has been rigid. That means the initial concentration row stays valid as the reaction proceeds, and only the chemical reaction changes the concentrations.

Now imagine replacing the rigid vessel with a cylinder containing a movable piston.

If the piston moves outwards, the gas expands. Even before the reaction has time to respond, the concentrations of all reacting gases decrease because the same amounts occupy a larger volume.

That means you must establish a new initial concentration row after the volume change.

Consider the equilibrium:

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

Suppose the system is initially at equilibrium with:

\[ [\ce{N2O4}]=0.300\ \mathrm{mol\,L^{-1}} \]

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

The equilibrium constant is:

\[ K_c=\frac{(0.200)^2}{0.300} =0.133\ \mathrm{mol\,L^{-1}} \]

Now double the container volume at constant temperature.

Immediately after expansion, both concentrations halve:

\[ [\ce{N2O4}]=0.150\ \mathrm{mol\,L^{-1}} \]

\[ [\ce{NO2}]=0.100\ \mathrm{mol\,L^{-1}} \]

Calculate the new reaction quotient:

\[ Q_c=\frac{(0.100)^2}{0.150} \]

\[ Q_c=0.0667\ \mathrm{mol\,L^{-1}} \]

Since:

\[ Q_c<K_c \]

the reaction shifts forwards, producing more nitrogen dioxide.

The important distinction is that changing the volume changes the concentrations immediately, while a chemical shift changes them according to the balanced equation.

At constant temperature, \(K_c\) itself does not change. The reaction proceeds until the new equilibrium mixture satisfies the same equilibrium constant.

Constant volume versus changing volume

There is another important point. At constant volume and temperature, pressure can still change as the reaction proceeds because the total number of gas moles can change.

So don’t assume that constant volume means constant pressure.

06A reliable method for HSC equilibrium calculations

Use this sequence whenever a question asks you to calculate equilibrium concentrations in a constant-volume vessel.

A few final checks are worth making before you submit your answer.

Check 1: Have you used the correct volume?

For a constant-volume vessel, use the same volume throughout the calculation. If a question describes an expansion or compression before equilibrium is re-established, calculate the new initial concentrations first.

Check 2: Have you used the correct coefficients?

For:

\[ \ce{2A(g) + B(g) <=> 3C(g)} \]

the changes must be \(-2x\), \(-x\), and \(+3x\), respectively.

Check 3: Are your equilibrium concentrations physically possible?

For example, if the initial concentration of A is 0.20 mol L\(^{-1}\), and its concentration decreases by \(2x\), then:

\[ 0.20-2x\geq0 \]

Therefore:

\[ x\leq0.10 \]

This constraint can eliminate a mathematical root before you even finish checking the equilibrium expression.

Check 4: Have you included the correct units?

Concentrations are expressed in mol L\(^{-1}\). The units of \(K_c\) depend on the equilibrium expression.

For example:

• \(\ce{N2O4(g) <=> 2NO2(g)}\): \(K_c\) has units of mol L\(^{-1}\).
• \(\ce{H2(g) + I2(g) <=> 2HI(g)}\): the concentration units cancel.

Follow the units required by the question and your course conventions.

Check 5: Does your answer satisfy the original equilibrium expression?

Substitute your final concentrations back into \(K_c\). Your calculated value should agree with the supplied equilibrium constant, allowing for rounding.

07Questions and solutions

The following questions progress from a straightforward ICE table to situations where you need to combine equilibrium calculations with changes to a gas system.

Question 1

Phosphorus pentachloride decomposes according to:

\[ \ce{PCl5(g) <=> PCl3(g) + Cl2(g)} \]

A 1.00 L rigid vessel initially contains 0.600 mol of \(\ce{PCl5}\), with no products present.

At equilibrium, the concentration of chlorine gas is measured to be 0.200 mol L\(^{-1}\).

Calculate:

a) The equilibrium concentration of each gas.

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

Solution 1

a) The equilibrium concentrations are 0.400 mol L\(^{-1}\) of \(\ce{PCl5}\), and 0.200 mol L\(^{-1}\) each of \(\ce{PCl3}\) and \(\ce{Cl2}\).

b) \(K_c=0.100\ \mathrm{mol\,L^{-1}}\).

Step 1: Write the initial concentrations.

The initial concentration of phosphorus pentachloride is:

\[ [\ce{PCl5}]_i=\frac{0.600}{1.00} =0.600\ \mathrm{mol\,L^{-1}} \]

Both products initially have zero concentration.

Step 2: Construct the ICE table.

Step 3: Use the measured equilibrium concentration.

\[ x=0.200 \]

Therefore:

\[ [\ce{PCl5}]_{\mathrm{eq}}=0.600-0.200 =0.400\ \mathrm{mol\,L^{-1}} \]

\[ [\ce{PCl3}]_{\mathrm{eq}}=0.200\ \mathrm{mol\,L^{-1}} \]

\[ [\ce{Cl2}]_{\mathrm{eq}}=0.200\ \mathrm{mol\,L^{-1}} \]

Step 4: Calculate \(K_c\).

The equilibrium expression is:

\[ K_c=\frac{[\ce{PCl3}][\ce{Cl2}]} {[\ce{PCl5}]} \]

Substitute:

\[ K_c=\frac{(0.200)(0.200)}{0.400} \]

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

The calculation is complete. Notice that the measured concentration of just one product was enough to determine all three equilibrium concentrations because the reaction coefficients fixed their relative changes.

Question 2

Hydrogen and iodine react according to:

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

A rigid vessel initially contains:

• \(0.500\ \mathrm{mol\,L^{-1}}\) of \(\ce{H2}\)
• \(0.300\ \mathrm{mol\,L^{-1}}\) of \(\ce{I2}\)
• \(0.100\ \mathrm{mol\,L^{-1}}\) of \(\ce{HI}\)

At the experimental temperature, \(K_c=4.00\).

Calculate the equilibrium concentration of each gas.

Solution 2

The equilibrium concentrations are 0.336 mol L\(^{-1}\) of \(\ce{H2}\), 0.136 mol L\(^{-1}\) of \(\ce{I2}\), and 0.428 mol L\(^{-1}\) of \(\ce{HI}\).

Step 1: Determine the direction of the reaction.

Initially:

\[ Q_c=\frac{(0.100)^2}{(0.500)(0.300)} \]

\[ Q_c=0.0667 \]

Since:

\[ Q_c<K_c \]

the reaction proceeds in the forward direction.

Step 2: Construct the ICE table.

Let \(x\) represent the decrease in hydrogen concentration.

Step 3: Substitute into the equilibrium expression.

\[ 4.00= \frac{(0.100+2x)^2} {(0.500-x)(0.300-x)} \]

Cross-multiply:

\[ (0.100+2x)^2 =4(0.500-x)(0.300-x) \]

Expand:

\[ 0.010+0.400x+4x^2 \]

\[ =0.600-3.200x+4x^2 \]

The \(4x^2\) terms cancel.

\[ 3.600x=0.590 \]

\[ x=0.1639 \]

Step 4: Calculate the equilibrium concentrations.

Hydrogen:

\[ [\ce{H2}]_{\mathrm{eq}} =0.500-0.1639 =0.336\ \mathrm{mol\,L^{-1}} \]

Iodine:

\[ [\ce{I2}]_{\mathrm{eq}} =0.300-0.1639 =0.136\ \mathrm{mol\,L^{-1}} \]

Hydrogen iodide:

\[ [\ce{HI}]_{\mathrm{eq}} =0.100+2(0.1639) =0.428\ \mathrm{mol\,L^{-1}} \]

The reaction produces more hydrogen iodide because the initial reaction quotient was smaller than the equilibrium constant.

An important detail is that the initial mixture already contained hydrogen iodide. You must include it in the initial row rather than treating every equilibrium problem as though it begins with reactants only.

Question 3

Nitrogen tetroxide dissociates according to:

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

A rigid vessel initially contains:

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

\[ [\ce{NO2}]=0.400\ \mathrm{mol\,L^{-1}} \]

At the experimental temperature:

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

Calculate the equilibrium concentration of each gas.

Your answer must identify the direction of the reaction and justify the rejection of any non-physical mathematical root.

Solution 3

The reaction shifts backwards. The equilibrium concentrations are approximately 0.230 mol L\(^{-1}\) of \(\ce{N2O4}\) and 0.339 mol L\(^{-1}\) of \(\ce{NO2}\).

Step 1: Calculate the initial reaction quotient.

\[ Q_c=\frac{(0.400)^2}{0.200} \]

\[ Q_c=0.800\ \mathrm{mol\,L^{-1}} \]

Since:

\[ Q_c>K_c \]

there is too much nitrogen dioxide relative to equilibrium. The reaction shifts backwards, forming more \(\ce{N2O4}\).

Step 2: Construct the ICE table.

Let \(x\) represent the increase in nitrogen tetroxide concentration.

Step 3: Substitute into \(K_c\).

\[ 0.500=\frac{(0.400-2x)^2}{0.200+x} \]

Cross-multiply:

\[ 0.500(0.200+x)=(0.400-2x)^2 \]

Expand:

\[ 0.100+0.500x =0.160-1.600x+4x^2 \]

Rearrange:

\[ 4x^2-2.100x+0.060=0 \]

Step 4: Solve the quadratic equation.

Using the quadratic formula:

\[ x=\frac{2.100\pm \sqrt{(2.100)^2-4(4)(0.060)}}{8} \]

\[ x=\frac{2.100\pm\sqrt{3.450}}{8} \]

This gives:

\[ x=0.0303 \]

or:

\[ x=0.495 \]

Step 5: Reject the non-physical root.

If \(x=0.495\):

\[ [\ce{NO2}]_{\mathrm{eq}} =0.400-2(0.495) \]

\[ =-0.590\ \mathrm{mol\,L^{-1}} \]

This is impossible. The vessel initially contained only 0.400 mol L\(^{-1}\) of nitrogen dioxide, so the reaction cannot consume more than that amount.

Therefore, reject \(x=0.495\).

Use:

\[ x=0.0303 \]

Step 6: Calculate the equilibrium concentrations.

Nitrogen tetroxide:

\[ [\ce{N2O4}]_{\mathrm{eq}} =0.200+0.0303 \]

\[ =0.230\ \mathrm{mol\,L^{-1}} \]

Nitrogen dioxide:

\[ [\ce{NO2}]_{\mathrm{eq}} =0.400-2(0.0303) \]

\[ =0.339\ \mathrm{mol\,L^{-1}} \]

The valid root gives non-negative concentrations and satisfies the supplied equilibrium constant.

The important lesson is that the quadratic formula does not know anything about chemistry. It produces mathematical roots, and you must decide which ones represent a possible physical mixture.

Question 4

The reaction:

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

has reached equilibrium in a vessel with an initial volume of 1.00 L.

At equilibrium:

\[ [\ce{N2O4}]=0.300\ \mathrm{mol\,L^{-1}} \]

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

The temperature remains constant, and the equilibrium constant is:

\[ K_c=0.1333\ \mathrm{mol\,L^{-1}} \]

The vessel is then expanded to 2.00 L using a movable piston.

Calculate:

a) The immediate concentrations after expansion.

b) The new reaction quotient and direction of the equilibrium shift.

c) The concentrations after the new equilibrium is established.

Solution 4

a) Immediately after expansion, the concentrations are 0.150 mol L\(^{-1}\) of \(\ce{N2O4}\) and 0.100 mol L\(^{-1}\) of \(\ce{NO2}\).

b) \(Q_c=0.0667\ \mathrm{mol\,L^{-1}}\), so the reaction shifts forwards.

c) The new equilibrium concentrations are approximately 0.133 mol L\(^{-1}\) of each gas.

Step 1: Calculate the concentrations immediately after expansion.

The volume doubles, but the amounts of each gas have not yet changed.

Therefore, each concentration halves:

\[ [\ce{N2O4}]_{\mathrm{new}} =\frac{0.300}{2} =0.150\ \mathrm{mol\,L^{-1}} \]

\[ [\ce{NO2}]_{\mathrm{new}} =\frac{0.200}{2} =0.100\ \mathrm{mol\,L^{-1}} \]

These are the new initial concentrations for the ICE table.

Step 2: Calculate the new reaction quotient.

\[ Q_c=\frac{(0.100)^2}{0.150} \]

\[ Q_c=0.0667\ \mathrm{mol\,L^{-1}} \]

Since:

\[ Q_c<K_c \]

the reaction shifts forwards, producing additional nitrogen dioxide.

Step 3: Construct the new ICE table.

Let \(x\) represent the increase in nitrogen dioxide divided by its stoichiometric coefficient of 2.

Step 4: Substitute into the equilibrium expression.

\[ 0.1333= \frac{(0.100+2x)^2}{0.150-x} \]

Using \(K_c=2/15\) before rounding:

\[ \frac{2}{15} =\frac{(0.100+2x)^2}{0.150-x} \]

Cross-multiply:

\[ 15(0.100+2x)^2=2(0.150-x) \]

Expand and simplify:

\[ 0.150+6x+60x^2=0.300-2x \]

\[ 60x^2+8x-0.150=0 \]

Step 5: Solve and check the roots.

Using the quadratic formula:

\[ x=\frac{-8\pm\sqrt{8^2-4(60)(-0.150)}}{120} \]

\[ x=\frac{-8\pm10}{120} \]

Therefore:

\[ x=0.0167 \]

or:

\[ x=-0.150 \]

The negative root would give:

\[ [\ce{NO2}]_{\mathrm{eq}} =0.100+2(-0.150) \]

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

This is impossible, so reject it.

Step 6: Calculate the new equilibrium concentrations.

\[ [\ce{N2O4}]_{\mathrm{eq}} =0.150-0.0167 \]

\[ =0.133\ \mathrm{mol\,L^{-1}} \]

\[ [\ce{NO2}]_{\mathrm{eq}} =0.100+2(0.0167) \]

\[ =0.133\ \mathrm{mol\,L^{-1}} \]

The new mixture has equal concentrations of the two gases, approximately 0.133 mol L\(^{-1}\).

The equilibrium constant has not changed. The volume change disturbed the original equilibrium, and the reaction shifted until the original \(K_c\) value was restored.

Question 5

A rigid 1.00 L vessel contains the following equilibrium mixture at constant temperature:

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

\[ [\ce{N2O4}]=0.300\ \mathrm{mol\,L^{-1}} \]

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

The system is at equilibrium.

Two additions are then made, in the following order:

1. Argon gas is added to the vessel. The temperature and volume remain constant.
2. After the system has settled, 0.100 mol of nitrogen dioxide is injected into the same vessel. The volume and temperature remain unchanged.

Assume ideal gas behaviour and that argon does not react.

Determine:

a) Whether adding argon shifts the equilibrium. Explain your reasoning.

b) The reaction quotient immediately after the nitrogen dioxide is added.

c) The direction of the subsequent equilibrium shift.

d) The final equilibrium concentrations of nitrogen tetroxide and nitrogen dioxide.

A student argues that both additions increase the total pressure, so both must shift the equilibrium towards the side with fewer gas molecules.

Explain precisely why this reasoning is incorrect.

Solution 5

Adding argon does not shift the equilibrium. Adding nitrogen dioxide does shift it backwards. The final concentrations are approximately 0.343 mol L\(^{-1}\) of \(\ce{N2O4}\) and 0.214 mol L\(^{-1}\) of \(\ce{NO2}\).

This question tests an important distinction: total pressure and the concentrations of the reacting gases are not interchangeable.

Step 1: Calculate the original equilibrium constant.

The initial mixture is already at equilibrium, so:

\[ K_c=\frac{(0.200)^2}{0.300} \]

\[ K_c=0.1333\ \mathrm{mol\,L^{-1}} \]

Step 2: Analyse the addition of argon.

Argon is an inert gas. It does not participate in the equilibrium reaction.

At constant temperature and volume, adding argon increases the total pressure, but does not change the number of moles of either reacting gas.

Using the ideal gas equation for each reacting gas:

\[ P_i=\frac{n_iRT}{V} \]

The partial pressures of \(\ce{N2O4}\) and \(\ce{NO2}\) remain unchanged because their amounts, the temperature, and the volume have not changed.

Their concentrations also remain unchanged.

Therefore:

\[ Q_c=K_c \]

The system remains at equilibrium. There is no net shift.

This is the crucial trap: Le Chatelier’s principle cannot be applied using total pressure alone. You must consider what actually happens to the reacting gases.

Step 3: Calculate the new initial concentrations after nitrogen dioxide is injected.

The vessel has a volume of 1.00 L.

The amount of nitrogen dioxide added is 0.100 mol.

Its concentration therefore increases by:

\[ \Delta[\ce{NO2}] =\frac{0.100}{1.00} \]

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

The nitrogen tetroxide concentration remains unchanged immediately after injection.

The new concentrations are:

\[ [\ce{N2O4}]_i=0.300\ \mathrm{mol\,L^{-1}} \]

\[ [\ce{NO2}]_i=0.200+0.100 \]

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

Step 4: Calculate the new reaction quotient.

\[ Q_c=\frac{(0.300)^2}{0.300} \]

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

Compare this with the equilibrium constant:

\[ Q_c=0.300>K_c=0.1333 \]

There is now too much nitrogen dioxide relative to equilibrium.

The reaction must shift backwards, producing more nitrogen tetroxide.

Step 5: Construct the ICE table.

Because the reaction shifts backwards, nitrogen tetroxide increases and nitrogen dioxide decreases.

Let \(x\) represent the increase in nitrogen tetroxide concentration.

Step 6: Substitute into the equilibrium expression.

\[ 0.1333= \frac{(0.300-2x)^2}{0.300+x} \]

Using the unrounded value \(K_c=2/15\):

\[ \frac{2}{15} = \frac{(0.300-2x)^2}{0.300+x} \]

Cross-multiply:

\[ 15(0.300-2x)^2=2(0.300+x) \]

Expand:

\[ 1.350-18x+60x^2 =0.600+2x \]

Rearrange:

\[ 60x^2-20x+0.750=0 \]

Step 7: Solve the quadratic equation.

Using the quadratic formula:

\[ x=\frac{20\pm\sqrt{(-20)^2-4(60)(0.750)}}{120} \]

\[ x=\frac{20\pm\sqrt{220}}{120} \]

This gives:

\[ x=0.0431 \]

or:

\[ x=0.290 \]

Step 8: Reject the non-physical root.

If \(x=0.290\):

\[ [\ce{NO2}]_{\mathrm{eq}} =0.300-2(0.290) \]

\[ =-0.280\ \mathrm{mol\,L^{-1}} \]

This is impossible because the reaction cannot consume more nitrogen dioxide than is available.

Reject this root and use:

\[ x=0.0431 \]

Step 9: Calculate the final equilibrium concentrations.

Nitrogen tetroxide:

\[ [\ce{N2O4}]_{\mathrm{eq}} =0.300+0.0431 \]

\[ =0.343\ \mathrm{mol\,L^{-1}} \]

Nitrogen dioxide:

\[ [\ce{NO2}]_{\mathrm{eq}} =0.300-2(0.0431) \]

\[ =0.214\ \mathrm{mol\,L^{-1}} \]

Therefore:

\[ \boxed{[\ce{N2O4}]_{\mathrm{eq}}=0.343\ \mathrm{mol\,L^{-1}}} \]

\[ \boxed{[\ce{NO2}]_{\mathrm{eq}}=0.214\ \mathrm{mol\,L^{-1}}} \]

The two additions have different effects despite both increasing total pressure.

• Argon increases total pressure without changing the reacting gas concentrations at constant volume. There is no equilibrium shift.
• Nitrogen dioxide increases the concentration of a reacting product. The reaction quotient increases above \(K_c\), so the system shifts backwards.

This is why you should calculate \(Q_c\) rather than automatically applying a pressure-based rule.

08Where to go next

The most important skill in a constant-volume equilibrium calculation is building an ICE table that accurately represents the reaction.

Once that becomes familiar, you can tackle more complicated equilibrium problems involving changes in pressure, volume, temperature, and concentration.

The next useful concept is Le Chatelier’s principle: predicting how an equilibrium responds to a disturbance. The reaction quotient gives you a precise way to check those predictions rather than relying on a memorised rule.

One final distinction: calculating an equilibrium composition is different from calculating the volume of gas produced by a reaction that goes to completion. For the latter, see tutorgum’s resources on Gas Volumes in Chemical Reactions for HSC Chemistry and How to Calculate CO₂ Volume from a Carbonate Reaction.