When Compression Shifts Equilibrium-and When It Doesn’t

You halve the volume of an equilibrium mixture at constant temperature and immediately write, “Pressure increases, so equilibrium shifts to the right.” The equation is balanced, your explanation sounds reasonable, and you can still lose the mark.

The problem is that increasing pressure does not automatically favour the right-hand side. You need to compare the total stoichiometric coefficients of the gaseous reactants and products. Sometimes compression shifts equilibrium. Sometimes it doesn’t shift it at all.

Before reading further, predict what happens when you compress this equilibrium mixture at constant temperature:

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

There are two gaseous coefficient units on the left and two on the right. Compression causes no equilibrium shift in this case. The partial pressures and concentrations of the gases increase immediately, but the reaction does not proceed preferentially in either direction.

That distinction is the key to getting these questions right.

01Diagnose the mistake: count gas coefficients, not gas species

The usual pressure rule from Le Chatelier’s principle is straightforward:

• Compression: equilibrium shifts towards the side with fewer moles of gas, provided the two sides have different gaseous coefficient totals.
• Expansion: equilibrium shifts towards the side with more moles of gas, provided the totals differ.
• Equal gaseous coefficient totals: compression or expansion alone causes no equilibrium shift.

The important word is coefficients.

Consider the equilibrium:

\[ \ce{N2(g) + 3H2(g) <=> 2NH3(g)} \]

A common mistake is to count two gases on the left and one on the right. That counts the different types of gas, not their stoichiometric amounts.

For one stoichiometric reaction event, the balanced equation specifies four gas-mole units on the left and two on the right.

The equation is like a shopping list: one unit of nitrogen plus three units of hydrogen makes two units of ammonia. Counting gas species instead of coefficients is like counting a shopping list containing three apples and four bananas as just two items.

The analogy has a limit, though. The coefficients describe mole ratios for the reaction, not the actual quantities of gas currently present in the vessel.

For the ammonia equilibrium, compression favours the right-hand side because it has fewer gas-mole units. For the hydrogen iodide equilibrium, neither side has fewer gas-mole units, so the pressure change does not favour either direction.

This is one focused application of Le Chatelier’s principle. For the broader topic, see tutorgum’s HSC Chemistry Module 5 Equilibrium: Complete Topic Summary.

02Why equal gaseous coefficients produce no shift

The rule is useful, but understanding why it works will help when an exam question changes the conditions slightly.

Imagine compressing an equilibrium mixture to half its original volume while keeping its temperature constant. Assume the gases behave ideally.

Immediately after compression, before any reaction occurs, the partial pressure of every reacting gas doubles.

For an equilibrium to remain undisturbed, the reaction quotient must still equal the equilibrium constant.

The reaction quotient, \(Q_p\), is calculated from the current partial pressures of the gases using the same expression as \(K_p\), the equilibrium constant expressed in terms of partial pressures. At equilibrium, \(Q_p=K_p\).

For the hydrogen iodide equilibrium,

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

the reaction quotient is

\[ Q_p=\frac{p_{\ce{HI}}^2} {p_{\ce{H2}}p_{\ce{I2}}} \]

Here, \(p_X\) means the partial pressure of gas \(X\).

After compression to half the original volume, all three partial pressures double. Therefore,

\[ Q_p'=\frac{(2p_{\ce{HI}})^2} {(2p_{\ce{H2}})(2p_{\ce{I2}})} \]

The numerator increases by a factor of four, and the denominator also increases by a factor of four. They cancel.

So \(Q_p'=Q_p=K_p\). There is no driving force for a net reaction in either direction.

Notice what this does not mean. It does not mean the pressure stays the same, or that the concentrations remain unchanged. They increase immediately after compression. It means the change leaves the equilibrium ratio unchanged.

The general rule

For any balanced gaseous equilibrium, calculate

\[ \Delta n_{\text{gas}} = \text{sum of product gas coefficients} - \text{sum of reactant gas coefficients} \]

Only gaseous species count. Do not include solids, liquids, or aqueous species in these totals.

If every reacting gas’s partial pressure is multiplied by a factor \(f\), the reaction quotient changes according to

\[ Q_p'=Q_p f^{\Delta n_{\text{gas}}} \]

Here, \(f\) is the factor by which each reacting gas’s partial pressure changes immediately after the disturbance. For compression to half the original volume, \(f=2\). For expansion to twice the original volume, \(f=\frac12\).

This equation explains all three cases:

• If \(\Delta n_{\text{gas}}<0\), compression makes \(Q_p<K_p\), so the reaction proceeds forwards, towards the products.
• If \(\Delta n_{\text{gas}}=0\), compression leaves \(Q_p\) unchanged, so there is no shift.
• If \(\Delta n_{\text{gas}}>0\), compression makes \(Q_p>K_p\), so the reaction proceeds backwards, towards the reactants.

Expansion reverses the first and third outcomes.

You do not need to calculate \(Q_p\) in every HSC response. Counting the balanced gaseous coefficients is usually enough to predict the direction. The quotient explains why that shortcut works and helps when a question combines several disturbances.

03A decision method you can reuse in an exam

When a question mentions a pressure or volume change, use this sequence rather than jumping straight to a direction.

Step 1: Check the equation is balanced.

You cannot count gaseous coefficients reliably until the equation is balanced.

Step 2: Count only the gaseous coefficients.

Add the coefficients of all gaseous reactants. Separately add the coefficients of all gaseous products.

For example, in \(\ce{N2(g) + 3H2(g) <=> 2NH3(g)}\), the totals are four on the left and two on the right.

Step 3: Identify what actually changed.

Was the pressure increased by compressing the vessel, or was an inert gas added at constant volume? These are different disturbances.

Step 4: Apply the appropriate rule.

Step 5: State what happens to \(K_p\).

If temperature remains constant, \(K_p\) does not change. A pressure change can alter the equilibrium composition, but it does not change the equilibrium constant.

That last step matters because questions often combine pressure and temperature changes. The pressure rule alone cannot solve the temperature part.

04Worked examples

05Edge cases that break memorised shortcuts

Only gaseous coefficients count

Consider

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

The left side has zero gaseous coefficient units, while the right side has one. The solids do not contribute to the gas-mole count.

Compression shifts the equilibrium to the left, towards \(\ce{CaCO3(s)}\), reducing the amount of gaseous \(\ce{CO2}\) through the reverse reaction. \(K_p\) remains constant if temperature is unchanged.

A rise in total pressure does not always cause a shift

Suppose argon is added to an equilibrium mixture in a rigid vessel at constant temperature.

The total pressure increases, but the partial pressures of the reacting gases remain unchanged under the ideal-gas model. Their reaction quotient therefore remains unchanged, and there is no equilibrium shift.

This is different from compression, which increases the partial pressures of the reacting gases themselves.

Temperature changes are different from pressure changes

At constant temperature, changing the volume can change the reaction quotient and cause the system to re-establish equilibrium. It does not change \(K_p\).

Changing temperature can change \(K_p\). For an endothermic forward reaction, heating favours the forward reaction and increases \(K_p\). For an exothermic forward reaction, heating favours the reverse reaction and decreases \(K_p\).

If a question says that a vessel is heated and its pressure rises, do not automatically apply the compression rule. The volume may be unchanged, and the temperature change itself affects the equilibrium constant.

06Exam-day self-check

Before writing your final equilibrium prediction, ask yourself:

• Have I balanced the equation?
• Have I counted coefficients rather than distinct gas species?
• Have I excluded solids, liquids, and aqueous species from the gas-mole totals?
• Was the pressure changed by compression or expansion, or by adding an inert gas?
• Did the temperature change? If so, have I considered the effect on \(K_p\)?
• Have I distinguished the direction of the equilibrium shift from the final pressure or concentration compared with its original value?

That final check becomes especially important in more demanding questions. A reaction can shift left after compression even though a product’s final partial pressure is still higher than it was before compression.

07Questions and solutions

Question 1

Consider the equilibrium

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

The system is compressed at constant temperature. Predict the direction of the equilibrium shift and state whether \(K_p\) changes.

Solution 1

The equilibrium shifts to the left, and \(K_p\) remains unchanged.

Only gaseous species count in the pressure comparison. There are zero gaseous coefficient units on the left and one on the right, so the reverse reaction is favoured by compression.

The partial pressure of \(\ce{CO2}\) initially increases, prompting the reverse reaction to consume some of that gas. Because the temperature remains constant, the equilibrium constant does not change.

Question 2

Consider the equilibrium

\[ \ce{2SO2(g) + O2(g) <=> 2SO3(g)} \]

The system is initially at equilibrium at constant temperature.

(a) The volume is reduced to half its original value. Predict the shift.

(b) In a separate experiment, argon is added to a rigid vessel at the same temperature and volume. Predict the shift.

(c) State what happens to \(K_p\) in each experiment.

Solution 2

The equilibrium shifts right in (a), does not shift in (b), and \(K_p\) remains unchanged in both experiments.

(a) Compression: The reactant-side gaseous coefficient total is \(2+1=3\), while the product-side total is \(2\). Compression favours the side with fewer gas-mole units, so the equilibrium shifts right, producing more \(\ce{SO3}\).

(b) Argon addition: Adding argon at constant volume increases total pressure but does not change the partial pressures of the reacting gases, assuming ideal-gas behaviour. The reaction quotient remains unchanged, so the equilibrium does not shift.

(c) Equilibrium constant: Neither disturbance changes temperature. Therefore, \(K_p\) remains constant in both cases.

The trap is treating every increase in total pressure as equivalent to compression. The equilibrium responds to changes in the reacting gases’ partial pressures, not simply to the pressure reading on a gauge.

Question 3

An equilibrium mixture follows

\[ \ce{CO(g) + H2O(g) <=> CO2(g) + H2(g)} \]

At constant temperature, the volume is halved. After this disturbance, additional \(\ce{CO}\) is injected while the volume and temperature are kept constant.

Predict the equilibrium response to each change. Explain what happens to \(K_p\), and why the first change does not prevent the second from causing a shift.

Solution 3

Compression causes no shift, but adding \(\ce{CO}\) causes a shift to the right. The equilibrium constant remains unchanged throughout.

First change: compression. The gaseous coefficient totals are two on each side. Halving the volume doubles all four partial pressures immediately, but the factors cancel in the reaction quotient:

\[ Q_p=\frac{p_{\ce{CO2}}p_{\ce{H2}}} {p_{\ce{CO}}p_{\ce{H2O}}} \]

The numerator and denominator both increase by a factor of four. Thus, \(Q_p\) remains equal to \(K_p\), and no net reaction occurs.

Second change: adding \(\ce{CO}\). At fixed volume and temperature, adding \(\ce{CO}\) increases its partial pressure. This increases the denominator of the reaction quotient, lowering \(Q_p\) below \(K_p\).

The forward reaction then consumes \(\ce{CO}\) and \(\ce{H2O}\), producing \(\ce{CO2}\) and \(\ce{H2}\), until equilibrium is restored.

The subtle point is that no shift does not mean nothing physical changed. Compression increased the concentrations and partial pressures, but it preserved the equilibrium ratio. Adding a reactant changes that ratio and creates a new reason for the reaction to proceed.

Question 4

The forward reaction in the equilibrium

\[ \ce{2SO2(g) + O2(g) <=> 2SO3(g)} \]

is exothermic.

The system is initially at equilibrium. First, its temperature is increased at constant volume, and the mixture is allowed to re-equilibrate. Then, at the new temperature, the vessel is compressed to half its volume and allowed to re-equilibrate again.

For each disturbance, predict the direction of the shift and state what happens to \(K_p\). Can you determine whether the final amount of \(\ce{SO3}\) is greater or smaller than its original amount using these qualitative rules alone?

Solution 4

Heating shifts the equilibrium left and decreases \(K_p\). The subsequent compression shifts it right, but the final amount of \(\ce{SO3}\) relative to the original amount cannot be determined from the qualitative rules alone.

First change: heating. The forward reaction is exothermic, so heat can be treated as a product. Increasing temperature favours the endothermic reverse reaction, shifting equilibrium left and decreasing \(K_p\).

The rise in pressure caused by heating at constant volume does not make this a compression question. The temperature has changed, so the temperature effect on equilibrium must be considered.

Second change: compression. The gaseous coefficient totals are three on the left and two on the right. At the new, constant temperature, compression favours the right-hand side, shifting equilibrium towards \(\ce{SO3}\).

This compression does not change \(K_p\). It remains at the lower value established by the temperature increase.

Final comparison: The two disturbances drive the equilibrium in opposite directions. The qualitative rules identify the direction of each response, but they do not quantify how much \(\ce{SO3}\) is consumed during heating or produced during compression. Without sufficient numerical information about the system and the temperature change, the final amount cannot be compared reliably with the original amount.

Do not assume the second shift simply cancels the first.

Question 5

An ideal-gas mixture is initially at equilibrium for

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

At the initial equilibrium,

\[ p_{\ce{N2O4}}=1.0\ \text{atm} \]

and

\[ p_{\ce{NO2}}=1.0\ \text{atm} \]

Use the pressure expression

\[ K_p=\frac{p_{\ce{NO2}}^2}{p_{\ce{N2O4}}} \]

so that \(K_p=1.0\ \text{atm}\) for this reaction under the stated convention.

The volume is suddenly halved at constant temperature, and the mixture is allowed to re-equilibrate.

(a) Predict the direction of the equilibrium shift.

(b) Determine the final partial pressures of both gases.

(c) Explain why the final partial pressure of \(\ce{NO2}\) can be greater than its original value, even though the equilibrium shifts to the left.

Solution 5

The equilibrium shifts left, but the final partial pressure of \(\ce{NO2}\) is \(1.50\ \text{atm}\), which is greater than its original value of \(1.0\ \text{atm}\).

This question tests whether you understand what the direction of an equilibrium shift actually describes.

Step 1: Calculate the immediate effect of compression.

Halving the volume doubles both partial pressures before any reaction occurs:

\[ p_{\ce{N2O4}}=2.0\ \text{atm} \]

\[ p_{\ce{NO2}}=2.0\ \text{atm} \]

The reaction quotient immediately after compression is

\[ Q_p=\frac{(2.0)^2}{2.0} =2.0\ \text{atm} \]

Since \(Q_p>K_p\), the system must shift left to restore equilibrium. The reverse reaction consumes \(\ce{NO2}\) and produces \(\ce{N2O4}\).

Step 2: Express the reverse reaction’s progress.

Let \(x\) be the pressure-equivalent extent of the reverse reaction, measured in atm. At the fixed final volume and temperature, partial-pressure changes are proportional to changes in the amounts of gas.

The stoichiometry tells us that one unit of \(\ce{N2O4}\) forms for every two units of \(\ce{NO2}\) consumed. Therefore, the final partial pressures are

\[ p_{\ce{N2O4}}=2.0+x \]

\[ p_{\ce{NO2}}=2.0-2x \]

Step 3: Apply the equilibrium constant.

At the new equilibrium, the same \(K_p\) applies because temperature is unchanged:

\[ \frac{(2.0-2x)^2}{2.0+x}=1.0 \]

Taking all pressure values numerically in atm, rearrange:

\[ \begin{aligned} (2-2x)^2&=2+x\\ 4-8x+4x^2&=2+x\\ 4x^2-9x+2&=0 \end{aligned} \]

Solving gives

\[ x=\frac{9\pm7}{8} \]

The two mathematical roots are \(x=0.25\) and \(x=2.0\). The second root would give a negative partial pressure of \(\ce{NO2}\), which is physically impossible. Therefore,

\[ x=0.25\ \text{atm} \]

Step 4: Calculate the final partial pressures.

For nitrogen tetroxide,

\[ \begin{aligned} p_{\ce{N2O4}}&=2.0+0.25\\ &=\boxed{2.25\ \text{atm}} \end{aligned} \]

For nitrogen dioxide,

\[ \begin{aligned} p_{\ce{NO2}}&=2.0-2(0.25)\\ &=\boxed{1.50\ \text{atm}} \end{aligned} \]

Check the result:

\[ K_p=\frac{(1.50)^2}{2.25} =1.0\ \text{atm} \]

The final pressures satisfy the equilibrium condition.

The tempting answer is that a left shift must leave the final \(\ce{NO2}\) partial pressure below its original value. That confuses two different comparisons.

The shift describes what happens after compression, relative to the immediately compressed state. During that left shift, the partial pressure of \(\ce{NO2}\) falls from \(2.0\) to \(1.50\ \text{atm}\). But compression had already raised it from \(1.0\) to \(2.0\ \text{atm}\).

The reverse reaction partly offsets the pressure increase; it does not necessarily undo it completely.

This is why a shift direction alone cannot tell you whether a particular gas’s final partial pressure or concentration is above or below its original value.

The next useful step is to connect these qualitative predictions to equilibrium calculations. Once you can distinguish changes in \(Q_p\) from changes in \(K_p\), you can determine not just which way a system shifts, but also the conditions the new equilibrium must satisfy.