How to Predict Temperature Shifts in Endothermic Equilibria

A sealed vessel contains a reversible reaction at equilibrium. The forward reaction is endothermic, so it absorbs energy. You raise the temperature of the vessel.

What happens next?

The forward and reverse reactions both become faster. However, they do not increase their rates by the same factor. The forward reaction becomes faster than the reverse reaction, so the mixture produces more products until a new equilibrium is established.

The direct rule is: when the forward reaction is endothermic, increasing temperature shifts equilibrium towards products. Decreasing temperature shifts it towards reactants.

The important part is understanding why the forward and reverse rates become unequal. Once you can explain that, you can predict temperature changes without relying on a memorised version of Le Chatelier’s principle.

01Start with the reaction system

Consider a general reversible reaction:

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

Assume the forward reaction is endothermic:

\[ \Delta H_{\mathrm{forward}}>0 \]

This means the forward reaction absorbs heat from its surroundings. The reverse reaction must therefore be exothermic.

Before predicting what heating does, establish exactly which part of the experiment is changing temperature.

Identify the system and surroundings

Imagine the reaction taking place inside a sealed, rigid glass vessel immersed in a temperature-controlled water bath.

There are two important boundaries:

• System: the reacting chemicals inside the vessel.
• Surroundings: the water bath, the vessel walls, and the external environment.

When the water bath is heated, thermal energy passes through the vessel walls and enters the reacting system.

No reactants or products enter or leave the sealed vessel. Because the vessel is rigid, its volume also remains constant.

This distinction is important because we want to isolate the effect of temperature.

Immediately after heating, the concentrations of A and B have not changed. The molecules have gained thermal energy, but the chemical reaction has not yet had time to change the mixture’s composition.

Also, heat is not literally a chemical reactant. We sometimes write an endothermic reaction as:

\[ \ce{A + heat <=> B} \]

This is useful energy bookkeeping. It indicates that the forward reaction absorbs energy, not that heat is a substance participating in the reaction.

If you need to revisit the difference between endothermic and exothermic reactions, see Endothermic and Exothermic Reactions for HSC Chemistry.

02Follow the forward and reverse reaction rates

Here is the key to predicting the temperature shift.

At equilibrium, the forward and reverse reactions are both occurring. Their rates are equal:

\[ r_f=r_r \]

Here:

• \(r_f\) is the forward reaction rate.
• \(r_r\) is the reverse reaction rate.

Equal rates mean that the concentrations of reactants and products remain constant.

They do not mean that the reactions have stopped.

What happens immediately when temperature increases?

Return to the original equilibrium mixture.

Before heating:

\[ r_f=r_r \]

Now increase the temperature.

The particles have greater average kinetic energy, and the reaction rates generally increase. Both the forward and reverse reactions become faster.

But here is the important detail:

The forward rate increases by a greater proportion than the reverse rate.

Consequently:

\[ r_f>r_r \]

The forward reaction is now occurring faster than the reverse reaction.

A net reaction occurs in the forward direction, consuming A and producing B.

As the concentrations change, the forward and reverse rates adjust. Eventually, they become equal again:

\[ r_f=r_r \]

The system has reached its new equilibrium.

Notice something important about the graph.

The forward reaction does not remain faster forever. It is faster only while the system is moving towards its new equilibrium.

At the new equilibrium, the forward and reverse rates are equal again.

Why does the forward rate increase more?

For a simple, single-step reversible reaction, we can understand this using activation energy.

Imagine an energy profile for the forward reaction.

Because the reaction is endothermic, the products have a higher energy than the reactants.

The activation energy of the forward reaction is therefore greater than that of the reverse reaction.

\[ E_{a,f}>E_{a,r} \]

Here, \(E_{a,f}\) and \(E_{a,r}\) represent the forward and reverse activation energies.

A higher activation energy means that a larger increase in temperature is needed to produce the same proportional change in the reaction rate.

The Arrhenius equation describes this temperature dependence:

\[ k=Ae^{-E_a/(RT)} \]

where:

• \(k\) is the reaction rate constant.
• \(A\) is the frequency factor.
• \(E_a\) is activation energy in \(\mathrm{J\,mol^{-1}}\).
• \(R\) is the gas constant, \(8.314\ \mathrm{J\,mol^{-1}\,K^{-1}}\).
• \(T\) is absolute temperature in kelvin.

For the simple one-step energy profile:

\[ E_{a,f}-E_{a,r}=\Delta H_{\mathrm{forward}} \]

Since the forward reaction is endothermic, its activation energy is greater.

Increasing temperature therefore produces a greater proportional increase in the forward rate constant.

Real reactions can involve several steps, so this energy-profile explanation is a simplified model rather than a universal description of every reaction mechanism. The overall equilibrium prediction remains the same.

What if the temperature decreases?

Now reverse the situation.

The system is initially at equilibrium, and you lower its temperature.

Both reaction rates generally decrease. However, the forward reaction is more temperature-sensitive because it is endothermic.

The forward rate decreases by a greater proportion than the reverse rate.

Therefore:

\[ r_r>r_f \]

The reverse reaction is now faster.

More reactants form, and the system moves towards a new equilibrium with a greater proportion of A.

The complete prediction is:

03A reliable method for predicting the temperature shift

When you encounter a temperature-change question in an HSC Chemistry exam, follow this sequence.

Step 1: Identify the direction of the endothermic reaction.

Check the sign of the forward enthalpy change.

For this topic:

\[ \Delta H_{\mathrm{forward}}>0 \]

Step 2: Identify the initial equilibrium condition.

Before the temperature changes:

\[ r_f=r_r \]

Step 3: Identify the temperature disturbance.

Determine whether thermal energy enters or leaves the reacting system.

• Heating means thermal energy enters the system.
• Cooling means thermal energy leaves the system.

Step 4: Compare the reaction rates immediately after the change.

For an endothermic forward reaction:

• Heating makes the forward reaction relatively faster.
• Cooling makes the reverse reaction relatively faster.

Step 5: State the net reaction and the new equilibrium.

Explain which substances increase and decrease.

Remember that the forward and reverse rates become equal again once the new equilibrium is established.

A contrasting case: what if the forward reaction is exothermic?

Consider:

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

Suppose the forward reaction is exothermic.

The reverse reaction is now endothermic.

If you increase the temperature, both rates generally increase, but the reverse rate increases by a greater proportion.

Therefore:

\[ r_r>r_f \]

The net reaction proceeds backwards, producing more \(\ce{SO2}\) and \(\ce{O2}\).

This is the opposite of the endothermic-forward case.

The distinction is not whether the temperature increases or decreases the reaction rates. It is which direction becomes relatively faster.

Three changes students often confuse

A catalyst does not produce the same effect as heating. It provides an alternative reaction pathway and speeds up the approach to equilibrium without changing the equilibrium composition.

Similarly, changing concentration can disturb the balance between reaction rates without changing the equilibrium constant.

Temperature is different because it changes the relative temperature dependence of the forward and reverse reactions.

04Worked examples

05Common mistakes to avoid

Before attempting exam questions, make sure you can distinguish these statements.

Mistake 1: Heating only increases the forward reaction rate.

Incorrect. Both forward and reverse rates generally increase when temperature increases. The forward rate increases by a greater proportion when the forward reaction is endothermic.

Mistake 2: The reverse reaction stops when the temperature increases.

Incorrect. The reverse reaction continues. It is simply slower than the forward reaction during the adjustment towards the new equilibrium.

Mistake 3: The forward reaction remains faster at the new equilibrium.

Incorrect. At the new equilibrium, the forward and reverse rates are equal again.

Mistake 4: An endothermic forward reaction means the system must cool down when heated.

Incorrect. Endothermic describes the energy absorbed by the forward chemical reaction. It does not mean the external heating process cannot increase the system’s temperature.

Mistake 5: Any increase in pressure caused by heating is equivalent to compressing the vessel.

Incorrect. Heating a gas in a rigid container increases its pressure without changing its volume. Physically compressing a gas changes its volume and concentrations. These are different disturbances and must be considered separately.

06Questions and solutions

The following questions progress from straightforward temperature predictions to situations where temperature interacts with other changes to the reaction system.

Question 1

Consider the reversible reaction:

\[ \ce{X(g) <=> Y(g)} \]

The forward reaction is endothermic.

The reaction mixture is initially at equilibrium inside a sealed, rigid container at \(320\ \mathrm{K}\). The temperature is increased to \(380\ \mathrm{K}\).

Explain:

a) What happens to the forward and reverse reaction rates immediately after heating?

b) Which direction does the net reaction proceed?

c) What happens to the concentrations of X and Y as the new equilibrium is established?

Solution 1

The equilibrium shifts towards Y because the forward reaction is endothermic.

Part a: Reaction rates

Immediately after heating, both reaction rates increase.

However, the forward rate increases by a greater proportion because the forward reaction absorbs heat.

Therefore:

\[ r_f>r_r \]

Part b: Net reaction

Since the forward reaction is faster, the net reaction proceeds from X to Y.

Part c: Concentrations

As the reaction proceeds:

• The concentration of X decreases.
• The concentration of Y increases.

Eventually, the forward and reverse rates become equal again.

The important distinction is that heating creates an initial difference between the rates. The reaction then changes the composition until dynamic equilibrium is restored.

Question 2

A reversible reaction is represented by:

\[ \ce{M(g) + heat <=> N(g)} \]

The system is initially at equilibrium at \(350\ \mathrm{K}\). Both reaction rates are:

\[ r_f=r_r=0.012\ \mathrm{mol\,L^{-1}\,s^{-1}} \]

The temperature is increased to \(420\ \mathrm{K}\).

Immediately after heating, the measured rates are:

\[ r_f=0.030\ \mathrm{mol\,L^{-1}\,s^{-1}} \]

\[ r_r=0.018\ \mathrm{mol\,L^{-1}\,s^{-1}} \]

Ten seconds later, the rates are:

\[ r_f=r_r=0.026\ \mathrm{mol\,L^{-1}\,s^{-1}} \]

All data are illustrative.

a) Calculate the initial net reaction rate immediately after heating.

b) Explain why the reverse reaction rate also increases, even though the reverse reaction is exothermic.

c) Explain why the forward rate decreases and the reverse rate increases between the immediate measurement and the ten-second measurement.

d) What does the ten-second measurement tell you about the system?

Solution 2

The reaction initially proceeds in the forward direction, and the measurements indicate that it has reached its new dynamic equilibrium after ten seconds.

Part a: Initial net reaction rate

The net rate is the forward rate minus the reverse rate:

\[ r_{\mathrm{net}}=r_f-r_r \]

Substituting:

\[ \begin{aligned} r_{\mathrm{net}}&=0.030-0.018\\ &=0.012\ \mathrm{mol\,L^{-1}\,s^{-1}} \end{aligned} \]

The positive result confirms that the net reaction proceeds towards N.

Part b: Why does the reverse rate increase?

The reverse reaction is exothermic, but this does not mean that heating decreases its reaction rate.

Increasing temperature generally increases the rates of both forward and reverse reactions because more particles have sufficient energy to overcome their activation energies.

The forward reaction is more temperature-sensitive in this case, so its rate increases by a greater proportion.

Part c: Why do the rates change over ten seconds?

The forward reaction initially occurs faster, consuming M and producing N.

As the concentration of M decreases, the forward rate decreases.

As the concentration of N increases, the reverse rate increases.

These changes continue until the two reaction rates become equal.

Part d: What does the final measurement establish?

The equality of the rates indicates that the system has reached its new dynamic equilibrium, assuming no other reactions or disturbances occur.

The rates are not zero:

\[ r_f=r_r=0.026\ \mathrm{mol\,L^{-1}\,s^{-1}} \]

Both reactions continue, but there is no net change in the concentrations.

Notice that the new equilibrium reaction rates are greater than the original rates of \(0.012\ \mathrm{mol\,L^{-1}\,s^{-1}}\). Reaching equilibrium does not mean returning to the original reaction conditions.

Question 3

Consider the equilibrium:

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

The forward reaction is endothermic.

The reaction takes place inside a cylinder fitted with a movable piston.

Two changes occur in sequence:

• Stage 1: The piston is locked in position, and the temperature is increased from \(300\ \mathrm{K}\) to \(380\ \mathrm{K}\). The system reaches its new equilibrium.
• Stage 2: The temperature is held constant at \(380\ \mathrm{K}\), and the piston compresses the gas to half its original volume.

Assume the forward and reverse reactions are elementary, with rate laws:

\[ r_f=k_f[\ce{N2O4}] \]

\[ r_r=k_r[\ce{NO2}]^2 \]

a) Explain the direction of the net reaction during Stage 1.

b) During Stage 2, determine how the forward and reverse rates change immediately after compression.

c) Identify the direction of the net reaction after compression and explain why it differs from the direction in Stage 1.

d) Explain why the pressure increase during Stage 1 should not be treated as a separate compression disturbance.

Solution 3

Stage 1 produces a net forward reaction, while Stage 2 produces a net reverse reaction.

Part a: Heating at constant volume

The forward reaction is endothermic.

When the temperature increases, both reaction rates increase, but the forward rate increases by a greater proportion.

Therefore:

\[ r_f>r_r \]

The net reaction proceeds towards \(\ce{NO2}\).

Part b: Rate changes during compression

The volume is reduced to half its original value.

Because the amount of each gas has not yet changed, the concentrations of both gases immediately double.

For the forward reaction:

\[ r_f=k_f[\ce{N2O4}] \]

Doubling the concentration doubles the forward rate:

\[ r_f'=2r_f \]

For the reverse reaction:

\[ r_r=k_r[\ce{NO2}]^2 \]

Doubling the concentration gives:

\[ r_r'=k_r(2[\ce{NO2}])^2 \]

\[ r_r'=4r_r \]

The forward rate doubles, while the reverse rate quadruples.

Part c: Direction after compression

Immediately before compression, the system was at equilibrium, so:

\[ r_f=r_r \]

After compression:

\[ r_r'=4r_r \]

\[ r_f'=2r_f \]

Since the original rates were equal:

\[ r_r'>r_f' \]

The net reaction therefore proceeds in the reverse direction, producing more \(\ce{N2O4}\).

This is also consistent with the pressure principle: the reverse reaction produces fewer gas particles, so compression favours the reverse direction.

The important point is that the two disturbances have opposite effects.

Heating favours the endothermic forward reaction, whereas compression favours the side with fewer gas particles.

Part d: Pressure increase during heating

During Stage 1, the piston is locked, so the volume remains constant.

Heating an ideal gas at constant volume increases its pressure, but this does not mean the gas has been compressed.

No volume change has occurred, and the concentrations remain unchanged at the instant of heating.

The initial equilibrium shift is therefore attributed to the temperature change.

Stage 2 is different because the piston physically reduces the volume, increasing the concentrations of both gases.

This question illustrates why you must identify what is actually changing rather than treating every pressure increase as the same disturbance.

Question 4

A reversible reaction is represented by:

\[ \ce{R(g) + heat <=> S(g)} \]

The reaction mixture is initially at equilibrium in a sealed, rigid vessel.

After increasing the temperature, the measured rates are:

\[ r_f=0.030\ \mathrm{mol\,L^{-1}\,s^{-1}} \]

\[ r_r=0.020\ \mathrm{mol\,L^{-1}\,s^{-1}} \]

Before the system reaches its new equilibrium, a catalyst is added.

Measurements show that the catalyst triples both instantaneous reaction rates:

\[ r_f=0.090\ \mathrm{mol\,L^{-1}\,s^{-1}} \]

\[ r_r=0.060\ \mathrm{mol\,L^{-1}\,s^{-1}} \]

a) Calculate the net reaction rate before the catalyst is added.

b) Calculate the net reaction rate immediately after the catalyst is added.

c) Explain whether the catalyst changes the direction of the net reaction.

d) Explain whether the catalyst changes the final equilibrium composition compared with the same temperature change without a catalyst.

Solution 4

The net reaction remains in the forward direction after the catalyst is added. The catalyst changes how quickly equilibrium is reached, not the final equilibrium composition.

Part a: Net rate before adding the catalyst

\[ r_{\mathrm{net}}=r_f-r_r \]

Substituting:

\[ \begin{aligned} r_{\mathrm{net}}&=0.030-0.020\\ &=0.010\ \mathrm{mol\,L^{-1}\,s^{-1}} \end{aligned} \]

The positive value indicates a net forward reaction.

Part b: Net rate after adding the catalyst

\[ r_{\mathrm{net}}=0.090-0.060 \]

\[ r_{\mathrm{net}}=0.030\ \mathrm{mol\,L^{-1}\,s^{-1}} \]

The net forward rate has tripled.

Part c: Direction of the reaction

The net reaction remains forward because:

\[ r_f>r_r \]

The catalyst accelerates both directions, but it does not reverse the existing net direction.

The forward reaction continues to produce S.

Part d: Final equilibrium composition

The catalyst does not change the equilibrium constant or the final equilibrium composition.

The forward reaction is endothermic, so the higher temperature establishes a new equilibrium with a greater proportion of S.

The catalyst simply allows the system to reach that equilibrium more quickly.

At the new equilibrium:

\[ r_f=r_r \]

A common mistake is to assume that accelerating the reverse reaction must undo the temperature shift. It does not. The catalyst accelerates the approach to equilibrium in both directions without changing where equilibrium lies.

Question 5

A chemist studies a simple reversible gas reaction:

\[ \ce{P(g) <=> Q(g)} \]

The forward reaction is not initially known to be endothermic or exothermic.

Assume that both reactions are first order:

\[ r_f=k_f[P] \]

\[ r_r=k_r[Q] \]

The reaction mixture is initially at equilibrium at \(350\ \mathrm{K}\), with:

\[ [P]=0.75\ \mathrm{mol\,L^{-1}} \]

\[ [Q]=0.25\ \mathrm{mol\,L^{-1}} \]

The following illustrative rate constants are measured:

The chemist performs the following experiment:

1. The mixture is heated rapidly from \(350\ \mathrm{K}\) to \(420\ \mathrm{K}\).
2. Before any measurable reaction progress occurs, \(0.25\ \mathrm{mol\,L^{-1}}\) of Q is injected into the vessel through a sealed injection port.
3. The vessel is resealed and maintained at constant volume and \(420\ \mathrm{K}\).

Assume the injection does not change the temperature, and no other reactions occur.

a) Verify that the mixture was initially at equilibrium at \(350\ \mathrm{K}\).

b) Using the rate constants, determine whether the forward reaction is endothermic or exothermic.

c) If the chemist had only heated the original mixture without injecting additional Q, which direction would the net reaction have taken?

d) Determine the immediate forward and reverse rates after the additional Q is injected. Which direction does the net reaction now take?

e) Calculate the final equilibrium concentrations of P and Q at \(420\ \mathrm{K}\).

f) Explain why the net reaction after the injection is opposite to the direction that heating alone would favour.

Solution 5

The forward reaction is endothermic. Heating alone would favour Q, but injecting additional Q changes the immediate rate balance, causing the net reaction to proceed in reverse.

This is the crucial distinction: the direction of the net reaction depends on both the temperature and the current concentrations.

Part a: Verify the initial equilibrium

At \(350\ \mathrm{K}\), calculate the forward rate:

\[ r_f=k_f[P] \]

\[ r_f=0.010\times0.75 \]

\[ r_f=0.0075\ \mathrm{mol\,L^{-1}\,s^{-1}} \]

Calculate the reverse rate:

\[ r_r=k_r[Q] \]

\[ r_r=0.030\times0.25 \]

\[ r_r=0.0075\ \mathrm{mol\,L^{-1}\,s^{-1}} \]

Therefore:

\[ r_f=r_r \]

The mixture is initially at equilibrium.

Part b: Determine whether the forward reaction is endothermic

Compare the relative changes in the forward and reverse rate constants.

The forward rate constant changes from:

\[ 0.010\rightarrow0.024\ \mathrm{s^{-1}} \]

Its multiplication factor is:

\[ \frac{0.024}{0.010}=2.4 \]

The reverse rate constant changes from:

\[ 0.030\rightarrow0.040\ \mathrm{s^{-1}} \]

Its multiplication factor is:

\[ \frac{0.040}{0.030}=1.33 \]

Both rate constants increase, but the forward rate constant increases by a greater proportion.

This is consistent with an endothermic forward reaction.

The equilibrium ratio also increases:

\[ \frac{k_f}{k_r}\bigg|_{350\,\mathrm{K}} = \frac{0.010}{0.030} =0.333 \]

At \(420\ \mathrm{K}\):

\[ \frac{k_f}{k_r}\bigg|_{420\,\mathrm{K}} = \frac{0.024}{0.040} =0.600 \]

For this simple first-order reversible model, the equilibrium condition gives:

\[ \frac{[Q]}{[P]}=\frac{k_f}{k_r} \]

The larger equilibrium ratio at the higher temperature confirms that the forward reaction is favoured by heating.

Part c: Determine the effect of heating alone

Immediately after heating, the concentrations are still:

\[ [P]=0.75\ \mathrm{mol\,L^{-1}} \]

\[ [Q]=0.25\ \mathrm{mol\,L^{-1}} \]

Using the new rate constants:

\[ r_f=0.024\times0.75 \]

\[ r_f=0.018\ \mathrm{mol\,L^{-1}\,s^{-1}} \]

For the reverse reaction:

\[ r_r=0.040\times0.25 \]

\[ r_r=0.010\ \mathrm{mol\,L^{-1}\,s^{-1}} \]

Therefore:

\[ r_f>r_r \]

The net reaction would proceed forwards, producing more Q.

Part d: Calculate the rates after injecting additional Q

The chemist injects:

\[ 0.25\ \mathrm{mol\,L^{-1}} \]

of Q.

The new concentrations are:

\[ [P]=0.75\ \mathrm{mol\,L^{-1}} \]

\[ [Q]=0.25+0.25 \]

\[ [Q]=0.50\ \mathrm{mol\,L^{-1}} \]

The forward rate is:

\[ r_f=0.024\times0.75 \]

\[ r_f=0.018\ \mathrm{mol\,L^{-1}\,s^{-1}} \]

The reverse rate is:

\[ r_r=0.040\times0.50 \]

\[ r_r=0.020\ \mathrm{mol\,L^{-1}\,s^{-1}} \]

Therefore:

\[ r_r>r_f \]

The net reaction now proceeds in the reverse direction, producing more P.

Part e: Calculate the final equilibrium concentrations

At the new equilibrium:

\[ r_f=r_r \]

Therefore:

\[ k_f[P]=k_r[Q] \]

Rearranging:

\[ \frac{[Q]}{[P]}=\frac{k_f}{k_r} \]

At \(420\ \mathrm{K}\):

\[ \frac{[Q]}{[P]}=\frac{0.024}{0.040} \]

\[ [Q]=0.60[P] \]

The total concentration after injection is:

\[ [P]+[Q]=0.75+0.50 \]

\[ [P]+[Q]=1.25\ \mathrm{mol\,L^{-1}} \]

Substituting:

\[ [P]+0.60[P]=1.25 \]

\[ 1.60[P]=1.25 \]

\[ [P]=0.78125\ \mathrm{mol\,L^{-1}} \]

Therefore:

\[ [Q]=1.25-0.78125 \]

\[ [Q]=0.46875\ \mathrm{mol\,L^{-1}} \]

The final equilibrium concentrations are:

\[ \boxed{[P]=0.781\ \mathrm{mol\,L^{-1}}} \]

\[ \boxed{[Q]=0.469\ \mathrm{mol\,L^{-1}}} \]

We can verify the result using the reaction rates.

Forward:

\[ r_f=0.024\times0.78125 \]

\[ r_f=0.01875\ \mathrm{mol\,L^{-1}\,s^{-1}} \]

Reverse:

\[ r_r=0.040\times0.46875 \]

\[ r_r=0.01875\ \mathrm{mol\,L^{-1}\,s^{-1}} \]

The rates are equal, confirming the new equilibrium.

Part f: Explain the apparent contradiction

Heating alone would favour the forward reaction because it is endothermic.

However, the injection of additional Q changes the concentrations without changing the temperature.

This increases the reverse reaction rate. In this case, the increase is large enough to make:

\[ r_r>r_f \]

The net reaction therefore moves backwards, even though the forward reaction remains endothermic.

There is no contradiction.

The temperature determines the new equilibrium ratio, while the injection changes the current composition relative to that ratio.

This is why you cannot predict the net direction by looking at temperature alone when another disturbance occurs at the same time.

06Connecting reaction rates to the equilibrium constant

You now have two ways to check a temperature-shift prediction.

The first is the rate approach:

• Heating an endothermic forward reaction makes the forward rate relatively greater.
• Cooling makes the reverse rate relatively greater.
• The reaction continues until the forward and reverse rates are equal.

The second is the equilibrium-constant approach.

For an endothermic forward reaction, increasing temperature increases the equilibrium constant, \(K_\mathrm{eq}\). This means the new equilibrium is more product-favouring.

The two approaches describe the same change from different perspectives.

The rate approach explains how the system moves towards equilibrium. The equilibrium constant describes the composition ratio at equilibrium.

For a more detailed treatment of the relationship between temperature and equilibrium composition, continue with How Temperature Changes Keq in Endothermic and Exothermic Reactions.

When answering an HSC question, aim to connect both ideas: identify which reaction rate becomes greater immediately after the temperature change, explain the resulting net reaction, and finish by describing the new dynamic equilibrium.