Does a catalyst change the position of equilibrium?

No. A catalyst lowers the activation energy for the forward and reverse reactions by the same amount, so both rates increase by the same factor. The system reaches equilibrium sooner, but the equilibrium concentrations and the value of K are identical to what they would be without the catalyst. If a question asks how to increase yield, a catalyst is never the answer — changing concentration, pressure, or temperature is.

Why does temperature change K when nothing else does?

Concentration and pressure changes only move the system along to a new set of concentrations that still satisfy the same ratio, so Q returns to the original K. Temperature changes the rate constants of the forward and reverse reactions by different amounts, so the ratio itself changes. For an exothermic reaction, heating lowers K; for an endothermic reaction, heating raises K.

How do I know whether a pressure change will cause a shift?

Count the moles of gas on each side using the coefficients, ignoring solids and liquids. If the counts differ, compression shifts the reaction toward the side with fewer gas moles and expansion shifts it toward the side with more. If the counts are equal, there is no shift. Also remember that adding an inert gas at constant volume changes total pressure but not the concentrations of the reacting gases, so it causes no shift.

What is the difference between Q and K?

They use the same formula, but K is the value at equilibrium while Q is the value at any instant. Comparing them predicts direction: if Q K there is too much product, so it runs in reverse; if Q = K the system is at equilibrium. This is the quantitative version of Le Chatelier's reasoning.

CHEM-9.4

Equilibrium & Le Chatelier's Principle

Learn how dynamic equilibrium works and use Le Chatelier's principle to predict shifts from concentration, pressure, volume, and temperature changes.

What you'll do in this lesson

A voice-first session with the Crimsora tutor on Equilibrium & Le Chatelier's Principle, then targeted practice and FRQs — with the tutor adapting to where you get stuck.

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What this lesson covers

Some reactions never truly finish. Seal nitrogen and hydrogen in a hot, pressurized vessel and ammonia forms — but ammonia also breaks apart at the same time. Eventually the two processes balance, and the amounts of everything in the container stop changing even though molecules are still reacting furiously. That balanced state is called dynamic equilibrium, and it describes everything from the carbonation in soda to the oxygen carried by your blood.

This lesson has two jobs. First, you will build an accurate picture of what equilibrium is: equal forward and reverse rates, not equal amounts and not a frozen system. Second, you will learn Le Chatelier's principle, a tool for predicting which way a reaction shifts when you disturb it by adding a substance, squeezing the container, or changing the temperature. By the end you should be able to look at any reversible reaction and reason out, with a stated cause, which direction it moves.

Dynamic Equilibrium: Equal Rates, Not Equal Amounts

A reversible reaction is written with a double arrow, as in

N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g)

. Both directions happen at once. At the moment you mix reactants, the forward rate is high (lots of reactant collisions) and the reverse rate is zero (no product yet). As reactants get used up, the forward rate falls; as product builds, the reverse rate rises. When the two rates become equal, the system is at dynamic equilibrium.

The word dynamic matters. Molecules keep colliding and converting in both directions forever — nothing stops. What stops changing are the measurable concentrations, because every ammonia molecule that decomposes is replaced by a new one forming. This is why equilibrium is sometimes called a steady state at the macroscopic level and a busy one at the molecular level.

The single most common misconception is that equilibrium means equal amounts of reactant and product. It does not. A system can sit at equilibrium with 99 percent products and 1 percent reactants, or the reverse. The position of equilibrium is described by the equilibrium constant

K

, which for a general reaction

aA + bB \rightleftharpoons cC + dD

K = \frac{[C]^c[D]^d}{[A]^a[B]^b}

A large

K

means products dominate at equilibrium; a small

K

means reactants dominate. Note that only substances whose concentration can change appear in

K

— pure solids and pure liquids are left out.

Equilibrium also requires a closed system. If ammonia gas escapes, or if a product precipitates and is filtered out, the reverse reaction can never keep up and the system will keep drifting forward instead of settling.

Le Chatelier's Principle and Concentration Changes

Le Chatelier's principle states that if a system at equilibrium is disturbed, the system shifts in the direction that partially counteracts the disturbance. Think of it as the reaction pushing back against whatever you did.

For concentration, the logic is direct. Add more of a substance and the system consumes some of it; remove a substance and the system makes more of it.

Consider

2SO_2(g) + O_2(g) \rightleftharpoons 2SO_3(g)

Change made	Direction of shift	Result
Add $SO_2$	Right (toward products)	More $SO_3$ , some added $SO_2$ consumed
Remove $SO_3$	Right	System replaces the lost $SO_3$
Add $SO_3$	Left (toward reactants)	More $SO_2$ and $O_2$ form
Remove $O_2$	Left	System regenerates $O_2$

Two cautions. First, the shift only partially undoes your change: if you add

SO_2

, the new equilibrium still has more

SO_2

than before, just less than immediately after you added it. Students often claim concentrations return to their original values — they do not.

Second, changing a concentration does not change

K

at constant temperature. Adding reactant temporarily makes the reaction quotient

Q

smaller than

K

; the reaction runs forward until

Q = K

again. Comparing

Q

K

is the quantitative version of Le Chatelier: if

Q < K

the reaction shifts right, if

Q > K

it shifts left, and if

Q = K

nothing happens.

Adding or removing a pure solid or pure liquid causes no shift, since those do not appear in the expression for

Q

K

Pressure, Volume, and Inert Gases

Pressure changes matter only when gases are involved, and only when the number of moles of gas differs between the two sides of the equation.

Decreasing the volume of a container raises the pressure. The system responds by shifting toward the side with fewer moles of gas, because fewer gas particles means lower pressure. Increasing the volume lowers the pressure, and the system shifts toward the side with more moles of gas.

Count carefully using coefficients. In

N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g)

, the left has

1 + 3 = 4

moles of gas and the right has

2

. Compressing the container shifts the reaction right, toward ammonia. This is exactly why industrial ammonia synthesis runs at very high pressure.

In

H_2(g) + I_2(g) \rightleftharpoons 2HI(g)

, both sides have 2 moles of gas. Squeezing this system raises all concentrations equally and causes no shift at all. Recognizing this case is where many students go wrong — they assume every pressure change forces a shift.

Only gases count in the mole tally. In

C(s) + CO_2(g) \rightleftharpoons 2CO(g)

, the solid carbon is ignored, so the gas count is 1 on the left and 2 on the right; compression shifts it left.

One more subtlety: adding an inert gas such as argon at constant volume raises the total pressure but does not change the partial pressure or concentration of any reacting species, so there is no shift. A pressure change only matters if it actually changes the concentrations of the gases in the reaction.

Temperature, Catalysts, and Building a Reliable Method

Temperature is the only disturbance that actually changes the value of

K

. To predict its effect, treat heat as a substance in the equation.

For an exothermic reaction, heat is a product:

N_2 + 3H_2 \rightleftharpoons 2NH_3 + \text{heat}

, with

\Delta H = -92

kJ. Raising the temperature is like adding a product, so the system shifts left and

K

decreases — less ammonia at equilibrium. Cooling shifts it right and

K

increases.

For an endothermic reaction, heat is a reactant. Heating shifts it right and increases

K

; cooling shifts it left. This is the mechanism behind color-change demonstrations such as the cobalt chloride equilibrium, where a hot tube turns blue and an ice bath turns it pink.

A catalyst speeds up the forward and reverse reactions by exactly the same factor. It helps the system reach equilibrium sooner but changes neither the position of equilibrium nor

K

. Saying a catalyst increases yield is a frequent error.

Disturbance	Does $K$ change?	Does the position shift?
Add or remove a reactant or product	No	Yes, unless it is a pure solid or liquid
Change volume or pressure	No	Only if moles of gas differ across the arrow
Change temperature	Yes	Yes
Add a catalyst	No	No

A reliable method: identify the disturbance, ask what the system must do to partially undo it, count moles of gas if pressure is involved, and write heat into the equation if temperature is involved. Then state the direction and one observable consequence.

Reading Equilibrium in Real Systems

Equilibrium reasoning explains a lot of ordinary chemistry. A sealed bottle of soda holds

CO_2(aq) \rightleftharpoons CO_2(g)

at equilibrium under high pressure. Opening the cap releases the gas above the liquid, dropping the partial pressure of

CO_2

; the system shifts to replace it, and dissolved gas escapes as fizz. Leaving the bottle open removes

CO_2

continuously, so equilibrium is never re-established and the soda goes flat.

In your bloodstream,

Hb + O_2 \rightleftharpoons HbO_2

shifts right in the oxygen-rich lungs and left in oxygen-poor tissues, delivering oxygen where it is needed. At high altitude, low oxygen pressure shifts the equilibrium left, which is why acclimatization takes time.

Industrially, the Haber process shows a real tension. High pressure favors ammonia, and low temperature also favors ammonia because the reaction is exothermic — but at low temperature the rate is impractically slow. Engineers compromise: moderate temperature near 450 degrees Celsius, very high pressure, an iron catalyst to restore speed, and continuous removal of ammonia to keep pulling the equilibrium right. Notice how this combines this lesson with reaction rates: equilibrium tells you where a reaction ends up, kinetics tells you how fast it gets there, and the two answers can point in opposite directions.

When you analyze any real system, always ask whether it is closed. Open flames, evaporating solvents, and escaping gases all prevent equilibrium, which is why reactions that produce a gas that bubbles away tend to run essentially to completion.

Key terms

Reversible reaction.: A reaction that can proceed in both the forward and reverse directions, written with a double arrow such as $\rightleftharpoons$ .
Dynamic equilibrium.: The state of a closed system in which the forward and reverse reaction rates are equal, so concentrations remain constant while molecular change continues.
Equilibrium constant ( $K$ ).: The ratio of product concentrations to reactant concentrations, each raised to its coefficient, at equilibrium; it depends only on temperature.
Reaction quotient ( $Q$ ).: The same ratio as $K$ but calculated at any moment; $Q < K$ means the reaction shifts right, $Q > K$ means it shifts left.
Le Chatelier's principle.: When a system at equilibrium is disturbed, it shifts in the direction that partially counteracts the disturbance.
Shift.: A temporary period in which one direction of the reaction outpaces the other until equal rates are re-established at a new set of concentrations.
Closed system.: A system that exchanges no matter with its surroundings, a requirement for reaching equilibrium.
Catalyst.: A substance that lowers activation energy and speeds both directions equally, shortening the time to reach equilibrium without changing $K$ or the equilibrium position.

Worked example

For the equilibrium

2SO_2(g) + O_2(g) \rightleftharpoons 2SO_3(g)

\Delta H = -198

kJ, predict the direction of shift and the effect on the amount of

SO_3

for each change: (a) adding more

O_2

, (b) compressing the container to half its volume, (c) raising the temperature, (d) adding a catalyst, (e) adding argon gas at constant volume.

Start by rewriting the equation with heat included, since the reaction is exothermic:

2SO_2(g) + O_2(g) \rightleftharpoons 2SO_3(g) + \text{heat}

. Also count gas moles: 3 on the left, 2 on the right.

(a) Adding

O_2

increases a reactant concentration. The system consumes some of it, so the shift is to the right and the amount of

SO_3

increases. In terms of the quotient, adding reactant makes

Q < K

, so the reaction runs forward until

Q = K

.

(b) Halving the volume doubles the pressure. The system shifts toward the side with fewer moles of gas, which is the right side (2 moles versus 3). More

SO_3

forms.

(c) Raising the temperature is like adding heat, and heat is a product here. The system shifts left to consume the added heat, so

SO_3

decreases. This change also lowers the value of

K

— it is the only one of the five that does.

(d) A catalyst increases the forward and reverse rates equally. Equilibrium is reached faster, but the position does not move and the amount of

SO_3

is unchanged.

(e) Argon does not react and, at constant volume, does not change the concentration or partial pressure of

SO_2

O_2

, or

SO_3

. Total pressure rises but there is no shift.

Summary: right, right, left, no shift, no shift.

Practice questions

For

H_2(g) + I_2(g) \rightleftharpoons 2HI(g)

at equilibrium, the volume of the container is suddenly decreased. What happens?

The equilibrium shifts right, producing more $HI$
The equilibrium shifts left, producing more $H_2$ and $I_2$
No shift occurs because both sides have the same number of moles of gas
No shift occurs because a volume change never affects a gaseous equilibrium

Answer: No shift occurs because both sides have the same number of moles of gas

Compression raises pressure, and the system would shift toward the side with fewer moles of gas to relieve it. Here the left side has

1 + 1 = 2

moles of gas and the right side has 2 moles, so neither side offers relief. All concentrations increase by the same factor,

Q

stays equal to

K

, and nothing shifts. The last choice is wrong because volume changes do matter whenever the gas mole counts differ, as in ammonia synthesis.

A student says, 'The reaction has reached equilibrium, so it has stopped and there must be equal amounts of reactants and products.' Identify the two errors and correct them.

Answer: Both claims are wrong: the reaction continues at the molecular level with equal forward and reverse rates, and the amounts of reactants and products are constant but generally unequal.

Error one is treating equilibrium as static. Molecules keep reacting in both directions; what is constant is concentration, because each direction converts material at the same rate. Error two is confusing equal rates with equal concentrations. The relative amounts are set by the equilibrium constant: a large

K

means products dominate, a small

K

means reactants dominate. A system with 95 percent products and 5 percent reactants is fully at equilibrium as long as the two rates match.

For

N_2O_4(g) \rightleftharpoons 2NO_2(g)

\Delta H = +57

kJ.

N_2O_4

is colorless and

NO_2

is brown. Predict the color change when a sealed tube of this mixture is placed in an ice bath, and explain why.

Answer: The tube becomes lighter in color (paler), because cooling shifts the endothermic reaction to the left toward colorless $N_2O_4$ .

Because

\Delta H

is positive, heat behaves as a reactant:

N_2O_4 + \text{heat} \rightleftharpoons 2NO_2

. Removing heat by cooling causes the system to shift in the direction that releases heat, which is the reverse (left) direction. Brown

NO_2

is consumed and colorless

N_2O_4

forms, so the tube fades. Placing the same tube in hot water darkens it. Temperature is also the only change here that alters the value of

K

itself; cooling makes

K

smaller for this endothermic reaction.

FAQ

Does a catalyst change the position of equilibrium?: No. A catalyst lowers the activation energy for the forward and reverse reactions by the same amount, so both rates increase by the same factor. The system reaches equilibrium sooner, but the equilibrium concentrations and the value of $K$ are identical to what they would be without the catalyst. If a question asks how to increase yield, a catalyst is never the answer — changing concentration, pressure, or temperature is.
Why does temperature change $K$ when nothing else does?: Concentration and pressure changes only move the system along to a new set of concentrations that still satisfy the same ratio, so $Q$ returns to the original $K$ . Temperature changes the rate constants of the forward and reverse reactions by different amounts, so the ratio itself changes. For an exothermic reaction, heating lowers $K$ ; for an endothermic reaction, heating raises $K$ .
How do I know whether a pressure change will cause a shift?: Count the moles of gas on each side using the coefficients, ignoring solids and liquids. If the counts differ, compression shifts the reaction toward the side with fewer gas moles and expansion shifts it toward the side with more. If the counts are equal, there is no shift. Also remember that adding an inert gas at constant volume changes total pressure but not the concentrations of the reacting gases, so it causes no shift.
What is the difference between $Q$ and $K$ ?: They use the same formula, but $K$ is the value at equilibrium while $Q$ is the value at any instant. Comparing them predicts direction: if $Q < K$ there is too little product, so the reaction runs forward; if $Q > K$ there is too much product, so it runs in reverse; if $Q = K$ the system is at equilibrium. This is the quantitative version of Le Chatelier's reasoning.

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