Le Chatelier's principle — pressure, temperature, concentration, catalyst effects: Question bank

The one picture behind every answer

The single tool behind every answer is comparing where the reaction is now with where it wants to be. We give these two positions names in the definition block just below, then use them everywhere.

Figure — Le Chatelier's principle — pressure, temperature, concentration, catalyst effects

Look at the number line above: $Q$ (blue/red) sits somewhere, $K$ is the fixed target (yellow). If $Q$ is to the left of $K$ (too few products), the reaction runs forward (blue arrow) until they meet. If $Q$ is to the right (too many products), it runs backward (red arrow). Every trap on this page is really asking: did the stress move $Q$ , or did it move $K$ ?

Why temperature is the odd one out — the van 't Hoff picture

Step 1 — connect $K$ to energy. The link between the standard free-energy change and $K$ is $Δ G^{\circ} = - R T ln K \Rightarrow ln K = - \frac{Δ G ^{\circ}}{R T} .$

Step 2 — substitute the master relation. Because $Δ G^{\circ} = Δ H^{\circ} - T Δ S^{\circ}$ (defined above), replace $Δ G^{\circ}$ and split the fraction term by term: $ln K = - \frac{Δ H ^{\circ} - T Δ S ^{\circ}}{R T} = - \frac{Δ H ^{\circ}}{R T} + \frac{T Δ S ^{\circ}}{R T} = - \frac{Δ H ^{\circ}}{R T} + \frac{Δ S ^{\circ}}{R} .$ The $T$ in $T Δ S^{\circ} / R T$ cancels, which is why the second term becomes temperature-free.

Step 3 — differentiate with respect to $T$ . Treat $Δ H^{\circ}, Δ S^{\circ}$ as roughly constant. The $Δ S^{\circ} / R$ term is constant so its derivative is $0$ . For the first term, $- \frac{Δ H ^{\circ}}{R} \cdot \frac{1}{T}$ , we need $\frac{d}{d T} (\frac{1}{T})$ . Since $\frac{1}{T} = T^{- 1}$ , the power rule gives $\frac{d}{d T} T^{- 1} = - 1 \cdot T^{- 2} = - \frac{1}{T ^{2}}$ . Therefore $\frac{d l n K}{d T} = - \frac{Δ H ^{\circ}}{R} \cdot (- \frac{1}{T ^{2}}) = \frac{d ln K}{d T} = \frac{Δ H ^{\circ}}{R T ^{2}} (van ’t Hoff, see [[v a n^{'} t H o f f E q u a t i o n]]) .$

The figure plots $ln K$ against $T$ . Because $R > 0$ and $T^{2} > 0$ , the sign of the slope is simply the sign of $Δ H^{\circ}$ : endothermic ( $Δ H^{\circ} > 0$ ) climbs (green), exothermic ( $Δ H^{\circ} < 0$ ) falls (red). This is the reason temperature is special — it moves $K$ itself, not just $Q$ .

True or false — justify

A catalyst increases the yield of product at equilibrium

False — a catalyst speeds forward and reverse equally, so

K = k_{f} / k_{r}

is unchanged; it changes how fast you reach the same equilibrium, not the position.

Adding an inert gas at constant volume shifts a gas equilibrium

False — at constant volume the partial pressures of the reacting gases are untouched, so

Q

stays equal to

K

and nothing shifts.

Adding an inert gas at constant pressure never shifts a gas equilibrium

False — to keep pressure constant the volume must expand, which dilutes every reacting gas and shifts the system toward the side with more moles of gas.

Increasing temperature always increases the reaction rate and the yield together

False — it always speeds the rate, but for an exothermic reaction it lowers

K

, so yield drops even as rate rises (exactly the Haber-process dilemma, see Haber Process).

Temperature is the only stress that actually changes the value of

K

True — concentration and pressure changes only move

Q

toward the same

K

, whereas temperature changes

K

itself through the van 't Hoff Equation.

For a reaction with equal moles of gas on both sides, compression causes no shift

True — if the sum of exponents in numerator and denominator are equal, changing volume scales

Q

and

K

identically, so

Q = K

still holds.

If you add more of a pure solid reactant, the equilibrium shifts forward

False — pure solids (and pure liquids) have constant activity

= 1

and never appear in

Q

K

, so adding more changes nothing.

"The system shifts to completely undo the stress"

False — it only partially counteracts the stress; the new equilibrium sits between the disturbed state and the original, never fully back.

A catalyst lowers the enthalpy change

Δ H

of the reaction

False — a catalyst lowers the Activation Energy of both directions; the energies of reactants and products (and hence

Δ H

) are unchanged.

Doubling the volume of

H_{2} + I_{2} ⇌ 2 H I

shifts it toward products

False — reactant side has

1 + 1 = 2

mol gas and product side has

2

mol gas; moles are equal, so a volume change produces no shift.

Spot the error

"Add

N_{2}

N_{2} + 3 H_{2} ⇌ 2 N H_{3}

, so

Q

rises above

K

and it shifts back."

Error: adding

N_{2}

enlarges the denominator of

Q = [N H_{3}]^{2} / ([N_{2}] [H_{2}]^{3})

, so

Q

falls below

K

and the system shifts forward.

"Heat is a product in an exothermic reaction, so raising

T

adds product and shifts forward."

Error: adding product shifts backward, not forward — raising

T

on an exothermic reaction shifts toward reactants and lowers

K

"Compress

N_{2} O_{4} ⇌ 2 N O_{2}

, so it shifts toward

N O_{2}

to relieve the pressure."

Error: compression favors the side with fewer moles; reactant

N_{2} O_{4}

is 1 mol vs 2 mol of

N O_{2}

, so it shifts toward

N_{2} O_{4}

M n O_{2}

catalyses

H_{2} O_{2}

decomposition, so it drives the reaction further to completion."

Error: it only accelerates reaching equilibrium; the final amounts (the position) are set by

K

, which the catalyst does not touch.

"Since

Δ G = Δ G^{\circ} + R T ln Q

, at equilibrium

Δ G^{\circ} = 0

Error: it is

Δ G

(not

Δ G^{\circ}

) that is zero at equilibrium;

Δ G^{\circ} = - R T ln K

and is generally nonzero.

"Removing product from

A ⇌ B

makes

Q > K

, so it shifts backward."

Error: removing product

B

shrinks the numerator, so

Q < K

, and the system shifts forward to replace the lost product.

"A liquid-only reaction shifts toward fewer molecules when we squeeze the flask."

Error: pressure/volume shifts apply only to gas moles; liquids and solids are essentially incompressible and their equilibria are unmoved by squeezing.

Why questions

Why does raising temperature change

K

but raising concentration does not?

Concentration only displaces

Q

from a fixed target

K

, while temperature alters the thermodynamics (

Δ G^{\circ}

, via van 't Hoff), redefining the target

K

itself.

Why does compression shift toward fewer moles of gas rather than more?

Compressing raises every partial pressure; the side with the larger total exponent in

Q

grows faster, unbalancing

Q

from

K

, and the system re-forms molecules on the fewer-mole side to reduce total pressure.

Why can a catalyst never change equilibrium position even though it "helps the reaction"?

It lowers the barrier for forward and reverse by the same amount, multiplying both

k_{f}

and

k_{r}

by the identical factor, leaving

K = k_{f} / k_{r}

— the whole equilibrium position — untouched.

Why does the Haber process deliberately avoid very high temperatures?

Ammonia synthesis is exothermic, so high

T

lowers

K

and yield; a compromise temperature keeps the rate workable while not sacrificing too much equilibrium product.

Why is the sign of

Δ H^{\circ}

enough to predict the direction of a temperature shift?

The van 't Hoff relation

d ln K / d T = Δ H^{\circ} / (R T^{2})

has

R, T^{2} > 0

, so the sign of

Δ H^{\circ}

alone decides whether

K

rises or falls with

T

, and hence the shift direction.

Why does a stress produce only a partial, not total, restoration of the original state?

The system stops moving the instant

Q

returns to

K

; because the imposed change is still partly present, that balance point is a new equilibrium different from the old one.

Edge cases

What happens to a solid-only or pure-liquid-only equilibrium when you halve the volume?

Nothing — condensed phases have fixed activity of 1 and appear in neither

Q

nor

K

, so there is no pressure-driven shift.

For

2 S O_{2} + O_{2} ⇌ 2 S O_{3}

, what does removing

O_{2}

do?

It shrinks the denominator of

Q

, raising

Q

above

K

, so the system shifts backward to regenerate

O_{2}

(and

S O_{2}

If a reaction has

Δ H^{\circ} = 0

, how does

K

respond to temperature?

d ln K / d T = 0

, so

K

is independent of temperature — the equilibrium position simply doesn't care how hot it gets.

A closed rigid container versus an open flexible one: does "add inert gas" behave the same?

No — rigid (constant volume): no shift, because partial pressures are unchanged; flexible (constant pressure): shift toward more gas moles, because expansion dilutes the reactants.

At the exact instant a catalyst is added to a system already at equilibrium, what shifts?

Nothing shifts — the system is already at

Q = K

; the catalyst only matters before equilibrium, cutting the time to arrive.

If you add product but

Q

is still less than

K

afterward, which way does it go?

Still forward — the direction depends only on the comparison

Q

versus

K

, not on which species you touched; as long as

Q < K

it moves toward products.

Compressing a gas equilibrium with more moles on the reactant side — which way?

Toward the product side, because that side has fewer gas moles and re-forming it lowers total molecule count, relieving the pressure stress.

Recall One-line survival kit

Concentration/pressure change $Q$ toward a fixed $K$ ; only temperature moves $K$ itself (van 't Hoff, slope $= Δ H^{\circ} / R T^{2}$ ); catalysts move neither — they just speed the trip.