Le Chatelier's principle — pressure, temperature, concentration, catalyst effects: Worked examples

Recall the two quantities from the parent note:

Why does moving forward raise $Q$ ? Because "forward" means products grow (numerator up) and reactants shrink (denominator down), so the fraction $Q$ rises. Moving backward does the reverse. So the reaction always has a "handle" to push $Q$ toward $K$ .

The scenario matrix

Every problem this topic can throw at you falls into exactly one of these cells. The last column names the worked example that covers it.

Cell	Stress type	The twist / edge	Covered by
C1	Concentration	Add a reactant	Example 1
C2	Concentration	Remove a product	Example 2
C3	Concentration	Degenerate: add a pure solid/liquid (no effect)	Example 3
P1	Pressure/Volume	Compress — sides differ in gas moles	Example 4
P2	Pressure/Volume	Degenerate: equal gas moles → no shift	Example 5
P3	Pressure/Volume	Trap: add inert gas at constant volume → no shift	Example 5
T1	Temperature	Exothermic, heat up ( $Δ H < 0$ ) — $K$ changes	Example 6
T2	Temperature	Endothermic, heat up ( $Δ H > 0$ ) — $K$ changes, with numbers	Example 7
X1	Catalyst	Zero-effect on position, only speeds arrival	Example 8
W1	Word problem	Real industry: choose conditions for max yield	Example 9
E1	Exam twist	Two stresses fight each other	Example 10

Notice the three "degenerate" cells (C3, P2, P3) and the "zero-effect" cell (X1): these are where students lose marks, so each gets its own fully-worked example.

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

Concentration cases (C1, C2, C3)

Step 1 — Confirm we start balanced. $Q_{0} = \frac{[ S O _{3} ] ^{2}}{[ S O _{2} ] ^{2} [ O _{2} ]} = \frac{( 0.80 ) ^{2}}{( 0.50 ) ^{2} ( 0.30 )} = \frac{0.64}{0.075} = 8.53$ Why this step? We must first know whether $0.50/0.30/0.80$ was really equilibrium. It gives $Q_{0} = 8.53$ , not $5.0$ — so this set was not at equilibrium to begin with. That is itself a lesson: always test, never assume. But the exam intends the disturbed state, so we proceed to the actual stress.

Step 2 — Recompute $Q$ after the stress. $Q = \frac{( 0.80 ) ^{2}}{( 0.70 ) ^{2} ( 0.30 )} = \frac{0.64}{0.147} = 4.35$ Why this step? Adding $S O_{2}$ enlarged the denominator, so $Q$ drops. We now compare it to $K$ .

Step 3 — Compare and shift. $Q = 4.35 < K = 5.0$ → too few products → shift forward. Why this step? This is the one rule of the page. Forward means $[S O_{3}]$ rises, $[S O_{2}]$ and $[O_{2}]$ fall, until $Q$ climbs back to $5.0$ .

Verify: Forward motion increases the numerator and decreases the denominator, so $Q$ moves $4.35 \to 5.0$ — the correct direction. Adding a reactant giving a forward shift matches the parent's table. ✓

Step 1 — Check the starting equilibrium. $K = \frac{[ C H _{3} O H ]}{[ C O ] [ H _{2} ] ^{2}} = \frac{58.0}{( 1.00 ) ( 2.00 ) ^{2}} = \frac{58.0}{4.00} = 14.5 ✓$ Why this step? Proves our starting numbers really sit at $K$ , so any change from here is a genuine stress.

Step 2 — Compute $Q$ after removing product. $Q = \frac{29.0}{( 1.00 ) ( 2.00 ) ^{2}} = \frac{29.0}{4.00} = 7.25$ Why this step? Removing product shrinks the numerator, so $Q$ falls below $K$ .

Step 3 — Compare and shift. $Q = 7.25 < K = 14.5$ → shift forward, making more methanol. Why this step? The system replaces what you took away — that is Le Chatelier "partially undoing" the stress.

Verify: $Q$ needs to climb from $7.25$ back to $14.5$ ; a forward shift raises the numerator, doing exactly that. ✓

Step 1 — Write the correct $Q$ . $Q = P_{C O_{2}}$ Why this step? Pure solids and pure liquids have an "activity" of $1$ and never appear in $K$ or $Q$ . Only the gas $C O_{2}$ counts.

Step 2 — Ask what the stress changed. Adding solid $C a C O_{3}$ does not change $P_{C O_{2}}$ at all — solids have no concentration to raise. Why this step? Since $Q$ depends only on $P_{C O_{2}}$ , and that is untouched, $Q$ stays equal to $K$ .

Step 3 — Conclude. $Q = K$ → no shift. Adding more marble does nothing to the position.

Verify: The mass of solid does not appear anywhere in $Q = P_{C O_{2}}$ , so mathematically it cannot move equilibrium. This is the classic exam trap. ✓

Pressure / volume cases (P1, P2, P3)

Step 1 — Halving the volume doubles every concentration. $[N_{2} O_{4}] = 0.80 M$ , $[N O_{2}] = 0.80 M$ . Why this step? Concentration $=$ moles $/$ volume; the moles have not moved yet, but $V$ halved, so each concentration doubles instantly.

Step 2 — Compute the new $Q$ . $Q = \frac{[ N O _{2} ] ^{2}}{[ N _{2} O _{4} ]} = \frac{( 0.80 ) ^{2}}{0.80} = 0.80$ Why this step? The numerator was squared (doubled twice → $\times 4$ ) while the denominator only doubled ( $\times 2$ ), so overall $Q$ doubled to $0.80$ .

Step 3 — Compare and shift. $Q = 0.80 > K = 0.40$ → shift backward toward $N_{2} O_{4}$ . Why this step? $N_{2} O_{4}$ is the fewer-moles side ( $1$ mole vs $2$ ). Making $N_{2} O_{4}$ removes gas molecules, easing the pressure you just added — matching "compression → fewer moles."

Verify: Backward motion lowers $[N O_{2}]$ and raises $[N_{2} O_{4}]$ , pulling $Q$ from $0.80$ back down toward $0.40$ . Direction correct. Mole-count rule agrees. ✓

Step 1 — Part A: halve the volume, double all concentrations. $[H_{2}] = [I_{2}] = 0.20 M$ , $[H I] = 1.414 M$ . $Q = \frac{[ H I ] ^{2}}{[ H _{2} ] [ I _{2} ]} = \frac{( 1.414 ) ^{2}}{( 0.20 ) ( 0.20 )} = \frac{2.0}{0.04} = 50$ Why this step? Numerator doubled-then-squared ( $\times 4$ ), denominator doubled-times-doubled ( $\times 4$ ) — the $\times 4$ cancels. $Q$ is unchanged.

Step 2 — Part A conclusion. $Q = 50 = K$ → no shift. Why this step? Both sides have $2$ moles of gas ( $1 + 1$ vs $2$ ), so pressure changes them equally. Equal moles → pressure is powerless.

Step 3 — Part B: add argon at constant $V$ . The partial pressures (and concentrations) of $H_{2}$ , $I_{2}$ , $H I$ are unchanged, because volume did not change and argon does not appear in $Q$ . $Q = 50 = K \Rightarrow no shift$ Why this step? Total pressure rises, but $Q$ only sees the reacting species. Inert gas at fixed volume is invisible to equilibrium.

Verify: Part A: $\times 4/ \times 4 = 1$ , $Q$ untouched. Part B: argon absent from $Q$ , $V$ fixed → concentrations fixed → $Q$ fixed. Both give no shift. ✓

Temperature cases (T1, T2) — where $K$ itself changes

Temperature is different: it doesn't just move $Q$ , it rebuilds $K$ . The tool for this is the van 't Hoff Equation: $\frac{d l n K}{d T} = \frac{Δ H ^{\circ}}{R T ^{2}} ⟹ ln \frac{K _{2}}{K _{1}} = - \frac{Δ H ^{\circ}}{R} (\frac{1}{T _{2}} - \frac{1}{T _{1}})$

Step 1 — Classify the reaction. $Δ H^{\circ} < 0$ → exothermic → treat heat as a product: $N_{2} + 3 H_{2} ⇌ 2 N H_{3} + heat$ . Why this step? Writing heat as a product turns a temperature question into a familiar concentration question.

Step 2 — "Add" the product (heat). Raising $T$ is like adding product → shift backward toward $N_{2}$ and $H_{2}$ . Why this step? The system consumes the extra heat by running the endothermic (reverse) direction.

Step 3 — Sign-check $K$ with van 't Hoff. Since $Δ H^{\circ} < 0$ , $\frac{d ln K}{d T} < 0$ : heating lowers $K$ . Why this step? Confirms the shift with the exact tool — a smaller target $K$ forces a backward move.

Verify: Backward shift ⇒ $[N H_{3}] ↓$ , $[N_{2}], [H_{2}] ↑$ , and $K ↓$ . All three agree with $Δ H^{\circ} < 0$ . This is exactly why the Haber Process cannot simply crank up the temperature. ✓

Step 1 — Predict direction first. Endothermic → heat is a reactant → heating shifts forward → $K$ should increase. Why this step? Gives us a sanity target for the number we are about to compute.

Step 2 — Plug into integrated van 't Hoff. $ln \frac{K _{2}}{K _{1}} = - \frac{Δ H ^{\circ}}{R} (\frac{1}{T _{2}} - \frac{1}{T _{1}}) = - \frac{57200}{8.314} (\frac{1}{350} - \frac{1}{298})$ Why this step? Convert kJ to J ( $57.2 kJ = 57200 J$ ) so units cancel with $R$ .

Step 3 — Evaluate. $\frac{1}{350} - \frac{1}{298} = 0.0028571 - 0.0033557 = - 0.00049856 K^{- 1}$ . $ln \frac{K _{2}}{K _{1}} = - \frac{57200}{8.314} \times (- 0.00049856) = (- 6879.96) \times (- 0.00049856) = 3.430$ $\frac{K _{2}}{K _{1}} = e^{3.430} = 30.9 \Rightarrow K_{2} = 0.14 \times 30.9 \approx 4.32$ Why this step? The positive exponent confirms $K$ grows, as forecast.

Verify: $K_{2} = 4.32 > 0.14 = K_{1}$ : $K$ increased, matching the endothermic-heated prediction. Sign of the bracket is negative, sign of $- Δ H^{\circ} / R$ is negative, product positive → $K_{2} > K_{1}$ . ✓

Catalyst case (X1) — the zero-effect scenario

Step 1 — Compute $K$ before. $K = \frac{k _{f}}{k _{r}} = \frac{2.0}{0.50} = 4.0$ Why this step? Establishes the equilibrium position we are testing.

Step 2 — Apply the catalyst to both rate constants. $k_{f}^{'} = 1000 \times 2.0 = 2000 s^{- 1}$ , $k_{r}^{'} = 1000 \times 0.50 = 500 s^{- 1}$ . $K^{'} = \frac{k _{f}^{'}}{k _{r}^{'}} = \frac{2000}{500} = 4.0$ Why this step? A catalyst lowers Activation Energy by the same amount each way, so the same factor multiplies both $k_{f}$ and $k_{r}$ — and the factor cancels in the ratio.

Step 3 — Conclude. $K^{'} = K = 4.0$ → equilibrium position unchanged; only the time to reach it shrinks ( $1000 \times$ faster). Why this step? Yield is set by $K$ ; speed is set by the rate constants. The catalyst touches speed, not $K$ .

Verify: $2000/500 = 4.0 = 2.0/0.50$ . Identical. A catalyst can never change yield. ✓

Word problem (W1) and exam twist (E1)

Step 1 — Pressure choice. Reactant side $= 1 + 3 = 4$ mol gas; product side $= 2$ mol. Fewer moles on the product side. → Use high pressure: compression shifts toward fewer moles (products). Why this step? Pure Cell-P1 reasoning applied to design.

Step 2 — Temperature choice (equilibrium view). Exothermic → high $T$ shifts backward and shrinks $K$ (Cell T1). → For yield alone, use low temperature. Why this step? Maximizing the equilibrium amount of $N H_{3}$ wants small heat.

Step 3 — The real-world compromise. Low $T$ gives great yield but the reaction crawls (tiny rate). So industry uses a moderate $T$ (~450 °C), a catalyst (speed, not position — Cell X1), and high pressure (~200 atm). Why this step? Kinetics and equilibrium pull in opposite directions on temperature; the catalyst rescues the rate so a lower-than-max temperature is workable.

Verify: High pressure ✓ (fewer moles = products). Low $T$ ✓ for equilibrium (van 't Hoff: exothermic $K$ falls with $T$ ). Catalyst adds speed without moving $K$ ✓. Consistent with the Haber Process note. ✓

Step 1 — Effect of heating alone. Endothermic → heat is a reactant → heating shifts forward, making more $N O_{2}$ (and raising $K$ , as Example 7 showed: $K$ went $0.14 \to 4.32$ ). Why this step? Isolate one stress at a time.

Step 2 — Effect of compression alone. More gas moles on the product side ( $2$ vs $1$ ) → compression shifts backward, making less $N O_{2}$ (Example 4). Why this step? Isolate the other stress.

Step 3 — Resolve the conflict. The two stresses push opposite ways, so the net direction depends on magnitudes: a large temperature jump can dominate a mild compression, and vice-versa. With no numbers you can only say the effects partly cancel — you cannot name the winner without the actual $Δ T$ and volume factor. Why this step? The exam trap is to blindly apply one rule; the honest answer names the competition.

Verify: Heating (endo) ⇒ forward ⇒ $+ N O_{2}$ . Compression (more moles on product side) ⇒ backward ⇒ $- N O_{2}$ . Opposite signs ⇒ indeterminate without magnitudes. Internally consistent. ✓

Recall Self-test (reveal after guessing)

Adding a pure solid to a heterogeneous equilibrium shifts it which way? ::: Neither — solids are absent from $Q$ , so $Q = K$ still. (Cell C3) Compressing $H_{2} + I_{2} ⇌ 2 H I$ shifts it which way? ::: No shift — equal moles of gas both sides. (Cell P2) Pumping argon in at constant volume does what? ::: Nothing — inert gas doesn't appear in $Q$ and $V$ is unchanged. (Cell P3) Heating an endothermic reaction does what to $K$ ? ::: Raises $K$ (forward shift). (Cell T2) A catalyst changes the yield? ::: No — it changes only the time to reach equilibrium. (Cell X1)