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

Before we start, one reminder in plain words. The whole game is comparing two numbers:

The picture below is the mental model for every single problem on this page — a number line with $K$ marked, and $Q$ as a bead that slides until it meets $K$ .

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

Level 1 — Recognition

Goal: read a stress and name the shift direction, no arithmetic.

Exercise 1.1

For $2 S O_{2} (g) + O_{2} (g) ⇌ 2 S O_{3} (g)$ , you inject extra $O_{2}$ . Which way does the equilibrium shift, and does $K$ change?

Recall Solution 1.1

WHAT the stress is: we added a reactant, $O_{2}$ . WHY it matters: the ratio $Q = \frac{[ S O _{3} ] ^{2}}{[ S O _{2} ] ^{2} [ O _{2} ]}$ has $[O_{2}]$ in the denominator. Making the denominator bigger makes $Q$ smaller, so now $Q < K$ . The shift: $Q < K$ means "too little product", so the system runs forward — it consumes some of the added $O_{2}$ and makes more $S O_{3}$ , sliding $Q$ back up to $K$ . Does $K$ change? No. Concentration stresses never change $K$ ; only temperature does. We just moved $Q$ ; $K$ sat still.

Exercise 1.2

For $N_{2} O_{4} (g) ⇌ 2 N O_{2} (g)$ , you compress the flask (smaller volume, higher pressure). Which side is favoured?

Recall Solution 1.2

WHAT to count: moles of gas on each side. Left: $1$ mole ( $N_{2} O_{4}$ ). Right: $2$ moles ( $N O_{2}$ ). WHY moles matter under pressure: squeezing the box crowds molecules together; the system relieves this by choosing the side with fewer gas molecules, because fewer molecules take up "less pressure room". The shift: fewer moles is the left side, so the equilibrium shifts backward toward $N_{2} O_{4}$ . $K$ unchanged (pressure is not a temperature stress).

Exercise 1.3

A reaction is exothermic, $Δ H < 0$ . You heat it up. Does $K$ increase, decrease, or stay the same?

Recall Solution 1.3

The trick: for an exothermic reaction, treat heat as a product: $reactants ⇌ products + heat$ WHAT heating does: adding heat is like adding product. Too much "product" ⇒ system runs backward. What happens to $K$ : since the resting ratio now sits with more reactants, $K$ itself decreases. Temperature is the only stress that actually changes $K$ (via the van 't Hoff equation).

Level 2 — Application

Goal: compute $Q$ , compare to $K$ , predict the shift.

Exercise 2.1

$2 S O_{2} (g) + O_{2} (g) ⇌ 2 S O_{3} (g)$ has $K_{c} = 5.0$ . Right now $[S O_{2}] = 0.70 M$ , $[O_{2}] = 0.30 M$ , $[S O_{3}] = 0.80 M$ . Is the system at equilibrium? If not, which way will it move?

Recall Solution 2.1

Step 1 — build $Q$ (WHAT): use current concentrations in the ratio. $Q = \frac{[ S O _{3} ] ^{2}}{[ S O _{2} ] ^{2} [ O _{2} ]} = \frac{( 0.80 ) ^{2}}{( 0.70 ) ^{2} ( 0.30 )} = \frac{0.64}{0.147} = 4.35$ Step 2 — compare (WHY): $Q = 4.35$ and $K = 5.0$ , so $Q < K$ . Step 3 — read the number line (WHAT IT LOOKS LIKE): on the picture, the $Q$ bead sits to the left of $K$ . To meet $K$ it must slide right → make more product. Answer: not at equilibrium; shifts forward, so $[S O_{3}]$ rises while $[S O_{2}]$ and $[O_{2}]$ fall until $Q$ climbs to $5.0$ .

Exercise 2.2

For $H_{2} (g) + I_{2} (g) ⇌ 2 H I (g)$ , $K_{c} = 50$ . A vessel holds $[H_{2}] = 0.20 M$ , $[I_{2}] = 0.20 M$ , $[H I] = 2.0 M$ . Predict the direction of net reaction.

Recall Solution 2.2

Step 1 — $Q$ : $Q = \frac{[ H I ] ^{2}}{[ H _{2} ] [ I _{2} ]} = \frac{( 2.0 ) ^{2}}{( 0.20 ) ( 0.20 )} = \frac{4.0}{0.04} = 100$ Step 2 — compare: $Q = 100 > K = 50$ . Too much product. Step 3 — shift: the $Q$ bead sits to the right of $K$ , so it slides left → the reaction runs backward, breaking down $H I$ into $H_{2}$ and $I_{2}$ until $Q$ falls to $50$ .

Exercise 2.3

$N_{2} (g) + 3 H_{2} (g) ⇌ 2 N H_{3} (g)$ . You remove some $N H_{3}$ as it forms (this is done industrially in the Haber process). Which way does equilibrium shift, and why is removing product a clever industrial move?

Recall Solution 2.3

WHAT happens to $Q$ : $N H_{3}$ is on top of $Q = \frac{[ N H _{3} ] ^{2}}{[ N _{2} ] [ H _{2} ] ^{3}}$ . Removing it shrinks the numerator, so $Q$ drops below $K$ . The shift: $Q < K$ ⇒ forward, making more $N H_{3}$ . WHY it's clever: by continuously siphoning off the product, you keep $Q$ permanently below $K$ , so the reaction never "finishes" — it keeps churning forward. You effectively harvest far more ammonia than a sealed vessel would ever allow.

Level 3 — Analysis

Goal: reason across competing effects and moles/pressure subtleties.

Exercise 3.1

Classify each reaction as pressure-sensitive (shifts on compression) or pressure-insensitive, and state the direction of shift on compression: (a) $N_{2} (g) + 3 H_{2} (g) ⇌ 2 N H_{3} (g)$ (b) $H_{2} (g) + I_{2} (g) ⇌ 2 H I (g)$ (c) $C (s) + C O_{2} (g) ⇌ 2 C O (g)$

Recall Solution 3.1

The rule: compression favours the side with fewer moles of gas. Count only gas moles.

(a) Left $= 1 + 3 = 4$ gas moles, right $= 2$ . Fewer on the right ⇒ compression shifts forward. Pressure-sensitive.
(b) Left $= 1 + 1 = 2$ , right $= 2$ . Equal ⇒ compression causes no shift. Pressure-insensitive.
(c) Carbon is a solid, so it doesn't count as gas. Left $= 1$ gas mole ( $C O_{2}$ ), right $= 2$ ( $C O$ ). Fewer on the left ⇒ compression shifts backward toward $C O_{2}$ . Pressure-sensitive. Key lesson: solids and liquids are invisible to pressure counting — only gas moles decide.

Exercise 3.2

For $N_{2} + 3 H_{2} ⇌ 2 N H_{3}$ ( $Δ H = - 92 kJ/mol$ , exothermic), you want maximum yield of $N H_{3}$ . Should you use high or low temperature? Then explain the real-world compromise the Haber process makes.

Recall Solution 3.2

Step 1 — pure equilibrium view: heat is a product (exothermic). To push forward (more $N H_{3}$ ) you must remove heat ⇒ use low temperature. Low $T$ also raises $K$ (van 't Hoff), so both the position and the constant favour product. Step 2 — the catch: at low $T$ the reaction is painfully slow — you would wait years to reach that lovely equilibrium. Step 3 — the compromise: the Haber process runs at a moderate temperature (~ $45 0^{\circ} C$ ) plus high pressure and an iron catalyst. This sacrifices some equilibrium yield for a reaction fast enough to be useful. It's a deliberate trade between how much (thermodynamics) and how fast (kinetics).

Exercise 3.3

A sealed flask holds $N_{2} O_{4} ⇌ 2 N O_{2}$ at equilibrium. You inject argon gas (inert, unreactive) at constant volume. Does the equilibrium shift?

Recall Solution 3.3

The trap in plain view: the total pressure went up when we added argon. Doesn't higher pressure shift toward fewer moles? WHY the answer is no: shifts depend on the partial pressures of the reacting gases, i.e. on their actual concentrations. At constant volume, adding argon does not change $[N_{2} O_{4}]$ or $[N O_{2}]$ — the reacting molecules are just as crowded as before. So $Q$ is untouched, $Q = K$ still holds. Answer: no shift. The extra total pressure is "dead weight" that never enters $Q$ . (Contrast: if you added argon at constant pressure, the volume must expand, diluting the real gases — and that would shift toward more moles.)

Level 4 — Synthesis

Goal: chain several effects and compute a new equilibrium quantitatively.

Exercise 4.1

$N_{2} O_{4} (g) ⇌ 2 N O_{2} (g)$ sits at equilibrium in a $2.0 L$ flask with $[N_{2} O_{4}] = 0.40 M$ and $[N O_{2}] = 0.20 M$ . (a) Find $K_{c}$ . (b) You now halve the volume to $1.0 L$ . Instantly (before any shift), what are the new concentrations and the new $Q$ ? Compare to $K$ and predict the direction — does it match the mole-counting rule?

Recall Solution 4.1

(a) Find $K_{c}$ (WHAT): $K_{c} = \frac{[ N O _{2} ] ^{2}}{[ N _{2} O _{4} ]} = \frac{( 0.20 ) ^{2}}{0.40} = \frac{0.040}{0.40} = 0.10$ (b) Halving the volume doubles every concentration (same moles in half the space): $[N_{2} O_{4}] = 0.80 M, [N O_{2}] = 0.40 M$ New $Q$ right after compression: $Q = \frac{( 0.40 ) ^{2}}{0.80} = \frac{0.16}{0.80} = 0.20$ Compare: $Q = 0.20 > K = 0.10$ . Too much product ⇒ shift backward toward $N_{2} O_{4}$ . Cross-check with mole counting: left has $1$ gas mole, right has $2$ ; compression favours the fewer-mole (left/backward) side. ✓ The two methods agree — the $Q$ calculation is just the quantitative proof of the mole-counting shortcut.

Exercise 4.2

Continue Exercise 4.1. After compression the system re-equilibrates in the $1.0 L$ flask. Using $K_{c} = 0.10$ , solve for the new equilibrium concentrations. (Let $x$ be the amount of $N O_{2}$ that reacts back, in mol/L.)

Recall Solution 4.2

Set up the change table (WHY $x$ ): starting from the just-compressed state $[N_{2} O_{4}] = 0.80$ , $[N O_{2}] = 0.40$ , some $N O_{2}$ converts back. Reverse reaction $2 N O_{2} \to N_{2} O_{4}$ : for every $2 x$ of $N O_{2}$ consumed, $x$ of $N_{2} O_{4}$ forms. $[N O_{2}] = 0.40 - 2 x, [N_{2} O_{4}] = 0.80 + x$ Apply $K_{c}$ : $0.10 = \frac{( 0.40 - 2 x ) ^{2}}{0.80 + x}$ Solve. Expand: $0.10 (0.80 + x) = (0.40 - 2 x)^{2}$ $0.080 + 0.10 x = 0.16 - 1.6 x + 4 x^{2}$ $4 x^{2} - 1.7 x + 0.080 = 0$ Quadratic formula, $x = \frac{1.7 \pm 1. 7 ^{2} - 4 ( 4 ) ( 0.080 )}{2 ( 4 )} = \frac{1.7 \pm 2.89 - 1.28}{8} = \frac{1.7 \pm 1.269}{8}$ . Two roots: $x = 0.371$ or $x = 0.0539$ . The first gives $[N O_{2}] = 0.40 - 0.742 < 0$ (impossible), so: $x = 0.0539$ New equilibrium: $[N O_{2}] = 0.40 - 2 (0.0539) = 0.292 M, [N_{2} O_{4}] = 0.80 + 0.0539 = 0.854 M$ Sanity check: $\frac{( 0.292 ) ^{2}}{0.854} = \frac{0.0854}{0.854} = 0.100 = K$ ✓. And note $[N O_{2}]$ dropped from its compressed $0.40$ toward $N_{2} O_{4}$ — exactly the backward shift we predicted.

Level 5 — Mastery

Goal: quantitative temperature effect via van 't Hoff, and multi-stress reasoning.

Exercise 5.1

For an exothermic reaction $Δ H^{\circ} = - 92, 000 J/mol$ , the equilibrium constant is $K_{1} = 6.0 \times 1 0^{- 2}$ at $T_{1} = 700 K$ . Predict $K_{2}$ at $T_{2} = 800 K$ using the integrated van 't Hoff equation. Use $R = 8.314 J mol^{- 1} K^{- 1}$ .

Recall Solution 5.1

WHY this tool: concentration and pressure stresses leave $K$ alone, but temperature changes $K$ itself. The van 't Hoff equation is precisely the bridge from a temperature change to the new $K$ . Its integrated form (assuming $Δ H^{\circ}$ roughly constant) is: $ln \frac{K _{2}}{K _{1}} = - \frac{Δ H ^{\circ}}{R} (\frac{1}{T _{2}} - \frac{1}{T _{1}})$ Step 1 — the temperature bracket: $\frac{1}{T _{2}} - \frac{1}{T _{1}} = \frac{1}{800} - \frac{1}{700} = 0.00125 - 0.0014286 = - 1.7857 \times 1 0^{- 4} K^{- 1}$ Step 2 — plug in ( $Δ H^{\circ} = - 92000$ ): $ln \frac{K _{2}}{K _{1}} = - \frac{- 92000}{8.314} \times (- 1.7857 \times 1 0^{- 4}) = (11066) \times (- 1.7857 \times 1 0^{- 4}) = - 1.976$ Step 3 — undo the log: $\frac{K _{2}}{K _{1}} = e^{- 1.976} = 0.1386$ $K_{2} = 0.1386 \times 6.0 \times 1 0^{- 2} = 8.3 \times 1 0^{- 3}$ Interpretation (WHAT IT MEANS): heating an exothermic reaction made $K$ smaller ( $0.060 \to 0.0083$ ). This is Le Chatelier in numbers: heat is a product, adding it drives the reaction backward, so the resting ratio holds less product. The qualitative rule and the equation tell one consistent story.

Exercise 5.2

The Haber reaction $N_{2} + 3 H_{2} ⇌ 2 N H_{3}$ is exothermic. An engineer proposes four independent changes to boost $N H_{3}$ yield. For each, state whether yield actually rises, and give the one-line reason. (a) Increase pressure. (b) Add a catalyst. (c) Lower temperature. (d) Continuously remove $N H_{3}$ .

Recall Solution 5.2

(a) Increase pressure — YES. Gas moles: $4$ left vs $2$ right; compression favours the fewer-mole product side ⇒ more $N H_{3}$ .
(b) Add a catalyst — NO (for yield). It speeds both directions equally, so $K$ and the equilibrium position are unchanged. It only makes the system reach that same yield faster. (Vital industrially, but not a yield booster.)
(c) Lower temperature — YES (for yield). Exothermic ⇒ heat is a product; removing heat pushes forward and raises $K$ . But it makes the reaction slow — the real plant compromises at ~ $45 0^{\circ} C$ .
(d) Remove $N H_{3}$ — YES. Shrinks the numerator of $Q$ , so $Q < K$ , driving continuous forward reaction. This is exactly why yield is harvested continuously. Master insight: only (b) fails to change the equilibrium position. The other three each move $Q$ or $K$ in the product-favouring direction — and the smart plant combines (a), (c-compromised), (d) with the catalyst (b) for speed.

Recall Quick self-test

Exothermic reaction heated — does $K$ rise or fall? ::: Falls (heat behaves as an added product). Adding inert gas at constant volume — shift? ::: No shift; no reactant concentration changed. Catalyst effect on $K$ ? ::: None; it changes only the speed to reach equilibrium. $Q = 100$ , $K = 50$ — direction? ::: Backward ( $Q > K$ ).

Related deeper notes: Chemical Equilibrium · Gibs Free Energy · Activation Energy · Haber Process.