Reversible reactions and dynamic equilibrium: Exercises

This page is a self-test ladder. Each problem is stated cleanly; the full worked solution hides inside a collapsible [!recall]- callout, so you can try first and reveal after. Difficulty climbs from L1 Recognition (spot the idea) to L5 Mastery (combine everything).

Before you start, recall the one relationship we derived in the parent note the parent topic — for the simplest reversible reaction $A ⇌ B$ :

Everything below is built from that single equation plus plain reasoning. See the prerequisite notes Rate of reaction and rate constants and Collision theory if "rate ∝ concentration" feels shaky.

LEVEL 1 — Recognition

Exercise 1.1

Which of these systems can reach dynamic equilibrium? For each, answer yes/no and say why.

(a) Wood burning in open air.
(b) $N_{2} O_{4} ⇌ 2 N O_{2}$ sealed in a glass tube.
(c) A cup of water evaporating on a windowsill.
(d) Water evaporating inside a sealed bottle half-full of water.

Recall Solution 1.1

The two requirements are: the process must be reversible, and the system must be closed (nothing escapes).

(a) No. Burning is irreversible (the reverse rate is negligible) and the CO₂/heat escape — open system.
(b) Yes. Reversible ( $⇌$ ) and sealed (closed). Concentrations settle when $r_{f} = r_{b}$ .
(c) No. Reversible in principle (liquid ⇌ vapour), but the beaker is open — vapour drifts away, so the back-condensation rate can never catch the evaporation rate. The water just keeps vanishing.
(d) Yes. Same evaporation, now closed. Vapour builds up until condensation rate = evaporation rate → constant amount of vapour above the water. That is a dynamic equilibrium.

Key: (c) vs (d) is the whole lesson — same chemistry, different container, opposite verdict.

Exercise 1.2

At equilibrium for a coloured system the flask's colour stops changing. A student says "the reaction has finished." True or false, and how would you prove them wrong without any new equipment except a colour meter over time?

Recall Solution 1.2

False. Constant colour means the net change is zero, not that molecules have stopped. Forward and backward reactions both continue and exactly cancel. A colour meter alone cannot prove it (colour is flat either way). The classic proof is an isotope-label experiment (see Ex. 3.2): inject a radioactive-tagged reactant and watch the label appear on the product side — motion revealed.

LEVEL 2 — Application

Exercise 2.1

For $A ⇌ B$ at $2 5^{\circ} C$ , $k_{f} = 6 s^{- 1}$ and $k_{b} = 2 s^{- 1}$ . Find $K_{c}$ and the equilibrium ratio $[B]_{e q} / [A]_{e q}$ .

Recall Solution 2.1

What we do: apply the derived tool directly. $K_{c} = \frac{k _{f}}{k _{b}} = \frac{6}{2} = 3.$ Since $K_{c} = [B]_{e q} / [A]_{e q}$ for $A ⇌ B$ , the ratio is also $3$ . Meaning: at equilibrium there are 3 B molecules for every 1 A. Products dominate because the forward reaction is intrinsically faster.

Exercise 2.2

Using the numbers of 2.1, you start with 4 mol of A in a 1 L flask and no B. How many mol of each are present at equilibrium?

Recall Solution 2.2

Step 1 — conservation. Every A that disappears becomes one B. Total moles stay $4$ : if $x$ mol of A convert, then $[A]_{e q} = 4 - x$ and $[B]_{e q} = x$ (volume 1 L, so mol = concentration numerically). Step 2 — impose the ratio. $\frac{[ B ] _{e q}}{[ A ] _{e q}} = 3 \Rightarrow \frac{x}{4 - x} = 3.$ Step 3 — solve. $x = 3 (4 - x) = 12 - 3 x \Rightarrow 4 x = 12 \Rightarrow x = 3$ . Answer: $[A]_{e q} = 1 mol$ , $[B]_{e q} = 3 mol$ . Check: $3/1 = 3 = K_{c}$ . ✓

Exercise 2.3

Repeat 2.2 but start with 4 mol B and no A. Predict the endpoint before calculating.

Recall Solution 2.3

Prediction: the equilibrium ratio depends only on $K_{c}$ (temperature), not on the starting side, so we should land on the same mixture: $[A] = 1, [B] = 3$ . Check by algebra. Let $y$ mol of B convert back to A. Then $[A]_{e q} = y$ , $[B]_{e q} = 4 - y$ . $\frac{4 - y}{y} = 3 \Rightarrow 4 - y = 3 y \Rightarrow y = 1.$ So $[A]_{e q} = 1, [B]_{e q} = 3$ — identical endpoint. "Same destination, different route."

LEVEL 3 — Analysis

Exercise 3.1

Below is a concentration–time sketch (Figure s01) for $N_{2} O_{4} ⇌ 2 N O_{2}$ starting from pure $N_{2} O_{4}$ . (a) Which curve is which? (b) Why does one fall faster near $t = 0$ than the other rises, yet both flatten together? (c) At the flat region, is $r_{f} = 0$ ?

Figure — Reversible reactions and dynamic equilibrium

Recall Solution 3.1

(a) The curve starting high and falling is $[N_{2} O_{4}]$ (the reactant we poured in). The curve starting at 0 and rising is $[N O_{2}]$ (the product, initially absent). (b) Look at the stoichiometry: 1 $N_{2} O_{4}$ makes 2 $N O_{2}$ . So $N O_{2}$ actually rises twice as fast (in mol) as $N_{2} O_{4}$ falls — the product curve is steeper in magnitude near $t = 0$ . Both flatten together because equilibrium is a single event: the instant $r_{f} = r_{b}$ , every concentration freezes simultaneously. (c) No — in the flat region $r_{f} = r_{b} \neq = 0$ . Both reactions run at full, equal speed; only the net rate is zero. The flat line hides furious two-way traffic.

Exercise 3.2

At equilibrium in $H_{2} + I_{2} ⇌ 2 H I$ , you inject a small amount of $I_{2}$ made of radioactive $^{131} I$ . Forecast what a Geiger counter finds in each species over the next hour if (i) equilibrium were static, versus (ii) equilibrium is dynamic. State which forecast real experiments confirm.

Recall Solution 3.2

(i) Static forecast: if no reaction occurred, the label would stay locked in $I_{2}$ forever. Only $I_{2}$ would read radioactive. (ii) Dynamic forecast: because HI is constantly being formed and torn apart, some $^{131} I$ gets built into HI molecules. Radioactivity spreads into HI (and, via reformed $I_{2}$ , distributes across the pool). Reality: the label appears in HI. This confirms dynamic equilibrium — molecules keep reacting even though total $[H_{2}], [I_{2}], [H I]$ stay constant. See also Effect of temperature on equilibrium for how the amounts would change if you heated it.

LEVEL 4 — Synthesis

Exercise 4.1

For $A ⇌ B$ , at $300 K$ you measure $K_{c} = 4$ . A catalyst is added. (a) State the new $K_{c}$ . (b) Sketch, in words, how the approach-to-equilibrium changes. (c) The temperature is then raised and $K_{c}$ becomes $9$ ; what does this tell you about $k_{f}$ vs $k_{b}$ changing?

Recall Solution 4.1

(a) $K_{c}$ is unchanged = 4. A catalyst speeds forward and backward reactions equally (it lowers the same activation-energy barrier both ways), so $k_{f} / k_{b}$ is untouched. It only reaches equilibrium faster. (See Catalysis and Le Chatelier's principle.) (b) Both concentration curves bend and flatten sooner, but they flatten at the same final values as without catalyst. Same destination, earlier arrival. (c) $K_{c} = k_{f} / k_{b}$ rose from 4 to 9 when heated. So heating raised $k_{f}$ more than it raised $k_{b}$ (their ratio grew). Both rate constants increase with temperature, but here the forward one increases proportionally more — consistent with an endothermic forward reaction (heat favours the direction that absorbs it). Detail belongs to Effect of temperature on equilibrium.

Exercise 4.2

A reaction $A ⇌ B$ has $K_{c} = 4$ . You start with $10 mol$ A and $0$ B in a 1 L vessel. (a) Find equilibrium amounts. (b) What fraction of A has converted (the "extent")?

Recall Solution 4.2

(a) Let $x$ mol convert: $[A] = 10 - x$ , $[B] = x$ . $\frac{x}{10 - x} = 4 \Rightarrow x = 40 - 4 x \Rightarrow 5 x = 40 \Rightarrow x = 8.$ So $[A]_{e q} = 2 mol$ , $[B]_{e q} = 8 mol$ . Check $8/2 = 4$ . ✓ (b) Fraction converted $= x /10 = 8/10 = 0.8 = 80%$ .

LEVEL 5 — Mastery

Exercise 5.1

Two students run the same reversible reaction $A ⇌ B$ at the same temperature.

Student 1 starts with 6 mol A, 0 mol B, and measures $[A]_{e q} = 2, [B]_{e q} = 4$ .
Student 2 starts with 0 mol A, 2 mol B.

Without repeating the whole calculation from scratch, predict Student 2's equilibrium amounts, then verify. Explain which single quantity guarantees your prediction is legitimate.

Recall Solution 5.1

Step 1 — extract the governing constant. From Student 1: $K_{c} = [B]_{e q} / [A]_{e q} = 4/2 = 2$ . Same temperature ⇒ Student 2 obeys the same $K_{c} = 2$ . That constant is the one quantity that transfers between the two runs. Step 2 — conserve Student 2's material. Total = 2 mol (all as B initially). Let $y$ mol of B convert to A: $[A]_{e q} = y$ , $[B]_{e q} = 2 - y$ . Step 3 — impose $K_{c}$ . $\frac{2 - y}{y} = 2 \Rightarrow 2 - y = 2 y \Rightarrow y = \frac{2}{3} .$ Answer: $[A]_{e q} = \frac{2}{3} \approx 0.667 mol$ , $[B]_{e q} = \frac{4}{3} \approx 1.333 mol$ . Check: $(4/3) / (2/3) = 2 = K_{c}$ . ✓ Why legitimate: $K_{c}$ depends only on temperature, so it is shared across both runs. The starting totals differ (6 vs 2), so the absolute amounts differ, but the ratio is locked at 2 for both.

Exercise 5.2 (capstone)

For $A ⇌ B$ , forward rate $r_{f} = k_{f} [A]$ and backward rate $r_{b} = k_{b} [B]$ , with $k_{f} = 3 s^{- 1}$ , $k_{b} = 1 s^{- 1}$ . You start with $[A]_{0} = 8$ , $[B]_{0} = 0$ (units mol L⁻¹). (a) Find $K_{c}$ and $[A]_{e q}, [B]_{e q}$ . (b) Compute the numerical value of $r_{f}$ and $r_{b}$ at equilibrium and confirm they are equal and nonzero. (c) In one sentence, connect part (b) to why the concentrations stay flat.

Recall Solution 5.2

(a) $K_{c} = k_{f} / k_{b} = 3/1 = 3$ . Conserve: $[A] = 8 - x$ , $[B] = x$ . $\frac{x}{8 - x} = 3 \Rightarrow x = 24 - 3 x \Rightarrow 4 x = 24 \Rightarrow x = 6.$ So $[A]_{e q} = 2$ , $[B]_{e q} = 6$ . Check $6/2 = 3$ . ✓ (b) $r_{f} = k_{f} [A]_{e q} = 3 \times 2 = 6 mol L^{- 1} s^{- 1}$ . $r_{b} = k_{b} [B]_{e q} = 1 \times 6 = 6 mol L^{- 1} s^{- 1}$ . Equal (both $6$ ) and clearly nonzero. ✓ (c) Because A is being consumed at exactly the same speed it is being regenerated ( $r_{f} = r_{b} = 6$ ), the net change in $[A]$ (and $[B]$ ) is zero — the concentrations sit frozen while both reactions roar on underneath. That is dynamic equilibrium, made numerical.

Recall One-line map of what you practised

L1 spotted when equilibrium is even possible (closed + reversible). L2–L3 applied and read the $k_{f} [A] = k_{b} [B]$ relation both algebraically and graphically. L4 tested robustness under catalysts, temperature and "extent." L5 fused kinetics and the constant into a single numeric picture.

Connections

Reversible reactions and dynamic equilibrium (index 2.6.1) — the parent this drills.
Equilibrium constant Kc and Kp — the constant every exercise leans on.
Rate of reaction and rate constants — source of $k_{f}, k_{b}$ .
Collision theory — why rate ∝ concentration.
Le Chatelier's principle & Effect of temperature on equilibrium — the L4 shifts.
Catalysis — the "no shift, faster arrival" result.