Reversible reactions and dynamic equilibrium: Worked examples

Before anything, let me re-earn every symbol so you can read line one cold.

Definition The symbols we use (all defined from zero)

$A ⇌ B$ — a reaction where $A$ turns into $B$ (the forward direction, read left→right) and $B$ turns back into $A$ (the backward direction, right→left). The double arrow $⇌$ means both happen at once.
$[A]$ — square brackets mean "concentration of": how many moles of $A$ sit in each litre of solution. Units: $mol dm^{- 3}$ (moles per litre; $dm^{3}$ is just a litre).
$r_{f} = k_{f} [A]$ — the forward rate: how fast $A$ is being eaten, per second. $k_{f}$ (the forward rate constant) is a fixed number at a fixed temperature that says "for each unit of $[A]$ , this much reacts per second". See Rate of reaction and rate constants.
$r_{b} = k_{b} [B]$ — the backward rate, same idea for $B$ .
$K_{c} = \frac{k _{f}}{k _{b}}$ — the equilibrium constant: the single number that pins where the reaction settles. Big $K_{c}$ → products win; small $K_{c}$ → reactants win. Quantified fully in Equilibrium constant Kc and Kp.

The scenario matrix

Every problem this topic can throw is one of these cells. Each example below is tagged with the cell it fills.

#	Case class	What makes it special	Example
C1	$K_{c} > 1$ (products favoured)	forward wins, $[B] > [A]$	Ex 1
C2	$K_{c} < 1$ (reactants favoured)	backward wins, $[A] > [B]$	Ex 2
C3	$K_{c} = 1$ (degenerate/balanced)	$[A] = [B]$ exactly	Ex 3
C4	Start from the other side (all product)	tests "same destination"	Ex 4
C5	Zero input on one side	$[B]_{0} = 0$ , limiting start	Ex 1, Ex 4
C6	Limiting values ( $K_{c} \to \infty$ , $K_{c} \to 0$ )	irreversible-like extremes	Ex 5
C7	Open system / escaping gas	equilibrium never reached	Ex 6
C8	Graph reading (concentration–time)	recognise the shape	Ex 7
C9	Real-world word problem	translate words → numbers	Ex 8
C10	Exam twist (catalyst / temperature)	$K_{c}$ change vs no change	Ex 9

We will use one master fact everywhere, so let me state it once as a tool.

The two-equation recipe used in almost every example:

Ratio: $[B]_{e q} = K_{c} [A]_{e q}$ .
Conservation: total moles $[A]_{e q} + [B]_{e q} = T$ (the total never changes, because one $A$ becomes one $B$ ).

Solve those two together: $[A]_{e q} = \frac{T}{1 + K _{c}}, [B]_{e q} = \frac{K _{c} T}{1 + K _{c}} .$

Why this shortcut? Substitute the ratio into the sum: $[A]_{e q} + K_{c} [A]_{e q} = T \Rightarrow [A]_{e q} (1 + K_{c}) = T$ . We will re-derive it live in Ex 1, then reuse it.

Worked Examples

Worked example Example 1 — Cell C1 & C5: products favoured, zero product at start

Reaction $A ⇌ B$ with $k_{f} = 4 s^{- 1}$ and $k_{b} = 1 s^{- 1}$ . We start with $3 mol dm^{- 3}$ of pure $A$ and no $B$ . Find $[A]_{e q}$ and $[B]_{e q}$ .

Forecast: $k_{f}$ is 4× bigger than $k_{b}$ , so the forward reaction is stronger. Guess: most of the $A$ becomes $B$ . Is it all? (Hold that thought — see Ex 5.)

Step 1 — Compute $K_{c}$ . $K_{c} = \frac{k _{f}}{k _{b}} = \frac{4}{1} = 4$ . Why this step? $K_{c}$ is the only number that decides the resting point; the individual $k$ 's are only needed to form their ratio.

Step 2 — Write the total (conservation). One $A$ → one $B$ , so total stays $T = 3 + 0 = 3 mol dm^{- 3}$ . Why this step? It gives the second equation we need; without it the ratio alone can't fix actual amounts.

Step 3 — Apply the ratio. $[B]_{e q} = K_{c} [A]_{e q} = 4 [A]_{e q}$ . Why this step? This is the master tool — equal rates.

Step 4 — Solve. Substitute into the total: $[A]_{e q} + 4 [A]_{e q} = 3 \Rightarrow 5 [A]_{e q} = 3 \Rightarrow [A]_{e q} = 0.6$ . Then $[B]_{e q} = 4 \times 0.6 = 2.4 mol dm^{- 3}$ . Why this step? Two equations, two unknowns — pure algebra.

Verify: Ratio $\frac{[ B ]}{[ A ]} = \frac{2.4}{0.6} = 4 = K_{c}$ ✓. Sum $0.6 + 2.4 = 3 = T$ ✓. Rates: $r_{f} = 4 (0.6) = 2.4$ , $r_{b} = 1 (2.4) = 2.4$ , equal ✓. Products dominate, matching the forecast.

Worked example Example 4 — Cell C4 & C5: same destination from the other side

Same constants as Ex 1 ( $K_{c} = 4$ ), but now start with $3 mol dm^{- 3}$ of pure $B$ and no $A$ .

Forecast: the parent note promised "same destination, different route." So I predict the same answer: $[A]_{e q} = 0.6$ , $[B]_{e q} = 2.4$ .

Step 1 — Which way does it run first? At the start $[B] = 3$ (large) and $[A] = 0$ , so $r_{b} = 1 (3) = 3$ but $r_{f} = 4 (0) = 0$ . Backward dominates first — the opposite of Ex 1. Why this step? To show the route is genuinely different even though the endpoint won't be. Step 2 — Total. One $B$ → one $A$ , so $T = 0 + 3 = 3$ , the same total as Ex 1. Step 3 — Ratio and solve. $K_{c}$ depends only on temperature, unchanged: $[A]_{e q} (1 + 4) = 3 \Rightarrow [A]_{e q} = 0.6$ , $[B]_{e q} = 2.4$ .

Verify: Identical to Ex 1 ✓. The system approached from the right instead of the left but landed on the same point — dynamic equilibrium is reachable from either direction (condition 4 in the parent).

Worked example Example 5 — Cell C6: the two limiting extremes

What happens as $K_{c} \to \infty$ (forward hugely favoured) and as $K_{c} \to 0$ (backward hugely favoured), starting from total $T = 3$ ?

Forecast: huge $K_{c}$ should look "irreversible forward" (almost all $B$ ); tiny $K_{c}$ should look "irreversible backward" (almost all $A$ ).

Step 1 — Push $K_{c}$ up. Use $[A]_{e q} = \frac{T}{1 + K _{c}}$ . As $K_{c} \to \infty$ , $[A]_{e q} \to \frac{3}{\infty} = 0$ and $[B]_{e q} \to 3$ . Why: the denominator explodes, squeezing $A$ to nothing. Step 2 — Numeric feel. At $K_{c} = 99$ : $[A]_{e q} = \frac{3}{100} = 0.03$ , $[B]_{e q} = 2.97$ . Nearly all product. Step 3 — Push $K_{c}$ down. As $K_{c} \to 0$ , $[A]_{e q} \to \frac{3}{1} = 3$ , $[B]_{e q} \to 0$ . At $K_{c} = 0.01$ : $[A]_{e q} = \frac{3}{1.01} \approx 2.970$ , $[B]_{e q} \approx 0.0297$ . Nearly all reactant.

Verify: Both limits match "irreversible" behaviour, connecting to the parent's point that irreversible reactions are just reversible ones with negligible reverse rate. At $K_{c} = 99$ : $2.97/0.03 = 99$ ✓. At $K_{c} = 0.01$ : $0.0297/2.970 \approx 0.01$ ✓.

Worked example Example 6 — Cell C7: open system, equilibrium never reached

$C a C O_{3} (s) ⇌ C a O (s) + C O_{2} (g)$ heated in an open crucible. Explain, using rates, why no equilibrium forms — and contrast with a sealed vessel.

Forecast: the parent says an escaping gas breaks equilibrium. So: open → never balances; sealed → balances.

Step 1 — Identify the escaping species. $C O_{2}$ is a gas and drifts out of the open crucible. Why this step? The backward reaction ( $C a O + C O_{2} \to C a C O_{3}$ ) needs $C O_{2}$ to collide with $C a O$ . Step 2 — Track the backward rate. $r_{b} \propto [C O_{2}]$ (loosely, its partial pressure). If $C O_{2}$ keeps leaving, $[C O_{2}]$ near the solid stays near 0, so $r_{b} \approx 0$ forever. Step 3 — Compare rates. Forward keeps decomposing $C a C O_{3}$ while backward can never rise to meet it: $r_{f} > r_{b}$ permanently, so $r_{f} \neq = r_{b}$ — the defining equality of equilibrium is never satisfied. Step 4 — Sealed contrast. In a closed vessel $C O_{2}$ builds up, $r_{b}$ climbs, and eventually $r_{f} = r_{b}$ : a real equilibrium with a fixed $C O_{2}$ pressure ( $K_{p}$ ; see Equilibrium constant Kc and Kp).

Verify: Matches the parent's "closed system required" condition. The whole difference is whether $[C O_{2}]$ can grow — a purely rate-based argument, no new maths.

Worked example Example 7 — Cell C8: reading the concentration–time graph

The figure below shows $A ⇌ B$ ( $K_{c} = 4$ , starting from $3 mol dm^{- 3}$ pure $A$ , as in Ex 1). Identify (a) where equilibrium begins, (b) the equilibrium concentrations, (c) where the two curves' slopes are steepest.

Forecast: curves start apart, bend toward each other, then go flat. Flat = equilibrium.

Step 1 — Find the flat region. Read where both curves stop changing (the pale-yellow "equilibrium reached" line). Why: constant concentration is the visual signature of $r_{f} = r_{b}$ . Step 2 — Read heights there. The blue $A$ curve settles at $0.6$ ; the pink $B$ curve settles at $2.4$ — matching Ex 1 exactly. Step 3 — Steepest slope is at $t = 0$ . Why: at the start $[A]$ is largest so $r_{f}$ is largest, and $[B] = 0$ so $r_{b} = 0$ ; the net rate (and thus the curve's steepness) is maximal, then it eases off as the rates close like scissors.

Verify: Heights $0.6$ and $2.4$ sum to $3$ (conservation) ✓ and ratio $4 = K_{c}$ ✓. The curves never cross here because $K_{c} \neq = 1$ (contrast Ex 3, where they would meet at $2.5$ ).

Worked example Example 8 — Cell C9: real-world word problem

An esterification runs as $A ⇌ B$ in a sealed flask. A chemist measures at equilibrium: $[A]_{e q} = 0.5 mol dm^{- 3}$ and $[B]_{e q} = 1.5 mol dm^{- 3}$ . (a) Find $K_{c}$ . (b) If she can measure $k_{b} = 2 s^{- 1}$ , find $k_{f}$ . (c) How much $A$ did she start with, given no $B$ at the start?

Forecast: more $B$ than $A$ , so $K_{c} > 1$ ; and $k_{f}$ should exceed $k_{b}$ .

Step 1 — $K_{c}$ from measured concentrations. $K_{c} = \frac{[ B ] _{e q}}{[ A ] _{e q}} = \frac{1.5}{0.5} = 3$ . Why: the master tool works backwards too — given the resting mix, read off $K_{c}$ . Step 2 — $k_{f}$ from $K_{c} = k_{f} / k_{b}$ . $k_{f} = K_{c} k_{b} = 3 \times 2 = 6 s^{- 1}$ . Why: rearrange the definition of $K_{c}$ . Step 3 — Starting amount via conservation. With no $B$ initially, all the $B$ present came from $A$ : start $[A]_{0} = [A]_{e q} + [B]_{e q} = 0.5 + 1.5 = 2 mol dm^{- 3}$ . Why: every mole of $B$ is a mole of $A$ that reacted.

Verify: $r_{f} = 6 (0.5) = 3$ , $r_{b} = 2 (1.5) = 3$ , equal ✓. $K_{c} = 6/2 = 3$ ✓. Start $2$ splits to $0.5 + 1.5$ ✓.

Worked example Example 9 — Cell C10: the exam twist (catalyst vs temperature)

Take the Ex 8 system at equilibrium ( $K_{c} = 3$ , $[A] = 0.5$ , $[B] = 1.5$ ). Predict the new equilibrium concentrations if the chemist (a) adds a catalyst, then separately (b) raises the temperature so that $K_{c}$ becomes $6$ (total kept at $2$ ).

Forecast: the parent warns catalysts don't move equilibrium — so (a) should give no change. Temperature genuinely changes $K_{c}$ , so (b) shifts the mix toward $B$ .

Step 1 — Catalyst (a). A catalyst multiplies both $k_{f}$ and $k_{b}$ by the same factor, so $K_{c} = k_{f} / k_{b}$ is unchanged. New concentrations $=$ old: $[A] = 0.5$ , $[B] = 1.5$ . Why: $K_{c}$ fixes the position; the catalyst only makes the system arrive faster. Step 2 — Temperature (b), new $K_{c} = 6$ . Use the recipe with $T = 2$ : $[A]_{e q} = \frac{2}{1 + 6} = \frac{2}{7} \approx 0.2857$ , $[B]_{e q} = \frac{6 \times 2}{7} = \frac{12}{7} \approx 1.7143$ . Why: only $K_{c}$ changed, so re-solve the same two equations with the new ratio. This is Le Chatelier's principle and Effect of temperature on equilibrium in numbers.

Verify: (a) unchanged ✓ ( $K_{c}$ still $1.5/0.5 = 3$ ). (b) $\frac{12/7}{2/7} = 6 = K_{c}$ ✓; sum $\frac{2}{7} + \frac{12}{7} = \frac{14}{7} = 2$ ✓. Temperature shifted product up (0.5→0.286 for A), catalyst did nothing — exactly the two contrasting outcomes the exam wants you to separate.

Recall Quick self-test on the matrix

Which cell has $[A] = [B]$ ? ::: C3, the case $K_{c} = 1$ . Why does an open crucible never reach equilibrium? ::: The gas escapes, so the backward rate stays near zero and can never equal the forward rate. Does a catalyst change $K_{c}$ ? ::: No — it speeds both directions equally; only temperature changes $K_{c}$ . Starting from all product instead of all reactant — same endpoint? ::: Yes, as long as the total and $K_{c}$ are the same (Ex 4). As $K_{c} \to \infty$ , what fraction is product? ::: Almost all — the reaction looks irreversible-forward.

Connections

Equilibrium constant Kc and Kp — where these $K_{c}$ numbers get their full quantitative treatment.
Rate of reaction and rate constants — the $k_{f}, k_{b}$ behind every ratio here.
Collision theory — why $r \propto [concentration]$ , the seed of the whole method.
Le Chatelier's principle — the qualitative shortcut for Ex 9(b).
Effect of temperature on equilibrium — why only temperature changed $K_{c}$ .
Catalysis — why the catalyst in Ex 9(a) moved nothing.