Reaction quotient Q vs K — direction of shift: Exercises

This is a practice ladder for the parent note Reaction quotient $Q$ vs $K$. Every problem has a fully worked solution hidden in a collapsible callout — cover it, try yourself, then reveal.

Before any problems, let us build every symbol we will use, from zero.

Definition Activity — what "put a

$1$ for solids and liquids" really means The activity of a substance is its "effective concentration" — a pure number (no units) that measures how much that substance participates in the ratio, relative to a standard reference state. For a dissolved species we approximate activity by its concentration in mol/L (divided by $1$ mol/L), and for a gas by its partial pressure in atm (divided by $1$ atm). But a pure solid or pure liquid never gets more or less "concentrated" — its molecules are packed at a fixed density that does not change as the reaction runs. So its activity is pinned at exactly $1$ . Anything multiplied or divided by $1$ is unchanged, so a pure solid/liquid simply drops out of both $K$ and $Q$ . That is why $Q$ and $K$ contain only the gases and dissolved species. See Activity and why pure solids-liquids are omitted.

Two housekeeping rules used in every problem: the exponents are the stoichiometric coefficients, and pure solids and pure liquids are left out (activity $= 1$ , defined just above).

Notation convention (used everywhere below). We always write the subscript: $Q_{c}$ / $K_{c}$ for concentration problems, and $Q_{p}$ / $K_{p}$ for pressure problems. When a solution occasionally shortens " $Q$ " or " $K$ " for readability, it means the subscripted version matching that problem — and we only ever compare $Q_{c}$ with $K_{c}$ , or $Q_{p}$ with $K_{p}$ , never a cross-mix.

The figures on this page

There is one figure, labelled s01 (the page's figure-numbering runs s01, s02, …; this page uses only s01). It is the "number line" you should keep open beside every solution — the text refers to it as "figure s01".

Figure — Reaction quotient Q vs K — direction of shift

Figure s01 — the $Q$ -vs- $K$ number line (used throughout). This is the mental picture for the whole page. Mark $K$ as the target dot. Wherever $Q$ lands, an arrow points toward $K$ : if $Q$ is to the left of $K$ (too few products), the reaction walks right (forward, teal arrow); if $Q$ is to the right of $K$ (too many products), it walks left (backward, orange arrow); if $Q$ sits exactly on $K$ (plum dot), it stays put. Keep this picture in mind — the solutions below say "read the number line (figure s01)" and mean this figure.

Level 1 — Recognition

Recall Solution 1.1

What we compare: $Q_{c}$ against $K_{c}$ (same units, same form — never cross $Q_{c}$ with $K_{p}$ ). $Q_{c} = 4, K_{c} = 12, Q_{c} < K_{c} .$ Read the number line (figure s01): $Q$ (at $4$ ) sits to the left of $K$ (at $12$ ). Too few products. Direction: forward (toward products). ✅

Recall Solution 1.2

$Q_{c} = K_{c}$ means the current ratio already equals the target ratio. By $Δ G = R T ln (Q_{c} / K_{c}) = R T ln 1 = 0$ , there is no driving force. On the number line (figure s01), $Q$ lands exactly on the $K$ dot (plum). Answer: the system is at equilibrium, no net shift. ✅

Recall Solution 1.3

$Q_{c} = 5$ is much greater than $K_{c} = 0.01$ : too many products. On the number line (figure s01), $Q$ sits far to the right of $K$ → backward arrow. $Q_{c} > K_{c} \Rightarrow$ backward (remake reactants). ✅

Level 2 — Application

Recall Solution 2.1

Step 1 — write the form. Coefficients become exponents: $Q_{c} = \frac{[ N H _{3} ] ^{2}}{[ N _{2} ] [ H _{2} ] ^{3}} .$ Step 2 — plug in current values. $Q_{c} = \frac{4 ^{2}}{2 \cdot 1 ^{3}} = \frac{16}{2} = 8.$ Step 3 — compare. $Q_{c} = 8 > K_{c} = 2$ ; on the number line (figure s01) $Q$ is right of $K$ . Direction: backward (too many products; remake $N_{2}$ and $H_{2}$ ). ✅

Recall Solution 2.2

For gases we use partial pressures, so this is a $Q_{p}$ compared with $K_{p}$ (matching units — see Relation between Kp and Kc). $Q_{p} = \frac{p _{S O_{3}}^{2}}{p _{S O_{2}}^{2} p _{O_{2}}} = \frac{4 ^{2}}{1 ^{2} \cdot 2} = \frac{16}{2} = 8.$ $Q_{p} = 8 = K_{p} = 8$ : on the number line (figure s01) $Q$ lands exactly on $K$ → the system is already at equilibrium → no net shift. ✅

Recall Solution 2.3

Step 1 — drop the pure solids. $C a C O_{3} (s)$ and $C a O (s)$ have activity $1$ (defined at the top of the page), so they never appear. Only the gas survives: $Q_{p} = p_{C O_{2}} = 0.2 atm .$ Step 2 — compare. $Q_{p} = 0.2 < K_{p} = 0.5$ : not enough $C O_{2}$ ; $Q$ is left of $K$ on the number line (figure s01). Direction: forward — more $C a C O_{3}$ decomposes to release $C O_{2}$ until $p_{C O_{2}}$ rises to $0.5$ atm. ✅

Level 3 — Analysis

Recall Solution 3.1

Why use $Δ G = R T ln (Q_{c} / K_{c})$ ? Because the sign of $Δ G$ is the direction: nature moves to make $Δ G < 0$ . $\frac{Q _{c}}{K _{c}} = \frac{1000}{10} = 100.$ $Δ G = R T ln \frac{Q _{c}}{K _{c}} = (8.314) (300) ln (100) .$ Now $ln (100) = 2 ln (10) \approx 2 \times 2.302585 = 4.60517$ . $Δ G \approx 2494.2 \times 4.60517 \approx + 11486 J/mol \approx + 11.49 kJ/mol .$ Read the sign: $Δ G > 0$ → forward is non-spontaneous → the reaction runs backward to lower $Q_{c}$ back toward $10$ . ✅

Recall Solution 3.2

$\frac{Q _{c}}{K _{c}} = \frac{0.1}{10} = 0.01.$ $Δ G = (8.314) (300) ln (0.01) = 2494.2 \times (- 4.60517) \approx - 11486 J/mol \approx - 11.49 kJ/mol .$ Symmetry check: $Q_{c} / K_{c}$ here ( $0.01$ ) is the reciprocal of Exercise 3.1's ( $100$ ), and $ln$ of a reciprocal just flips the sign — so $Δ G$ has the same magnitude but opposite sign. Neat. $Δ G < 0$ → forward is spontaneous → forward. ✅

Recall Solution 3.3

$Δ G = R T ln (Q_{c} / K_{c}) = 0$ requires $ln (Q_{c} / K_{c}) = 0$ , i.e. $Q_{c} / K_{c} = 1$ , i.e. $Q_{c} = K_{c} = 10.$ Physical meaning: the current ratio has climbed (or fallen) until it equals the target ratio. The driving force vanishes; there is no net shift. This is the definition of equilibrium. ✅

Level 4 — Synthesis

Recall Solution 4.1

(a) $Q_{c} = \frac{[ B ] ^{2}}{[ A ]} = \frac{0 ^{2}}{1} = 0.$ (b) $Q_{c} = 0 < K_{c} = 4$ → forward. (c) With only reactants, $Q$ starts at the far left of the number line (figure s01), at $Q = 0$ . Since $K = 4$ is to its right, the only way to reach the target is to make products, driving $Q$ up from $0$ toward $4$ . A pure-reactant start always gives $Q_{c} = 0$ , so a reaction with any products in its equilibrium ( $K_{c} > 0$ ) must initially go forward. ✅

Recall Solution 4.2

(a) $Q_{c} = \frac{[ N O _{2} ] ^{2}}{[ N _{2} O _{4} ]} = \frac{0. 2 ^{2}}{0.1} = \frac{0.04}{0.1} = 0.4.$ $Q_{c} = 0.4 > K_{c} = 0.36$ → backward (toward $N_{2} O_{4}$ ). Barely over — a gentle backward push. (b) Halving the volume doubles each concentration: $[N_{2} O_{4}] = 0.2$ , $[N O_{2}] = 0.4$ . $Q_{c}^{'} = \frac{0. 4 ^{2}}{0.2} = \frac{0.16}{0.2} = 0.8.$ $Q_{c}^{'} = 0.8 > K_{c} = 0.36$ → still backward, and now the ratio $Q_{c} / K_{c}$ is larger ( $0.8/0.36 \approx 2.22$ vs $0.4/0.36 \approx 1.11$ ), so the backward push is stronger. Le Chatelier agreement: compression favours the side with fewer gas moles. The left side ( $N_{2} O_{4}$ , $1$ mole of gas) has fewer moles than the right ( $2 N O_{2}$ ). So compression should shift toward $N_{2} O_{4}$ = backward. The $Q_{c}$ calculation and Le Chatelier's principle give the same answer — as they must. ✅

Level 5 — Mastery

Recall Solution 5.1

(a) $Q_{c} = \frac{[ C ]}{[ A ] [ B ]} = \frac{0}{2 \cdot 2} = 0 < K_{c} = 4$ → forward. Products will form, so $x > 0$ . (b) Going forward by $x$ : $[A] = 2 - x$ , $[B] = 2 - x$ , $[C] = x$ . At equilibrium: $K_{c} = \frac{x}{( 2 - x ) ^{2}} = 4.$ Why the next step? We want $x$ out of the denominator so we can get a plain polynomial. To do that we multiply both sides by $(2 - x)^{2}$ — the same quantity on each side keeps the equation balanced, and on the left it cancels the denominator, leaving $x$ alone: $x = 4 (2 - x)^{2} .$ Now expand the square $(2 - x)^{2} = 4 - 4 x + x^{2}$ : $x = 4 (4 - 4 x + x^{2}) = 16 - 16 x + 4 x^{2} .$ Bring everything to one side (subtract $x$ from both sides) to get the standard quadratic form $a t^{2} + b t + c = 0$ : $4 x^{2} - 17 x + 16 = 0.$ Solve with the quadratic formula $x = \frac{- b \pm b ^{2} - 4 a c}{2 a}$ with $a = 4$ , $b = - 17$ , $c = 16$ : $x = \frac{17 \pm 1 7 ^{2} - 4 \cdot 4 \cdot 16}{2 \cdot 4} = \frac{17 \pm 289 - 256}{8} = \frac{17 \pm 33}{8} .$ $33 \approx 5.7446$ , giving $x \approx \frac{17 - 5.7446}{8} \approx 1.4069$ or $x \approx \frac{17 + 5.7446}{8} \approx 2.843$ . Reject the second root: $x$ cannot exceed the starting $2$ mol/L (you can't consume more $A$ than you have — that would make $[A]$ negative). So $x \approx 1.407 mol/L .$ (c) Then $[A] = [B] = 2 - 1.407 = 0.593$ , $[C] = 1.407$ . $Q_{c} = \frac{1.407}{0.59 3 ^{2}} = \frac{1.407}{0.3516} \approx 4.00 = K_{c} . ✓$ ✅

Recall Solution 5.2

(a) Start from $Δ G^{\circ} = - R T ln K_{c}$ and solve for $ln K_{c}$ (divide both sides by $- R T$ ). Convert $Δ G^{\circ} = - 5.0$ kJ/mol to $- 5000$ J/mol so the units match $R$ : $ln K_{c} = - \frac{Δ G ^{\circ}}{R T} = - \frac{- 5000}{( 8.314 ) ( 298 )} = \frac{5000}{2477.6} \approx 2.0181.$ Undo the $ln$ by exponentiating (apply $e^{(\cdot)}$ to both sides): $K_{c} = e^{2.0181} \approx 7.52.$ (b) Now use the driving-force formula with the current ratio $Q_{c} = 3$ : $Δ G = R T ln \frac{Q _{c}}{K _{c}} = (8.314) (298) ln \frac{3}{7.52} .$ $3/7.52 \approx 0.3989$ , and $ln (0.3989) \approx - 0.9191$ : $Δ G \approx 2477.6 \times (- 0.9191) \approx - 2277 J/mol \approx - 2.28 kJ/mol .$ Read the sign: $Δ G < 0$ → forward. (This matches the quick check: $Q_{c} = 3 < K_{c} \approx 7.52$ , so on the number line (figure s01) $Q$ is left of $K$ → forward.) ✅

Recall Solution 5.3

(a) $Q_{c} = \frac{[ C ]}{[ A ] [ B ]} = \frac{2}{0 \cdot 0}$ . Here the denominator is zero, so $Q_{c}$ is formally undefined — you cannot literally divide by zero. What we really mean is a limiting case: as $[A]$ and $[B]$ shrink toward $0$ with some product present, the ratio grows without bound, so $Q_{c} \to \infty$ . Physically: there are essentially only products and (almost) no reactants, so $Q_{c}$ is enormously larger than the target $K_{c} = 4$ . That forces the reaction backward. (A pure-product start is the mirror image of a pure-reactant start, where $Q_{c} = 0$ .) (b) Going backward by $y$ : $[C] = 2 - y$ , $[A] = y$ , $[B] = y$ . $K_{c} = \frac{2 - y}{y ^{2}} = 4 \Rightarrow 2 - y = 4 y^{2} \Rightarrow 4 y^{2} + y - 2 = 0.$ (Same move as 5.1: multiply both sides by $y^{2}$ to clear the denominator, then bring all terms to one side.) $y = \frac{- 1 \pm 1 + 32}{8} = \frac{- 1 \pm 33}{8} .$ Take the positive root: $y = \frac{- 1 + 5.7446}{8} \approx 0.5931 mol/L .$ Final: $[A] = [B] \approx 0.593$ , $[C] = 2 - 0.593 \approx 1.407$ mol/L. Beautiful sanity check: these are the exact same equilibrium concentrations as Exercise 5.1 — because equilibrium at a fixed $T$ depends only on $K_{c}$ and the total atoms present, not on which side you started from. Whether you approach from pure reactant (5.1) or pure product (5.3), you land on the same point. ✅

Connections

Equilibrium constant K (Kc and Kp)
Gibbs free energy and spontaneity
Le Chatelier's principle
Relation between Kp and Kc
Activity and why pure solids-liquids are omitted
Position of equilibrium vs rate of reaction