Reaction quotient Q vs K — direction of shift: Worked examples

Prerequisites we lean on: the parent rule, Equilibrium constant K (Kc and Kp), Gibbs free energy and spontaneity, Activity and why pure solids-liquids are omitted, and Relation between Kp and Kc.

The scenario matrix

Before working anything, let's list every case class this topic contains. Each worked example below is tagged with the cell it fills.

#	Case class	What's special about it	Example
C1	$Q < K$ , plain concentrations	forward shift	Ex 1
C2	$Q > K$ , plain concentrations	backward shift	Ex 2
C3	$Q = K$ exactly	no net shift, $Δ G = 0$	Ex 3
C4	Degenerate: only reactants ( $Q = 0$ )	limiting value, must go forward	Ex 4
C5	Degenerate: only products ( $Q \to \infty$ )	limiting value, must go backward	Ex 5
C6	Reaction with a pure solid/liquid	activity $= 1$ , omit it	Ex 6
C7	Gas reaction using $Q_{p}$	compare with $K_{p}$ , not $K_{c}$	Ex 7
C8	$Δ G$ number-crunch	get magnitude + sign of $Δ G$	Ex 8
C9	Exam twist: same $Q, K$ but reversed reaction	shows direction flips	Ex 9
C10	Word problem (real scenario)	translate story → $Q$ vs $K$	Ex 10

The number line below is the map every example lives on: $Q$ slides left or right of $K$ , and that side decides the direction.

Figure — Reaction quotient Q vs K — direction of shift

Worked examples

Worked example Example 5 — C5: degenerate, only products (a

limit, not a division) Same ammonia reaction, $K_{c} = 0.5$ . Now you start with only $N H_{3}$ ; no $N_{2}$ or $H_{2}$ . Direction?

Forecast: nothing in the denominator — enormous $Q$ — backward?

Look at $Q_{c} = \frac{[ N H _{3} ] ^{2}}{[ N _{2} ] [ H _{2} ] ^{3}}$ . With $[N_{2}] = [H_{2}] = 0$ the denominator is $0$ , so $Q_{c}$ is strictly undefined — you cannot literally divide by zero. Why this step? Honesty: $Q$ has no finite value here. We must treat it as a limit instead.
Ask what happens as the reactants approach zero: let $[N_{2}], [H_{2}] \to 0^{+}$ . The denominator shrinks toward $0$ while the top stays positive, so $Q_{c} \to + \infty$ . Why this step? The limit tells us the trend: $Q$ grows without bound as we near pure product — the right edge of the number line.
Compare the limit with $K_{c}$ : for any tiny-but-nonzero reactant amounts, $Q_{c}$ is already far greater than $K_{c} = 0.5$ . Why this step? $Q > K$ means $Δ G > 0$ for forward, so the reverse reaction is spontaneous.
Answer: must shift backward — pure product can only decompose to reactants (which also immediately makes the denominator nonzero, so $Q$ becomes a normal finite number again). ✅

Verify: $[N_{2}], [H_{2}] \to 0^{+} lim ln \frac{Q}{K} = + \infty \Rightarrow Δ G \to + \infty$ : forward impossible, backward forced.

Worked example Example 6 — C6: pure solid present (

Activity and why pure solids-liquids are omitted) $C a C O_{3} (s) ⇌ C a O (s) + C O_{2} (g)$ , $K_{p} = 0.2 atm$ . Current $p_{C O_{2}} = 0.05$ atm, and both solids are present. Direction?

Forecast: two solids on the equation — do they count?

Build $Q_{p}$ : omit the pure solids $C a C O_{3}$ and $C a O$ (activity $= 1$ ), leaving $Q_{p} = p_{C O_{2}}$ . Why this step? A pure solid's activity is fixed at $1$ ; it never changes the ratio, so it drops out — exactly as in $K$ .
Plug in: $Q_{p} = 0.05$ . Why this step? Only the gas contributes.
Compare: $Q_{p} = 0.05 < K_{p} = 0.2$ . Why this step? Below target → forward.
Answer: shifts forward — more $C a C O_{3}$ decomposes, raising $p_{C O_{2}}$ toward $0.2$ . ✅

Verify: $ln (0.05/0.2) = ln (0.25) < 0 \Rightarrow Δ G < 0$ forward. If you had wrongly included the solids you'd have divided by $1$ s — same answer here, but the habit fails when you try to "use up" a solid's amount.

Worked example Example 9 — C9: exam twist, reverse the reaction

Given $A ⇌ B$ has $K = 3$ and current $Q = 6$ , we found it shifts backward (toward $A$ ). Twist: the exam reverses the equation to $B ⇌ A$ . What are the new $K^{'}$ , $Q^{'}$ , and direction?

Forecast: reversing an equation inverts its $K$ … does the physical direction change?

For the reversed reaction, $K^{'} = \frac{1}{K} = \frac{1}{3}$ and $Q^{'} = \frac{1}{Q} = \frac{1}{6}$ . Why this step? Reversing a reaction swaps products and reactants, so its quotient is the reciprocal (from Equilibrium constant K (Kc and Kp)).
Compare: $Q^{'} = \frac{1}{6} \approx 0.1667 < K^{'} = \frac{1}{3} \approx 0.3333$ . Why this step? Now $Q^{'} < K^{'}$ , so the reversed reaction runs forward (i.e. $B \to A$ ).
Answer: both descriptions agree — the physical motion is "make more $A$ ." "Backward for $A ⇌ B$ " = "forward for $B ⇌ A$ ." ✅

Verify: ratio invariant: $Q / K = 6/3 = 2$ and $Q^{'} / K^{'} = (1/6) / (1/3) = 1/2$ . Both give $Δ G$ of equal magnitude, opposite sign — the same physical shift. $ln 2 = - ln (1/2)$ ✓.

Worked example Example 10 — C10: real-world word problem

A blood-gas buffer models $C O_{2} + H_{2} O ⇌ H^{+} + H C O_{3}^{-}$ with $K_{c} = 4 \times 1 0^{- 7}$ . A patient's sample reads $[C O_{2}] = 1.0 \times 1 0^{- 3}$ , $[H^{+}] = 5 \times 1 0^{- 8}$ , $[H C O_{3}^{-}] = 2 \times 1 0^{- 2}$ (mol/L; water omitted as pure liquid). Which way does the reaction drift, and does that raise or lower $[H^{+}]$ ?

Forecast: with $K$ so tiny, the system barely wants products; guess backward (consuming $H^{+}$ )?

Omit $H_{2} O$ (pure liquid, activity $1$ ): $Q_{c} = \frac{[ H ^{+} ] [ H C O _{3}^{-} ]}{[ C O _{2} ]}$ . Why this step? Same omission rule as solids (Activity and why pure solids-liquids are omitted).
Plug in: $Q_{c} = \frac{( 5 \times 1 0 ^{- 8} ) ( 2 \times 1 0 ^{- 2} )}{1.0 \times 1 0 ^{- 3}} = \frac{1 \times 1 0 ^{- 9}}{1 \times 1 0 ^{- 3}} = 1 \times 1 0^{- 6}$ . Why this step? Multiply top ( $1 0^{- 9}$ ), divide by bottom ( $1 0^{- 3}$ ) → subtract exponents.
Compare: $Q_{c} = 1 \times 1 0^{- 6} > K_{c} = 4 \times 1 0^{- 7}$ . Why this step? Since $Q > K$ , $Δ G > 0$ for forward, so the reverse reaction is spontaneous.
Answer: shifts backward — $H^{+}$ and $H C O_{3}^{-}$ recombine, so $[H^{+}]$ drops (blood becomes slightly less acidic). This is the buffer resisting acidity, echoing Le Chatelier's principle. ✅

Verify: $Q / K = 1 0^{- 6} / (4 \times 1 0^{- 7}) = 2.5 > 1 \Rightarrow Δ G > 0$ forward, backward confirmed; and backward consumes $H^{+}$ → lower acidity ✓.

Reaction quotient Q vs K — direction of shift (index 2.6.4) — parent rule
Equilibrium constant K (Kc and Kp)
Gibbs free energy and spontaneity
Relation between Kp and Kc
Activity and why pure solids-liquids are omitted
Le Chatelier's principle
Position of equilibrium vs rate of reaction

Worked examples — Reaction quotient Q vs K — direction of shift

The scenario matrix

Worked examples

The matrix, filled

Active recall

Connections