Practise this lesson
Four printable worksheets that build from the foundations up to exam-style questions — start at whatever level suits you.
In 1955, biochemist Theodor Bücher at the University of Münster monitored the glycolysis pathway in living muscle cells and found that the [product]/[reactant] ratios in the cell were consistently far from the thermodynamic Keq values — the cell maintained Q ≠ Keq to keep reactions running continuously. His 1955 paper introduced the concept of "mass action ratio" (what we now call Q) as distinct from the equilibrium constant.
"Q and Keq use the same expression — so Q and Keq must always be equal."
What is wrong with this statement? If Q and Keq have identical algebraic forms, why can they have different values? Write your reasoning before continuing.
Know
- The definition of reaction quotient Q and how it differs from Keq
- The Q vs Keq decision rule: Q > Keq shifts left, Q < Keq shifts right, Q = Keq means no net change
- How to calculate Q using the same algebraic expression as Keq but with current concentrations
Understand
- Why Q is a snapshot of current concentrations while Keq describes the equilibrium destination
- How adding or removing species changes Q but not Keq
- Why Q is useful for predicting physiological processes like blood CO₂ regulation
Can Do
- Calculate Q from given concentrations and compare to Keq to predict direction of shift
- Apply the 5-step Q procedure to exam-style problems
- Explain how Q changes as a system approaches equilibrium
Module 5 — Key Formulas: Lesson 12
Misconceptions to Fix
Keq tells you where the system will end up — Q tells you where the system is right now. Together they let you predict which direction the system must travel to reach equilibrium.
The reaction quotient Q has the same algebraic expression as Keq — products raised to stoichiometric powers in the numerator, reactants in the denominator, solids and pure liquids excluded.
The only difference is the concentrations used:
- Keq uses equilibrium concentrations — measured when the system has reached equilibrium.
- Q uses current concentrations — whatever the concentrations are at this particular moment, whether at equilibrium or not.
When the system is at equilibrium, Q = Keq by definition. At any other moment, Q ≠ Keq. Q is a snapshot; Keq is the destination.
Memory aid: if Q is too low, the system needs to go up (right, more products) to reach Keq. If Q is too high, the system needs to come down (left, back to reactants).
Q vs Keq — three relationships and the direction of shift each predicts
Q uses the same expression as Keq but with current (not equilibrium) concentrations. Three rules: Q < Keq → too few products → shift RIGHT; Q > Keq → too many products → shift LEFT; Q = Keq → at equilibrium, no net shift. Q = 0 from pure reactants; Q = ∞ from pure products.
Copy the Q snapshot definition and the three comparison rules into your notes before the check below.
For the reaction N₂(g) + 3H₂(g) ⇌ 2NH₃(g) at 400°C, Q = 0.12 and Keq = 0.50. Which statement is correct?
We just saw that Q is a snapshot of the current product-to-reactant ratio compared to Keq as the destination. That raises a question: how do you actually calculate Q from a set of given concentrations, and how do you turn that number into a direction prediction? This card answers it → by laying out the five-step Q calculation procedure with a worked SO₃ example.
Q calculations follow the same mechanics as Keq calculations — but you must use the concentrations specified in the problem, not equilibrium values.
5-Step Procedure:
- Write the Keq expression (same expression as Q)
- Identify the current concentrations given — NOT equilibrium concentrations unless stated
- Substitute current concentrations → calculate Q
- Compare Q to the given Keq
- State the direction: Q < Keq → right; Q > Keq → left; Q = Keq → at equilibrium
Example: $\text{2SO}_2(g) + \text{O}_2(g) \rightleftharpoons \text{2SO}_3(g)$, Keq = 270 at 700°C.
Current: [SO₂] = 0.40 mol/L, [O₂] = 0.30 mol/L, [SO₃] = 0.20 mol/L.
$$Q = \frac{[\text{SO}_3]^2}{[\text{SO}_2]^2[\text{O}_2]} = \frac{(0.20)^2}{(0.40)^2(0.30)} = \frac{0.0400}{0.04800} = 0.833$$
Q = 0.833 < Keq = 270 → system shifts RIGHT → more SO₃ forms at the expense of SO₂ and O₂.
Five-step Q calculation: (1) write Keq expression; (2) identify current (not equilibrium) concentrations; (3) substitute and calculate Q; (4) compare Q to Keq; (5) state direction — Q < Keq → RIGHT, Q > Keq → LEFT. Only use concentrations explicitly labelled "equilibrium concentrations" in Keq directly.
Pause — write the five-step Q procedure into your notes before the check below.
For PCl₃(g) + Cl₂(g) ⇌ PCl₅(g), Keq = 65.0. Current concentrations: [PCl₃] = 0.250, [Cl₂] = 0.150, [PCl₅] = 0.600 mol/L. What is Q and which direction does the system shift?
We just saw the five-step procedure for calculating Q and comparing it to Keq to predict shift direction. That raises a question: when a disturbance (adding or removing a species) hits a system already at equilibrium, how does Q explain what happens — and why does LCP always give the right answer? This card answers it → by showing how each disturbance changes Q's numerator or denominator, forcing a predictable shift.
Q is the mathematical tool that converts a qualitative LCP prediction into a quantitative check — it explains precisely why adding a species disturbs equilibrium and in which direction.
When a species is added to a system already at equilibrium (Q = Keq), the addition immediately changes the concentrations. The new Q is instantly different from Keq — the system is no longer at equilibrium and must shift.
| Disturbance | Effect on Q expression | Q vs Keq | Direction of shift | LCP consistent? |
|---|---|---|---|---|
| Add reactant | Denominator increases → Q decreases | Q < Keq | RIGHT | Yes ✓ |
| Add product | Numerator increases → Q increases | Q > Keq | LEFT | Yes ✓ |
| Remove reactant | Denominator decreases → Q increases | Q > Keq | LEFT | Yes ✓ |
| Remove product | Numerator decreases → Q decreases | Q < Keq | RIGHT | Yes ✓ |
Q provides the mathematical underpinning for every LCP prediction. The rule "add reactant → shift right" is just a verbal description of what happens to Q when the denominator increases. Q is the mechanism behind LCP.
Adding a reactant → denominator of Q increases → Q decreases → Q < Keq → shift RIGHT. Adding a product → numerator of Q increases → Q increases → Q > Keq → shift LEFT. Q is the mathematical mechanism behind every LCP prediction — adding or removing a species instantly changes Q, and the shift is Q's journey back to Keq.
Add the four disturbance→Q→direction entries to your notes before the diagram and check below.
A system H₂(g) + I₂(g) ⇌ 2HI(g) is at equilibrium. More I₂ is added. Which correctly explains what happens to Q?
We just saw that disturbances change Q's numerator or denominator, driving the system to shift until Q returns to Keq. That raises a question: how does this Q mechanism operate in a real biological system — and what does it mean when blood chemistry is described as an equilibrium? This card answers it → by tracing how CO₂ buildup during exercise changes Q, drops pH, and triggers faster breathing to restore equilibrium.
Your respiratory system is a Q vs Keq controller — it continuously adjusts your breathing rate to keep the Q value of a specific blood equilibrium as close as possible to Keq, maintaining blood pH within the narrow range compatible with life.
Carbon dioxide in blood participates in a critical equilibrium:
$$\text{CO}_2(aq) + \text{H}_2\text{O}(l) \rightleftharpoons \text{H}^+(aq) + \text{HCO}_3^-(aq)$$
Normal blood pH 7.35–7.45 requires this ratio [H⁺][HCO₃⁻]/[CO₂] to equal a specific Keq (Ka for carbonic acid ≈ 4.3 × 10⁻⁷).
During intense exercise:
- Cells produce CO₂ rapidly → [CO₂(aq)] increases
- Denominator of Q increases → Q decreases → Q < Keq
- Equilibrium shifts RIGHT → more H⁺ and HCO₃⁻ produced → blood pH drops
- Chemoreceptors in brainstem detect pH drop → increase breathing rate and depth
- Exhaling CO₂ removes it from blood → [CO₂(aq)] decreases → Q increases back toward Keq
- Equilibrium shifts LEFT → [H⁺] decreases → pH recovers
Blood CO₂ mechanism: exercise → [CO₂(aq)] increases → denominator of Q for CO₂ + H₂O ⇌ H⁺ + HCO₃⁻ increases → Q decreases < Keq → shifts right → [H⁺] rises → pH drops → faster breathing → exhale CO₂ → [CO₂] falls → Q rises back to Keq → shifts left → pH recovers. It is a Q-driven shift, not neutralisation.
Pause — write the exercise-CO₂ mechanism chain into your notes before the check below.
During exercise, CO₂ builds up in blood. For CO₂(aq) + H₂O(l) ⇌ H⁺(aq) + HCO₃⁻(aq), what immediately happens to Q?
We just saw how Q explains the blood CO₂ pH recovery — a biological system that continuously drives Q back to Keq. That raises a question: what is Q doing mathematically as a reaction proceeds from scratch — how does it actually change from its starting value until it reaches Keq? This card answers it → by showing Q's journey from 0 (pure reactants) or ∞ (pure products) continuously converging on Keq, and connecting that journey to the ICE table variable x.
Q is not fixed — it changes continuously as the reaction proceeds, always moving toward Keq. This dynamic behaviour is what the ICE table calculation tracks, step by step.
When a reaction begins from non-equilibrium conditions:
- Q < Keq initially: system shifts right → product concentrations increase (numerator of Q increases), reactant concentrations decrease (denominator decreases) → Q increases progressively until Q = Keq.
- Q > Keq initially: system shifts left → reactant concentrations increase (denominator increases), product concentrations decrease (numerator decreases) → Q decreases progressively toward Keq.
| Starting conditions | Q value | vs Keq | Direction |
|---|---|---|---|
| Pure reactants only | Q = 0 (no products → numerator = 0) | Q < any positive Keq | Always RIGHT initially |
| Pure products only | Q = ∞ (no reactants → denominator = 0) | Q > any finite Keq | Always LEFT initially |
| At equilibrium | Q = Keq | Equal | No net shift |
Q changes continuously during approach to equilibrium: from pure reactants Q = 0 → always shifts right; from pure products Q = ∞ → always shifts left. As the reaction proceeds Q converges monotonically on Keq. ICE table variable x = the composition change required to take Q from its initial value to Keq — ICE and Q are two views of the same journey.
Write the Q-journey concept (Q = 0 → right; Q = ∞ → left; x = Q-to-Keq distance) into your notes before the check below.
A container is filled with only NH₃(g) for the reaction N₂(g) + 3H₂(g) ⇌ 2NH₃(g). What is the value of Q and which direction will the reaction go?
The equilibrium $\text{N}_2(g) + 3\text{H}_2(g) \rightleftharpoons 2\text{NH}_3(g)$ has Keq = 0.500 at 400°C. A reaction mixture contains [N₂] = 0.200 mol/L, [H₂] = 0.300 mol/L, and [NH₃] = 0.150 mol/L. (a) Calculate Q. (b) Compare Q to Keq and predict the direction of shift. (c) Describe what happens to Q as the system moves toward equilibrium.
Calculate Q
$$Q = \frac{[\text{NH}_3]^2}{[\text{N}_2][\text{H}_2]^3} = \frac{(0.150)^2}{(0.200)(0.300)^3}$$
Numerator: $(0.150)^2 = 0.02250$
Denominator: $(0.200)(0.300)^3 = (0.200)(0.02700) = 0.005400$
$Q = 0.02250 / 0.005400 = \mathbf{4.17}$
Compare Q to Keq and Predict
Q = 4.17 > Keq = 0.500. The current ratio of products to reactants is greater than the equilibrium ratio — there are too many products relative to equilibrium.
The system shifts LEFT (reverse direction) — NH₃ decomposes back to N₂ and H₂ until equilibrium is established.
Q During Approach
As the system shifts left, [NH₃] decreases (numerator of Q decreases) and [N₂] and [H₂] increase (denominator increases). Q decreases progressively from 4.17 toward 0.500. When Q = 0.500 = Keq, equilibrium is established.
The equilibrium $\text{H}_2(g) + \text{I}_2(g) \rightleftharpoons 2\text{HI}(g)$ has Keq = 54.3 at 430°C. The system is at equilibrium with [H₂] = [I₂] = 0.020 mol/L and [HI] = 0.148 mol/L. Verify this is at equilibrium, then predict what happens when 0.030 mol/L of HI is added. Calculate the new Q and state the direction of shift.
Verify Equilibrium
$$Q = \frac{[\text{HI}]^2}{[\text{H}_2][\text{I}_2]} = \frac{(0.148)^2}{(0.020)(0.020)} = \frac{0.02190}{0.000400} = 54.8 \approx 54.3 \checkmark$$
(Small rounding difference confirms equilibrium.)
After Adding 0.030 mol/L HI
New [HI] = 0.148 + 0.030 = 0.178 mol/L. [H₂] and [I₂] are momentarily unchanged at 0.020 mol/L each.
Calculate New Q
$$Q = \frac{(0.178)^2}{(0.020)(0.020)} = \frac{0.03168}{0.000400} = 79.2$$
Q = 79.2 > Keq = 54.3 → system shifts LEFT → HI decomposes to form more H₂ and I₂ until Q decreases back to 54.3.
Checking Your Understanding
-
Define the reaction quotient Q and explain how it differs from Keq.
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Identify one common misconception about Q and Keq. Explain why it is incorrect.
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Describe how adding a product to a system at equilibrium changes Q and the subsequent direction of shift.
Applying Your Knowledge
Q1. For the reaction $\text{PCl}_3(g) + \text{Cl}_2(g) \rightleftharpoons \text{PCl}_5(g)$, Keq = 65.0 at 250°C. Current concentrations are: [PCl₃] = 0.250 mol/L, [Cl₂] = 0.150 mol/L, [PCl₅] = 0.600 mol/L. Which statement correctly describes the system?
Q2. A student adds more N₂ to the Haber process equilibrium $\text{N}_2(g) + 3\text{H}_2(g) \rightleftharpoons 2\text{NH}_3(g)$. Which correctly explains the direction of shift using Q?
Q3. At the start of the reaction $\text{A}(g) + \text{B}(g) \rightleftharpoons 2\text{C}(g)$, a flask contains only A and B (no C). What is Q, and in which direction will the reaction proceed?
SAQ 1 (4 marks). The equilibrium $2\text{SO}_2(g) + \text{O}_2(g) \rightleftharpoons 2\text{SO}_3(g)$ has Keq = 280 at 700°C. A reaction mixture at 700°C contains [SO₂] = 0.350 mol/L, [O₂] = 0.220 mol/L, and [SO₃] = 0.510 mol/L. (a) Calculate Q. (b) Predict and justify the direction of shift. (c) Describe how Q changes as the system reaches equilibrium.
SAQ 2 (3 marks). Explain, using Q and Le Chatelier's Principle, what happens when a product is removed from a system at equilibrium. Include the effect on Q, the comparison to Keq, and the direction of the resulting shift.
Reveal Answers
SAQ 1: (a) $Q = (0.510)^2 / [(0.350)^2 (0.220)] = 0.2601 / 0.02695 = 9.65$. (b) Q = 9.65 < Keq = 280 → system shifts RIGHT. There are too few products relative to the equilibrium ratio; the forward reaction is favoured to produce more SO₃ at the expense of SO₂ and O₂. (c) As the system shifts right, [SO₃] increases (numerator increases) and [SO₂], [O₂] decrease (denominator decreases). Q progressively increases from 9.65 toward 280 until Q = Keq and equilibrium is re-established.
SAQ 2: Removing a product decreases the numerator of Q (since products appear in the numerator of the Q expression). This causes Q to decrease below Keq (Q < Keq). The system is no longer at equilibrium — by Le Chatelier's Principle, the system responds to oppose the change by producing more of the removed product. The equilibrium shifts RIGHT (forward) until Q increases back to equal Keq.
For a reaction at 25 °C, Keq = 50. A non-equilibrium mixture gives Q = 120. Predict: which direction will the reaction shift, and why?
How close was your prediction?
Excellent Q vs K reasoning — this is high-frequency HSC content.
Always compare Q to K: Q > K means excess products → shift left. Q < K means excess reactants → shift right.
Complete the Learn phase to unlock practice questions.
A student claims: "Q is a useful concept but it adds unnecessary complexity — Le Chatelier's Principle already predicts the direction of shift for any disturbance, so Q is redundant." Evaluate this claim. In your response, discuss: (1) what Q tells us that LCP does not; (2) a specific example where Q provides quantitative information beyond LCP; (3) whether Q and LCP ever give conflicting predictions and why or why not. (6 marks)
At the start of this lesson you were asked: "Q and Keq use the same expression — so Q and Keq must always be equal." Recall Bücher's 1955 finding: living muscle cells deliberately maintain Q ≠ Keq to keep metabolic reactions running. You now know why this is possible — Q is calculated using current concentrations (a snapshot), while Keq uses equilibrium concentrations. They use the same algebraic form but different inputs — Q = Keq only when the concentrations used happen to be the equilibrium concentrations.