Keq tells you where a reaction is heading. Q tells you where it is right now. Put the two side by side and you can predict, without any experiment, which way any mixture must shift to reach equilibrium.
Today's hook, A chemist is handed a sealed flask of an unknown mixture and asked which way it will react. No measurements over time, no waiting. One calculation and a comparison answers it, and that calculation is the reaction quotient.
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You’re here
Get oriented
Warm up first
Three quick questions from earlier lessons. Pulling old material back to mind before you learn something new makes the new material stick better, so this is not busywork.
Worksheets
Practise this lesson
Four printable worksheets that build from the foundations up to exam-style questions, start at whatever level suits you.
Biochemists measuring the reactions of living cells kept finding the same odd result: the product-to-reactant ratios inside a working cell sit a long way from the values thermodynamics predicts for equilibrium. A cell that reached equilibrium would be a dead cell, so living systems hold that ratio away from Keq on purpose. Chemists gave the measured ratio its own name, the mass action ratio, which is what we now call Q.
"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 the lever an industrial reactor is actually run on, while Keq stays fixed
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
Q = [products]n / [reactants]musing CURRENT concentrations (not equilibrium)
Q < Keq → system shifts RIGHT (more products needed)
Q > Keq → system shifts LEFT (too many products)
Q = Keq → system is at equilibrium, no net shift
Q = 0 at start from pure reactants → always shifts right initially
Q = ∞ at start from pure products → always shifts left initially
Key Terms, scan these before reading
Reaction quotient (Q)
The ratio of product to reactant concentrations at any instant, using the same form as Keq.
Q vs Keq comparison
If Q < Keq: reaction proceeds forward (more products form); Q > Keq: reverse; Q = Keq: at equilibrium.
Predicting shift direction
Comparing Q to Keq allows prediction of which direction the reaction must proceed to reach equilibrium.
Non-equilibrium state
Any mixture where Q ≠ Keq; concentrations are still changing.
Effect of dilution on Q
Diluting all species simultaneously may increase or decrease Q depending on the mole ratio in Keq.
Reaction progress
As a reaction proceeds from any starting point, Q changes until it equals Keq at equilibrium.
Misconceptions to Fix
✗ Wrong: Q and Keq are calculated using different formulas.
✓ Right: Q and Keq use the exact same algebraic expression. The difference is that Q uses current concentrations (a snapshot), while Keq uses equilibrium concentrations. Q tells you where the system is now; Keq tells you where it will end up.
Predict then reveal+8 XP
1 · Predict
2 · Reveal
3 · Compare
For a reaction at 25 °C, Keq = 50. A non-equilibrium mixture gives Q = 120. Predict: which direction will the reaction shift, and why?
50%
Actual answer
Shifts in the reverse direction (left), toward reactants.
Q > Keq (120 > 50) means there are too many products relative to equilibrium. The system converts products back into reactants until Q = Keq = 50. Rule: Q > K → reverse; Q < K → forward; Q = K → at equilibrium.
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.
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Q compared with Keq
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Q vs Keq: The Snapshot vs The Destination
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.
Q < Keq→Too few products relative to equilibrium → shift RIGHT → Q increases toward Keq
Q > Keq→Too many products relative to equilibrium → shift LEFT → Q decreases toward Keq
Q = Keq→At equilibrium, no net shift
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).
Common Error
"Q and Keq are equal when the same concentrations are used." This is circular and wrong. Q = Keq only when the concentrations used happen to be the equilibrium concentrations. Q can be calculated at any moment, it only equals Keq at equilibrium. The Think First misconception is wrong precisely because Q and Keq use the same expression but different inputs.
Q vs Keq, three relationships and the direction of shift each predicts
Exam Tip
When explaining equilibrium shifts, always state the direction (left or right) and justify using Le Chatelier's Principle language, simply stating the direction alone will not earn full marks.
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?
Interactive, Q vs Keq Predictor
Use the tool. A mixture of N2, H2 and NH3 gives Q = 0.25 while Keq = 0.060. Which way does the system shift, and why?
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Calculating Q
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Calculating Q and Comparing to Keq
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
Q = 0.833 < Keq = 270 → system shifts RIGHT → more SO₃ forms at the expense of SO₂ and O₂.
Must Know
When a problem says "the following concentrations are present in a reaction mixture", these are current concentrations for Q, not equilibrium concentrations. Only use concentrations labelled "equilibrium concentrations" or "concentrations at equilibrium" for a direct Keq substitution.
Common Error
Students calculate Q correctly but state the wrong direction, most commonly "Q < Keq therefore shifts left." Wrong. Q < Keq means the products side is deficient, the system shifts RIGHT to produce more products. Always write: Q < Keq → RIGHT; Q > Keq → LEFT.
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?
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Q when a system is disturbed
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Q Applied to Disturbances: Adding/Removing Species
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.
Must Know (HSC)
When a question asks you to "use Q to explain the effect of adding a reactant to an equilibrium system," your answer must: (1) calculate Q after the addition showing the denominator increase; (2) compare Q to Keq; (3) state the direction of shift. Do not just state LCP without Q when the question specifically asks for Q.
Insight
This is why Q is more powerful than LCP for quantitative problems. LCP tells you direction; Q tells you direction AND by how much the system is displaced from equilibrium (the larger |Q − Keq|, the greater the displacement). In IQ4, Qsp compared to Ksp will tell you not just whether a precipitate forms, but how much precipitate forms.
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.
Read the last entry in each zone carefully. Adding or removing a species moves Q while Keq stays put. A temperature change is the opposite case: Q is momentarily unchanged and Keq is what moves. Both can leave you with Q ≠ Keq, but by opposite mechanisms, and only one of them is a concentration change.
A system H₂(g) + I₂(g) ⇌ 2HI(g) is at equilibrium. More I₂ is added. Which correctly explains what happens to Q?
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Q in an industrial reactor
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Running an Industrial Reactor on 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 is that mechanism used deliberately, by someone who wants more product rather than just a prediction? This card answers it → by following how an ammonia plant keeps Q permanently below Keq so the forward reaction never stops.
An industrial reactor is a Q machine. The operator cannot change Keq at a fixed temperature, so every lever they pull, removing product, recycling reactants, is really a lever on Q, and the plant runs profitably precisely because Q is never allowed to reach Keq.
The Haber process synthesises ammonia in a sealed reactor:
with $Q = \dfrac{[\text{NH}_3]^2}{[\text{N}_2][\text{H}_2]^3}$, exactly the Keq expression evaluated on whatever the reactor happens to contain right now.
How an operator uses Q:
Sample the reactor and calculate Q from the measured concentrations.
Q < Keq tells the operator the mixture is still net-producing NH₃, so there is yield left to collect.
Q = Keq tells them the reactor has stalled: no further net product, no matter how long they wait.
Condensing the ammonia out lowers [NH₃], the numerator of Q falls, so Q drops below Keq again and the forward reaction restarts.
Recycling the unreacted N₂ and H₂ back into the feed raises the denominator, pushing Q lower still.
The size of the gap matters as well as its sign. A mixture with Q far below Keq is a long way from equilibrium and converts quickly; as Q climbs toward Keq the net forward rate fades. This is why plants pull ammonia out continuously instead of waiting for one batch to finish.
Same reasoning, different system
The identical Q argument works on any equilibrium, including one in solution. Dissolved carbon dioxide sits in the equilibrium CO₂(aq) + H₂O(l) ⇌ H⁺(aq) + HCO₃⁻(aq), so Q = [H⁺][HCO₃⁻]/[CO₂]. Working muscle releases CO₂ into the blood, which raises the denominator, so Q falls below Keq and the system shifts right, producing more H⁺ and HCO₃⁻. Breathing the CO₂ back out lowers [CO₂] again, Q rises back toward Keq, and the system shifts left. Note that this is a Q-driven shift, not a neutralisation. The acid-base chemistry of that equilibrium itself, including how H⁺ relates to pH and how Ka is defined, is Module 6 work; here it is only an example of reading Q against Keq.
Must Know
For a "use Q to explain why the product is removed continuously" question, structure your answer as: (1) identify which species changes; (2) state whether that species sits in the numerator or the denominator; (3) state the direction of the Q change; (4) compare Q to Keq; (5) state the direction of shift and the effect on yield.
Common Error
"Removing the ammonia increases Keq, so more forms." Keq is fixed at a fixed temperature and no concentration change can move it. Removing ammonia changes Q, not Keq: the numerator falls, Q drops below the unchanged Keq, and the system shifts right. Only a temperature change moves Keq itself.
Industrial Q control: condense NH₃ out → numerator of Q for N₂ + 3H₂ ⇌ 2NH₃ falls → Q drops below Keq → shifts right → more NH₃ forms. Recycling unreacted N₂ and H₂ raises the denominator and lowers Q further. Keq never changes at constant temperature, only Q does.
Pause, write the remove-product to Q to shift-direction chain into your notes before the check below.
In an ammonia reactor at equilibrium, N₂(g) + 3H₂(g) ⇌ 2NH₃(g), the NH₃ is condensed and drained off. What immediately happens to Q?
Cross-lesson links: The Q and Keq distinction introduced here is tested quantitatively in the ICE table problems from L10 and L11. The industrial Q control in Card 4 builds directly on the Haber process yield reasoning from L07. Q changes during the approach to equilibrium (Card 5) give the conceptual foundation for the ICE table change-row direction. In IQ4 the same comparison returns as Qsp against Ksp, which you use in L18 to decide whether a precipitate forms.
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How Q changes on the way to equilibrium
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Q Changes During Approach to Equilibrium
We just saw how an ammonia plant holds Q below Keq on purpose so the forward reaction never stalls. 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 = 0 at the start from pure reactants → system always shifts right regardless of Keq. Q = ∞ from pure products → system always shifts left. These extreme cases explain why the approach to equilibrium always starts in the appropriate direction.
Insight
The ICE table calculation is, at its heart, a Q-to-Keq journey. The variable x represents how far Q must travel from its initial value to reach Keq. Solving for x is equivalent to asking: "how much does the composition need to change for Q to equal Keq?" This is why ICE table problems and Q calculations are in consecutive lessons, they are two perspectives on the same question.
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?
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Worked examples and activities
Worked Examples
Worked Example 1 Analyse
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.
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.
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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.
Answer: (a) Q = 4.17. (b) Q > Keq → shift LEFT; NH₃ decomposes. (c) Q decreases from 4.17 toward 0.500 as equilibrium is approached. ✓
Worked Example 2 Analyse/Evaluate
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.
Q = 79.2 > Keq = 54.3 → system shifts LEFT → HI decomposes to form more H₂ and I₂ until Q decreases back to 54.3.
Answer: Original system verified at equilibrium (Q = 54.8 ≈ 54.3). After adding HI: Q = 79.2 > Keq = 54.3 → shift LEFT. HI decreases; H₂ and I₂ increase until Q = 54.3 at new equilibrium. ✓
Activities
🔎 Activity 1, Spot + Fix
Checking Your Understanding
Answer the questions below to check your understanding of the key concepts from this lesson.
Define the reaction quotient Q and explain how it differs from Keq.
Identify one common misconception about Q and Keq. Explain why it is incorrect.
Describe how adding a product to a system at equilibrium changes Q and the subsequent direction of shift.
🔬 Activity 2, Apply + Analyse
Applying Your Knowledge
Use what you have learned to solve the problems below. Show your reasoning clearly.
Question A
Explain how Q can be applied to a real-world context such as controlling the yield of an industrial ammonia reactor.
Question B
Compare and contrast Q and Keq using specific examples, explaining when they are equal and when they differ.
Question C
Predict what would happen to Q and the direction of shift if a catalyst were added to an equilibrium system, and justify your prediction.
✓
Practice questions
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Multiple Choice
+5 XP
A fresh set drawn from this lesson's question bank, with feedback shown immediately. +5 XP per correct · +25 XP all correct
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Short Answer
(4 marks) 1. 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.
(3 marks) 2. 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.
(5 marks) 3. At 25°C, the equilibrium $\text{CO}(g) + 3\text{H}_2(g) \rightleftharpoons \text{CH}_4(g) + \text{H}_2\text{O}(g)$ has Keq = 5.67 × 10⁶. A mixture contains [CO] = 0.250, [H₂] = 0.700, [CH₄] = 0.0400, [H₂O] = 0.0400 mol/L. (a) Write the Q expression. (b) Calculate Q. (c) In which direction will the reaction proceed? (d) As equilibrium is established, describe how Q changes.
C
Calculation practice
Written calculation questions in the style and mark range of the HSC paper. Set your working out in full, because method marks are awarded for the steps as well as the final value.
AnalyseBand 5(4 marks) A reaction vessel contains PCl₅, PCl₃ and Cl₂ at concentrations 0.0100, 0.0200 and 0.0300 mol/L respectively, for the system PCl₅(g) ⇌ PCl₃(g) + Cl₂(g) where Keq = 0.0400 at this temperature. Calculate the reaction quotient Q, compare it with Keq, and predict the direction in which the system will shift to reach equilibrium. Justify your prediction.
ApplyBand 3(2 marks) For the system CO(g) + H₂O(g) ⇌ CO₂(g) + H₂(g), a mixture contains [CO] = 0.0500, [H₂O] = 0.0500, [CO₂] = 0.100 and [H₂] = 0.100 mol/L. Calculate Q for this mixture.
Show all answers
Short Answer Model Answers
SA1 (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.
SA2 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.
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)
How did your thinking change?
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 the living-cell measurements: a working cell holds its mass action ratio a long way from Keq, and an ammonia plant does exactly the same thing deliberately by draining the product away. You now know why that is possible, Q is calculated using current concentrations (a snapshot), while Keq uses equilibrium concentrations. They use the same algebraic form but different inputs, so Q = Keq only when the concentrations you put in happen to be the equilibrium ones.
