Every breath you take changes the Q value of a carbon dioxide equilibrium in your blood. Your brainstem monitors that Q vs Keq gap continuously — and adjusts your breathing rate to close it.
Use the PDF for classwork, homework or revision. It includes key ideas, activities, questions, an extend task and success-criteria proof.
"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.
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.
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:
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 calculations follow the same mechanics as Keq calculations — but you must use the concentrations specified in the problem, not equilibrium values.
5-Step Procedure:
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.
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.
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:
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:
| 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 |
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.
$$Q = \frac{[\text{NH}_3]^2}{[\text{N}_2][\text{H}_2]^3} = \frac{(0.150)^2}{(0.200)(0.300)^3}$$
Numerator: $(0.150)^2 = 0.02250$
Denominator: $(0.200)(0.300)^3 = (0.200)(0.02700) = 0.005400$
$Q = 0.02250 / 0.005400 = \mathbf{4.17}$
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.
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.
$$Q = \frac{[\text{HI}]^2}{[\text{H}_2][\text{I}_2]} = \frac{(0.148)^2}{(0.020)(0.020)} = \frac{0.02190}{0.000400} = 54.8 \approx 54.3 \checkmark$$
(Small rounding difference confirms equilibrium.)
New [HI] = 0.148 + 0.030 = 0.178 mol/L. [H₂] and [I₂] are momentarily unchanged at 0.020 mol/L each.
$$Q = \frac{(0.178)^2}{(0.020)(0.020)} = \frac{0.03168}{0.000400} = 79.2$$
Q = 79.2 > Keq = 54.3 → system shifts LEFT → HI decomposes to form more H₂ and I₂ until Q decreases back to 54.3.
Q is calculated with the same expression as Keq but using the current (not necessarily equilibrium) concentrations. If Q < Keq, the reaction shifts right (toward products). If Q > Keq, it shifts left (toward reactants). If Q = Keq, the system is already at equilibrium and no net shift occurs. This is a powerful predictive tool before any equilibrium concentrations are measured.
Look back at what you wrote in the Think First section. What has changed? What did you get right? What surprised you?
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?
Answer questions on Reaction Quotient Q before your opponents cross the line. Fast answers = faster car. Pool: lessons 1–12.
Lesson 12 complete — Reaction Quotient Q