Unit 7 · Topic 7.10 Beta

Reaction Quotient and Le Châtelier's Principle

Every Le Châtelier shift can be justified by comparing Q with K.

Practice 6: Argumentation

Question set for this topic

Part 1 · Hook

Why this matters

Two students explain why adding hydrogen makes more ammonia. One writes “the system shifts to relieve the stress.” The other writes “adding H₂ makes Q less than K, so the net reaction proceeds forward until Q equals K.” Only the second earns the point. This lesson is about writing the second answer every time.

Part 2 · Before you start

What this builds on

Part 3 · Prerequisite check

Quick check before you start

1. A mixture has Q < K. Which way is the net reaction?

  1. Forward
  2. Reverse
  3. No net reaction
  4. It depends on the rate
Show the answer

Q < K means too little product, so products form.

  • Correct: Forward:
  • Reverse:
  • No net reaction:
  • It depends on the rate:

2. Adding a product to an equilibrium mixture makes it shift

  1. toward reactants
  2. toward products
  3. nowhere
  4. toward whichever side has more moles
Show the answer

Le Châtelier: the system partly uses up what was added.

  • Correct: toward reactants:
  • toward products:
  • nowhere:
  • toward whichever side has more moles:

3. Which change alters K for a given reaction?

  1. Changing the temperature
  2. Adding a reactant
  3. Halving the volume
  4. Adding argon
Show the answer

Only temperature changes K.

  • Correct: Changing the temperature:
  • Adding a reactant:
  • Halving the volume:
  • Adding argon:

Part 4 · See it

See it first

Four boxes in a row joined by arrows. At equilibrium, Q equals K. A disturbance makes Q not equal to K. Compare Q with K: if Q is less than K the net reaction goes forward, if greater it goes in reverse. New equilibrium: Q equals K again with new amounts. A note says a concentration or volume change moves Q while K stays fixed, and a temperature change moves K while Q stays fixed. A model justification reads: adding H2 makes Q less than K, so the net reaction proceeds forward until Q equals K.
Concentration and volume changes move Q; temperature moves K. Either way Q ≠ K, and the system reacts until Q = K. LevlPrep original diagram.

Part 5 · Step by step

How it works, step by step

  1. A concentration or the volume is changedQ changes while K stays the same
  2. The temperature is changedK changes while Q stays the same at that instant
  3. Q is now less than Kthe net reaction goes forward
  4. Q is now greater than Kthe net reaction goes in reverse
  5. The net reaction changes the concentrationsQ moves until Q = K again at a new equilibrium

Part 6 · Key ideas

Key ideas

  • Concentration and volume changes move Q; K stays fixed.
  • Temperature changes move K; Q is unchanged at that instant.
  • Then compare: Q < K net forward; Q > K net reverse; until Q = K.
  • A volume change shifts the equilibrium only if the gas exponents on the top and bottom of Q differ.

Part 7 · Misconception

A common mistake

The wrong idea: Adding a reactant increases K, and that is why more product forms.

What actually happens: Adding a reactant decreases Q (more on the bottom), so Q < K, and the net reaction goes forward until Q = K. K is unchanged; only temperature changes it.

Part 8 · Check yourself

Check yourself

Exam-style questions. Anything you miss goes into your review queue.

Data table

Q right after three changes

For N2O4(g) ⇌ 2 NO2(g) at a fixed temperature, Kc = 4.6 × 10−3. A flask at equilibrium is changed in three separate experiments, each at constant temperature. The table gives the concentrations immediately after each change, before any reaction.

Concentrations and Q immediately after each change
Situation[N2O4] (M)[NO2] (M)Qc
Before any change (at equilibrium)0.04000.01364.6 × 10−3
A: NO2 added0.04000.03002.3 × 10−2
B: volume doubled0.02000.006802.3 × 10−3
C: N2O4 removed0.02000.01369.2 × 10−3

1. Which statement correctly justifies the shift after change A?

  1. Q > K, so the net reaction goes in reverse.
  2. K increases because NO2 was added, so the system shifts forward.
  3. The reverse reaction speeds up, so the system shifts toward NO2.
  4. Q < K, so the net reaction goes forward.
Show the answer

After a concentration change, compare the new Q with the unchanged K. Q > K: net reverse.

  • Correct: Q > K, so the net reaction goes in reverse.: Right: adding NO₂ raised Q to 2.3 × 10⁻² while K stayed 4.6 × 10⁻³.
  • K increases because NO2 was added, so the system shifts forward.: K does not change with concentration. The added NO₂ changes Q.
  • The reverse reaction speeds up, so the system shifts toward NO2.: A faster reverse reaction makes N₂O₄, not NO₂. And a justification should compare Q with K.
  • Q < K, so the net reaction goes forward.: Q = 2.3 × 10⁻² is larger than K = 4.6 × 10⁻³, not smaller.

2. After change B (volume doubled), which way does the system shift and why?

  1. Forward: halving each concentration halves Q.
  2. No shift, because both concentrations were halved equally.
  3. In reverse, because a larger volume favors fewer gas moles.
  4. Forward, because doubling the volume doubles K.
Show the answer

Volume changes alter every concentration, and Q changes whenever the gas coefficients differ. Q < K means net forward, toward more gas molecules, consistent with Le Châtelier.

  • Correct: Forward: halving each concentration halves Q.: Right: Q = [NO₂]²/[N₂O₄] has two factors on top and one on the bottom; halving all of them halves Q, to 2.3 × 10−3.
  • No shift, because both concentrations were halved equally.: Equal halving cancels only when the exponents on top and bottom match. Here they are 2 and 1.
  • In reverse, because a larger volume favors fewer gas moles.: A larger volume favors more gas moles. The products side has 2 mol gas versus 1.
  • Forward, because doubling the volume doubles K.: Volume does not change K; it changes Q.

3. After change C, what happens to [NO2] as the system returns to equilibrium?

  1. It decreases, because Q > K and the net reaction runs in reverse.
  2. It increases, because Q > K and the net reaction runs forward.
  3. It stays the same, because NO2 was not changed.
  4. It increases, because the removed N2O4 is replaced by NO2.
Show the answer

Q = (0.0136)² / 0.0200 = 9.2 × 10−3 > K, so the net reaction goes in reverse: NO₂ is used to form N₂O₄.

  • Correct: It decreases, because Q > K and the net reaction runs in reverse.: Right: removing N₂O₄ lowered the denominator, so Q rose above K; the reverse reaction uses NO₂ to re-form N₂O₄.
  • It increases, because Q > K and the net reaction runs forward.: Q > K means net reverse, which uses NO₂.
  • It stays the same, because NO2 was not changed.: Q no longer equals K, so every concentration changes until it does.
  • It increases, because the removed N2O4 is replaced by NO2.: Replacing N₂O₄ means making N₂O₄, which uses NO₂.

Particle view

Adding X₂ to an equilibrium mixture

EquilibriumRight after adding X₂

Key: X2Y2XY

For X2(g) + Y2(g) ⇌ 2 XY(g), the left container is at equilibrium. Three X2 molecules are then injected at constant temperature and volume.

4. Using particle counts (the gas moles are equal on both sides), calculate K for this reaction.

Type a number.

Show the answer

K = (XY)² / ((X₂)(Y₂)) = 4² / (2 × 2) = 4.0.

  • Answer: 4.0

5. Which justification of the shift after X2 is added would earn full credit?

  1. Q = 4²/(5 × 2) = 1.6 < K = 4.0, so net forward until Q = K.
  2. Adding X2 to the container increases K, so more XY forms.
  3. The system wants to use up the extra X2, so it shifts right.
  4. Adding X2 speeds up the forward reaction, so it shifts right.
Show the answer

A full-credit justification compares Q with K: here Q = 1.6 < K = 4.0, so the net reaction goes forward.

  • Correct: Q = 4²/(5 × 2) = 1.6 < K = 4.0, so net forward until Q = K.: Right: claim (direction), evidence (the value of Q from the counts) and reasoning (comparison with the unchanged K).
  • Adding X2 to the container increases K, so more XY forms.: K does not change with concentration. This answer would lose the point.
  • The system wants to use up the extra X2, so it shifts right.: Systems do not "want" anything; a justification needs Q compared with K.
  • Adding X2 speeds up the forward reaction, so it shifts right.: This is rate language alone. On the exam, a shift is justified with Q vs K.

6. For N2(g) + 3 H2(g) ⇌ 2 NH3(g) at equilibrium, the volume is suddenly halved at constant temperature. By what factor does Q change at that instant? (Give Qnew / Qold.)

Type a number.

Show the answer

Each concentration doubles. Qnew = (2[NH₃])² / ((2[N₂])(2[H₂])³) = 4/16 × Qold = 0.25 Qold. Q < K, so the net reaction goes forward, toward fewer gas moles.

  • Answer: 0.25

7. Which statements are valid justifications on a free-response question? Select all that apply.

  1. "Adding H2 makes Q < K, so the net reaction proceeds forward until Q = K."
  2. "Raising the temperature increases K for this endothermic reaction, so Q < K and the reaction shifts forward."
  3. "Adding H2 increases K, so more product forms."
  4. "The reaction shifts right to relieve the stress."
Show the answer

A justification names what changed (Q, or K for a temperature change) and compares Q with K to give the direction.

  • Correct: "Adding H2 makes Q < K, so the net reaction proceeds forward until Q = K.": Right: names the change in Q and compares it with K.
  • Correct: "Raising the temperature increases K for this endothermic reaction, so Q < K and the reaction shifts forward.": Right: identifies that K changes with temperature and uses the comparison.
  • "Adding H2 increases K, so more product forms.": K does not change with concentration.
  • "The reaction shifts right to relieve the stress.": This restates Le Châtelier without evidence or reasoning; readers do not award it.

Part 9 · Summary

Summary

Every Le Châtelier shift can be justified by comparing Q with K. Changing a concentration, the volume or the pressure changes Q but not K; changing the temperature changes K but not Q at that instant. If Q is then less than K the net reaction goes forward, if greater it goes in reverse, until Q equals K again. Full-credit answers name what changed, give the comparison and the direction.

Part 10 · Up next

What comes next

Part 11 · Connections

Connections