Unit 7 · Topic 7.3 Beta

Reaction Quotient and Equilibrium Constant

The equilibrium constant expression puts products over reactants, each raised to its coefficient, and leaves out pure solids and liquids; K c uses concentrations and K p partial pressures.

Practice 1: Models and RepresentationsPractice 5: Mathematical Routines

Question set for this topic

Part 1 · Hook

Why this matters

A chemist mixes three gases in a steel tank and needs to know, before opening a valve, whether the tank will make more product or eat some up. Watching rates is slow and hard. Instead, she plugs today’s concentrations into one formula and compares the answer with a number from a table. Thirty seconds of arithmetic, and she knows which way the reaction will go.

Part 2 · Before you start

What this builds on

Part 3 · Prerequisite check

Quick check before you start

1. At equilibrium, how do the forward and reverse rates compare?

  1. They are equal
  2. The forward rate is larger
  3. The reverse rate is larger
  4. Both are zero
Show the answer

Equal rates keep the concentrations constant.

  • Correct: They are equal:
  • The forward rate is larger:
  • The reverse rate is larger:
  • Both are zero:

2. If the reverse rate is greater than the forward rate, what happens to the product concentrations?

  1. They decrease
  2. They increase
  3. They stay constant
  4. They become equal to the reactant concentrations
Show the answer

A faster reverse reaction turns products back into reactants overall.

  • Correct: They decrease:
  • They increase:
  • They stay constant:
  • They become equal to the reactant concentrations:

3. What is the partial pressure of a gas in a mixture?

  1. The pressure that gas alone would exert in the same volume
  2. The total pressure divided by the number of gases
  3. The pressure of the gas before it was mixed
  4. The pressure of the gas at STP
Show the answer

Each gas contributes its own share of the total pressure (Dalton’s law).

  • Correct: The pressure that gas alone would exert in the same volume:
  • The total pressure divided by the number of gases:
  • The pressure of the gas before it was mixed:
  • The pressure of the gas at STP:

Part 4 · See it

See it first

A number line for Q starting at Q = 0 (no products) on the left and increasing to the right. K is marked in the middle. To the left of K, Q is less than K and an arrow points right toward K: the net reaction goes forward, products form and Q rises. To the right of K, Q is greater than K and an arrow points left toward K: the net reaction goes in reverse, reactants form and Q falls. At K, Q equals K and there is no net change.
A mixture always reacts in the direction that brings Q to K: forward when Q is less than K, in reverse when Q is greater. LevlPrep original diagram.

Part 5 · Step by step

How it works, step by step

  1. The expression for K is written from the balanced equationproducts over reactants, each raised to its coefficient, with pure solids and liquids left out
  2. Today’s concentrations or pressures are put into the same expressionyou get Q, a snapshot of where the mixture is now
  3. Q is less than Kthe forward rate is larger, so products form and Q rises toward K
  4. Q is greater than Kthe reverse rate is larger, so reactants form and Q falls toward K
  5. Q reaches Kthe system is at equilibrium and the concentrations stop changing

Part 6 · Key ideas

Key ideas

  • K = products over reactants, each to the power of its coefficient, at equilibrium. Leave out pure solids and liquids.
  • Kc uses molar concentrations in brackets; Kp uses partial pressures (no brackets).
  • Q has the same expression, filled in with the amounts present now.
  • Q < K: net forward. Q > K: net reverse. Q = K: at equilibrium.

Part 7 · Misconception

A common mistake

The wrong idea: If a mixture contains more product than reactant, it will react in reverse.

What actually happens: Direction comes only from Q compared with K. A flask full of SO₃ can still react forward if its Q is below K, as flask 1 does with Q = 80.0 and K = 280.

Part 8 · Check yourself

Check yourself

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

Data table

Four mixtures of SO₂, O₂ and SO₃

For 2 SO2(g) + O2(g) ⇌ 2 SO3(g), Kc = 2.80 × 102 at a certain temperature. Four sealed flasks are prepared at that temperature with the starting concentrations below.

Starting concentrations in each flask
Flask[SO2] (M)[O2] (M)[SO3] (M)
10.1000.05000.200
20.02000.04000.0800
30.05000.02000.118
40.3000.1000

1. Calculate the reaction quotient, Qc, for flask 1.

Type a number.

Show the answer

Qc = [SO₃]² / ([SO₂]²[O₂]) = (0.200)² / ((0.100)² × 0.0500) = 0.0400 / 0.000500 = 80.0.

  • Answer: 80.0

2. For flask 2, Qc = 400. Which way does the net reaction go as flask 2 approaches equilibrium?

  1. In reverse, because Q > K: there is too much product for equilibrium.
  2. Forward, because Q > K means the forward reaction is faster.
  3. Forward, because flask 2 holds more SO3 than SO2.
  4. No net reaction, because Q and K are both greater than 1.
Show the answer

Compare Q with K. If Q > K, the mixture has too many products relative to reactants, so the net reaction goes in reverse, lowering Q until Q = K.

  • Correct: In reverse, because Q > K: there is too much product for equilibrium.: Right: Q = 400 > K = 280. To lower Q, products must turn back into reactants until Q = K.
  • Forward, because Q > K means the forward reaction is faster.: This flips the rule. Q > K means the ratio of products to reactants is already too high, so the net reaction runs in reverse.
  • Forward, because flask 2 holds more SO3 than SO2.: Direction comes from comparing Q with K, not from comparing amounts in the flask.
  • No net reaction, because Q and K are both greater than 1.: Being on the same side of 1 means nothing. The flask is at equilibrium only when Q equals K.

3. Which flask is already at equilibrium when it is prepared?

  1. Flask 3
  2. Flask 1
  3. Flask 2
  4. Flask 4
Show the answer

A mixture is at equilibrium when Q = K. Only flask 3 has Q ≈ 280.

  • Correct: Flask 3: Right: Q = (0.118)² / ((0.0500)² × 0.0200) = 278, equal to K = 280 within the precision of the data.
  • Flask 1: Q = 80.0 for flask 1, well below K, so it will react forward.
  • Flask 2: Q = 400 for flask 2, above K, so it will react in reverse.
  • Flask 4: Flask 4 has no SO₃, so Q = 0. It can only react forward.

4. What happens in flask 4 after it is prepared?

  1. SO3 forms, because Q = 0 is less than K.
  2. Nothing, because there is no SO3 to react.
  3. SO2 forms, because Q is less than K.
  4. Q is undefined, because [SO3] is zero.
Show the answer

A mixture with reactants and no products has Q = 0, which is smaller than any K, so it reacts forward.

  • Correct: SO3 forms, because Q = 0 is less than K.: Right: with no product, Q = 0 < K, so the net reaction must go forward.
  • Nothing, because there is no SO3 to react.: The forward reaction needs SO₂ and O₂, which are present. Only the reverse reaction lacks a reactant.
  • SO2 forms, because Q is less than K.: Q < K means net forward, which uses SO₂. It cannot form SO₂.
  • Q is undefined, because [SO3] is zero.: Q can be calculated: zero on top gives Q = 0. Only a zero on the bottom would be a problem.

5. A student says flask 1 will react in reverse because it contains the most SO3. Which response is best?

  1. Disagree: Q = 80.0 is less than K, so flask 1 reacts forward and makes more SO3.
  2. Agree: the flask with the most product reacts in reverse.
  3. Disagree: flask 1 is at equilibrium because it has more SO3 than SO2.
  4. Agree: Q = 80.0 is greater than 1, so the net reaction goes in reverse.
Show the answer

The ratio, not a single concentration, decides direction. Flask 1 has Q = 80.0 < K = 280, so more SO₃ forms until Q = K.

  • Correct: Disagree: Q = 80.0 is less than K, so flask 1 reacts forward and makes more SO3.: Right: the comparison that matters is Q with K, and Q is below K.
  • Agree: the flask with the most product reacts in reverse.: The amount of product alone does not decide direction. What matters is the whole ratio Q compared with K.
  • Disagree: flask 1 is at equilibrium because it has more SO3 than SO2.: Q = 80.0 is not 280, so flask 1 is not at equilibrium.
  • Agree: Q = 80.0 is greater than 1, so the net reaction goes in reverse.: Q is compared with K, not with 1. Here Q < K, so the net reaction goes forward.

6. Which is the equilibrium constant expression for CaCO3(s) ⇌ CaO(s) + CO2(g)?

  1. Kc = [CO2]
  2. Kc = [CaO][CO2] / [CaCO3]
  3. Kc = [CaO] / [CaCO3]
  4. Kc = 1 / [CO2]
Show the answer

Pure solids and pure liquids do not appear in K or Q, because their concentration is fixed by their density. Only gases and dissolved species are included.

  • Correct: Kc = [CO2]: Right: the two pure solids are left out, leaving only the gas.
  • Kc = [CaO][CO2] / [CaCO3]: Pure solids are left out of K: their "concentration" does not change however much solid is present.
  • Kc = [CaO] / [CaCO3]: This keeps the two solids and drops the gas, the opposite of the rule.
  • Kc = 1 / [CO2]: CO₂ is a product, so it goes on top.

7. For H2(g) + I2(g) ⇌ 2 HI(g), a mixture holds [H2] = 0.0150 M, [I2] = 0.0300 M and [HI] = 0.0900 M. Calculate Qc.

Type a number.

Show the answer

Qc = [HI]² / ([H₂][I₂]) = (0.0900)² / (0.0150 × 0.0300) = 0.00810 / 0.000450 = 18.0.

  • Answer: 18.0

Part 9 · Summary

Summary

The equilibrium constant expression puts products over reactants, each raised to its coefficient, and leaves out pure solids and liquids; Kc uses concentrations and Kp partial pressures. The reaction quotient Q has the same expression filled in with the current amounts. If Q is less than K the net reaction goes forward, if Q is greater than K it goes in reverse, and when Q equals K the system is at equilibrium.

Part 10 · Up next

What comes next

Part 11 · Connections

Connections