Unit 7 · 7–9% of the exam Beta

Unit 7 practice test: Equilibrium

12 exam-style questions from the unit, free, with an explanation for every option. Four answer choices each, and data sets that share one table, graph or particle diagram, as on the real exam.

Covers: Introduction to Equilibrium; Direction of Reversible Reactions; Reaction Quotient and Equilibrium Constant; Calculating the Equilibrium Constant; Magnitude of the Equilibrium Constant; Properties of the Equilibrium Constant; Calculating Equilibrium Concentrations; Representations of Equilibrium; Introduction to Le Châtelier's Principle; Reaction Quotient and Le Châtelier's Principle; Introduction to Solubility Equilibria; Common-Ion Effect.

Experimental setup

Two students solve the same problem

For 2 H2S(g) ⇌ 2 H2(g) + S2(g), Kc = 1.0 × 10−6 at a certain temperature. A flask starts with 0.10 M H2S only.

Student P writes the Equilibrium row as [H2S] = 0.10 − 2x, [H2] = 2x, [S2] = x, assumes 2x is small next to 0.10, and writes (2x)2(x) / (0.10)2 = 1.0 × 10−6.

Student Q writes the same Equilibrium row, then drops x everywhere it appears, including the top: (0)2(0) / (0.10)2 = 1.0 × 10−6, and says there is no solution.

1. Which student used the small-x approximation correctly?

  1. Student P, who drops x just from (0.10 − 2x).
  2. Student Q, who sets each x term to zero because x is small.
  3. Both, because dropping x anywhere gives the same approximation.
  4. Neither, because the approximation works for two-species reactions.
Show the answer

The approximation replaces (initial − x) by (initial) when x is tiny. It never replaces x itself by zero.

  • Correct: Student P, who drops x just from (0.10 − 2x).: Right: 0.10 − 2x ≈ 0.10 is fine, but x terms standing alone must stay, or nothing is left to solve for.
  • Student Q, who sets each x term to zero because x is small.: Setting x to zero on top makes Q = 0, which can never equal K. Only x inside a sum or difference may be dropped.
  • Both, because dropping x anywhere gives the same approximation.: Dropping x from "0.10 − 2x" changes the value by a few percent; dropping it from "(2x)²x" changes it to zero. They are not the same.
  • Neither, because the approximation works for two-species reactions.: The number of species does not matter. What matters is whether x is small next to the number it is subtracted from.

2. How should Student P check that the approximation was justified?

  1. Compare 2x with 0.10 M; it is under 5%.
  2. Check that x is positive.
  3. Check that K is less than 1.
  4. Compare x with K; x should be smaller than K.
Show the answer

The 5% rule: the amount subtracted (here 2x) should be under about 5% of the starting concentration. If it is, the approximation was justified.

  • Correct: Compare 2x with 0.10 M; it is under 5%.: Right: 2x = 2.7 × 10−3 M, which is 2.7% of 0.10 M.
  • Check that x is positive.: A positive x is needed, but it does not show the approximation is valid.
  • Check that K is less than 1.: A small K makes the approximation likely to work, but you still verify it with the answer.
  • Compare x with K; x should be smaller than K.: x and K are different kinds of quantity; the test compares x with the starting concentration.

3. Why was it reasonable to expect x to be small before solving?

  1. K = 1.0 × 10−6 is very small.
  2. The flask starts with no H2 or S2.
  3. H2S has a coefficient of 2 in the balanced equation.
  4. The starting concentration, 0.10 M, is less than 1 M.
Show the answer

When K is much smaller than the starting concentration terms, little reactant is used, so x is small relative to the initial amount.

  • Correct: K = 1.0 × 10−6 is very small.: Right: a tiny K means a reactant-favored equilibrium, so the change from the starting amount is small.
  • The flask starts with no H2 or S2.: Starting with no products tells you the direction (forward), not how far the reaction goes.
  • H2S has a coefficient of 2 in the balanced equation.: Coefficients appear in the change row but do not decide whether x is small.
  • The starting concentration, 0.10 M, is less than 1 M.: The size of the concentration alone does not decide it; the ratio of K to the concentration does.

Experimental setup

Making ammonia at high pressure

Ammonia forms from nitrogen and hydrogen: N2(g) + 3 H2(g) ⇌ 2 NH3(g). At the temperature of a reactor, Kp = 1.5 × 10−5. At one moment the gas in the reactor has these partial pressures: N2, 10.0 atm; H2, 30.0 atm; NH3, 1.50 atm.

4. Which expression is Kp for this reaction?

  1. PNH₃2 / (PN₂ × PH₂3)
  2. [NH3]2 / ([N2][H2]3)
  3. 2 PNH₃ / (PN₂ × 3 PH₂)
  4. (PN₂ × PH₂3) / PNH₃2
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Kp has the same form as Kc, with partial pressures in place of concentrations: products over reactants, each to the power of its coefficient.

  • Correct: PNH₃2 / (PN₂ × PH₂3): Right: partial pressures of products over reactants, each raised to its coefficient.
  • [NH3]2 / ([N2][H2]3): This is Kc. Square brackets mean molar concentrations; Kp uses partial pressures.
  • 2 PNH₃ / (PN₂ × 3 PH₂): Coefficients become exponents, not multipliers.
  • (PN₂ × PH₂3) / PNH₃2: This is upside down: products go on top.

5. Qp for the reactor gas is 8.33 × 10−6. What happens to the partial pressure of NH3 as the gas approaches equilibrium?

  1. It increases, because Qp < Kp.
  2. It decreases, because Qp < Kp.
  3. It stays the same, because both values are much less than 1.
  4. It decreases, because there is far more H2 than NH3.
Show the answer

Qp < Kp means too little product for equilibrium, so the net reaction goes forward and PNH₃ rises until Qp = Kp.

  • Correct: It increases, because Qp < Kp.: Right: 8.33 × 10⁻⁶ < 1.5 × 10⁻⁵, so the net reaction goes forward and makes NH₃.
  • It decreases, because Qp < Kp.: Q < K means net forward, which makes NH₃. This mixes up the rule.
  • It stays the same, because both values are much less than 1.: Q and K are compared with each other, not with 1. They differ by almost a factor of 2.
  • It decreases, because there is far more H2 than NH3.: Direction comes from comparing Q with K, not from which gas has the higher pressure.

6. A student calculates Qp using the moles of each gas in the reactor in place of the partial pressures. What is wrong with this approach?

  1. Kp uses partial pressures, so Qp has to as well.
  2. Moles are fine, because Qp depends on the ratio and not the quantity used.
  3. Moles are fine, as long as the student uses the coefficients as exponents.
  4. Qp should use the concentrations in brackets, as Kc does.
Show the answer

Q is only meaningful next to the K of the same form: Qc with Kc (molarities), Qp with Kp (partial pressures in atm).

  • Correct: Kp uses partial pressures, so Qp has to as well.: Right: Q and K can only be compared if they are built from the same kind of quantity.
  • Moles are fine, because Qp depends on the ratio and not the quantity used.: With exponents that do not cancel (2 on top, 4 in total on the bottom), changing the quantity changes the value of Q.
  • Moles are fine, as long as the student uses the coefficients as exponents.: The exponents are right, but moles are not pressures. The result cannot be compared with Kp.
  • Qp should use the concentrations in brackets, as Kc does.: Concentrations belong in Qc, compared with Kc. For Kp, use partial pressures without brackets.

7. Solid calcium carbonate decomposes when heated: CaCO3(s) ⇌ CaO(s) + CO2(g). A sample is heated in an open dish. Why does this system not reach equilibrium?

  1. CO2 escapes, so it rarely meets CaO to react back.
  2. Solids are not able to take part in reversible reactions.
  3. The forward reaction is too slow to reach equilibrium at any temperature.
  4. Heating makes the reverse reaction faster than the forward reaction.
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A reversible reaction reaches equilibrium only in a closed system, where products stay available to react back. An open dish lets CO₂ escape.

  • Correct: CO2 escapes, so it rarely meets CaO to react back.: Right: equilibrium needs a closed system. With the gas leaving, CO₂ can hardly ever collide with CaO to re-form CaCO₃.
  • Solids are not able to take part in reversible reactions.: Solids can react in both directions; in a sealed container this same reaction does reach equilibrium.
  • The forward reaction is too slow to reach equilibrium at any temperature.: Speed is not the issue: the heated solid does decompose. The problem is that a product leaves the system.
  • Heating makes the reverse reaction faster than the forward reaction.: If the reverse reaction were faster, CaCO₃ would form, but the open dish loses CO₂ and the solid keeps decomposing.

8. In a sealed flask, the forward rate of 2 SO2(g) + O2(g) ⇌ 2 SO3(g) is 5.0 × 10−3 M/s and the reverse rate is 2.0 × 10−3 M/s. Which statement is correct?

  1. [SO3] is increasing.
  2. [SO3] is decreasing.
  3. [SO3] is constant.
  4. [O2] is increasing.
Show the answer

When the forward rate is greater, the net reaction goes forward: reactants decrease and products increase.

  • Correct: [SO3] is increasing.: Right: the forward reaction, which makes SO₃, is faster than the reverse reaction, which uses it.
  • [SO3] is decreasing.: That would need the reverse reaction to be faster. Here the forward rate is the larger one.
  • [SO3] is constant.: Constant concentrations need equal rates. 5.0 × 10⁻³ and 2.0 × 10⁻³ M/s are not equal.
  • [O2] is increasing.: O₂ is used by the faster forward reaction, so it is decreasing.

9. For N2(g) + 3 H2(g) ⇌ 2 NH3(g) at a certain temperature, the equilibrium partial pressures are N2 0.50 atm, H2 1.5 atm and NH3 0.12 atm. Calculate Kp.

Type a number.

Show the answer

Kp = PNH₃² / (PN₂ × PH₂³) = (0.12)² / (0.50 × (1.5)³) = 0.0144 / 1.6875 = 8.5 × 10−3.

  • Answer: 0.0085

10. For a reaction at 25 °C, K = 6.0 × 10−9. Which describes the equilibrium mixture?

  1. Mostly reactants, with a very small amount of products
  2. Mostly products, with a very small amount of reactants
  3. Equal amounts of reactants and products
  4. No products whatsoever
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K ≪ 1: reactant-favored. Some product is present at equilibrium, but very little.

  • Correct: Mostly reactants, with a very small amount of products: Right: K far below 1 means the reaction barely goes forward before it reaches equilibrium.
  • Mostly products, with a very small amount of reactants: That would need K far above 1.
  • Equal amounts of reactants and products: K near 1 gives comparable amounts; 10⁻⁹ is very far from 1.
  • No products whatsoever: A small K still means some product forms. K is never zero.

11. For 2 SO2(g) + O2(g) ⇌ 2 SO3(g), Kc = K. What is Kc for SO2(g) + ½ O2(g) ⇌ SO3(g)?

  1. K1/2
  2. K ÷ 2
  3. 2K
  4. 1/K
Show the answer

Multiplying an equation by n raises K to the n. Here n = ½, so K becomes √K.

  • Correct: K1/2: Right: halving every coefficient raises K to the power ½.
  • K ÷ 2: Coefficients are exponents in K, so halving them takes a square root, not half.
  • 2K: This would follow from treating coefficients as multipliers, which they are not.
  • 1/K: That is the reversed reaction, not the halved one.

12. For 2 SO2(g) + O2(g) ⇌ 2 SO3(g), ΔH° = −198 kJ/molrxn, which change increases the value of K?

  1. Cooling the mixture
  2. Adding more O2
  3. Decreasing the volume
  4. Adding a catalyst
Show the answer

Only temperature changes K. For an exothermic reaction (heat is a product), lowering the temperature increases K.

  • Correct: Cooling the mixture: Right: for an exothermic reaction, cooling favors the products and increases K.
  • Adding more O2: This shifts the equilibrium forward but does not change K.
  • Decreasing the volume: This shifts toward fewer gas moles (products), but K stays the same.
  • Adding a catalyst: A catalyst changes neither K nor the equilibrium position.

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Practice has every question in the unit, with feedback after each one; the notes for every topic are free. Other units: Unit 1 · Unit 2 · Unit 3 · Unit 4 · Unit 5 · Unit 6 · Unit 7 · Unit 8 · Unit 9.