Equilibrium: Unit Diagnostic (Chemistry Unit 7)

What this diagnostic is for

Use this page to find which Unit 7 topics to revisit. Answer every question before opening any answer, then use the table at the end.

These are original Marlbridge practice questions, not past exam questions. They are not calibrated, so your total gives no predicted score.

Calculator and data. A scientific calculator is allowed. Concentrations are in M, partial pressures in atm; Q and K have no units. Every K value and lettered substance is invented. Temperature is constant unless stated.

Question 1 (multiple choice · 7.1)

A rigid, sealed flask starts with colourless gas A, which forms brown gas B: A(g) ⇌ 2B(g). Which observation does not show that equilibrium has been reached?

(A) The brown colour stays exactly the same for an hour.
(B) The total pressure in the flask stays the same for an hour.
(C) At one moment, the partial pressures of A and B are equal.
(D) The partial pressure of A is the same in two readings 10 minutes apart.

Answer and explanation

Answer: (C). Equilibrium amounts are constant, not equal. One reading cannot show that nothing is changing.

(A) A constant colour means a constant amount of B.
(B) Each A gives two B, so a constant total pressure means no net reaction.
(D) A constant partial pressure is direct evidence.

If you missed this: read the Topic 7.1 study guide.

Question 2 (multiple choice · 7.2)

A sealed flask starts with only C, the product of the reversible reaction A + B ⇌ C. Which statement is true at every moment before equilibrium is reached?

(A) Reverse rate > forward rate, and the reverse rate is decreasing.
(B) Forward rate > reverse rate, because the forward reaction must happen first.
(C) The rates are equal, but concentrations are still changing.
(D) Reverse rate > forward rate, and the reverse rate is increasing.

Answer and explanation

Answer: (A). The forward rate starts at zero. As C is used up, the reverse rate falls and the forward rate rises until they meet.

(B) The larger rate sets the direction.
(C) Equal rates is equilibrium.
(D) [C] is falling, so the reverse rate falls.

If you missed this: read the Topic 7.2 study guide.

Question 3 (multiple choice · 7.3)

Zinc is placed in silver nitrate solution and equilibrium is reached:

Zn(s) + 2Ag⁺(aq) ⇌ Zn²⁺(aq) + 2Ag(s)

Which is the expression for Kc?

(A) [Zn²⁺] / [Ag⁺]²
(B) [Zn²⁺][Ag]² / ([Zn][Ag⁺]²)
(C) [Zn²⁺] / (2[Ag⁺])
(D) [Ag⁺]² / [Zn²⁺]

Answer and explanation

Answer: (A). Products over reactants, each raised to its coefficient. The solids Zn and Ag are left out.

(B) includes the solids.
(C) multiplies by the coefficient instead of raising to it.
(D) is the expression for the reverse reaction.

If you missed this: read the Topic 7.3 study guide.

Question 4 (short answer · 7.4)

A rigid flask starts with the fictional gases G at 0.600 atm and H at 0.900 atm, and no GH₂:

G(g) + 2H(g) ⇌ GH₂(g)

The total pressure falls and then stays at 1.100 atm.

(a) Calculate the equilibrium partial pressure of each gas. (b) Calculate Kp. (c) The experiment is repeated at the same temperature with G at 1.200 atm. Will Kp differ? Explain.

Worked answer

(a) Let x atm of G react. H falls by 2x and GH₂ rises by x, so the total falls by 2x: 1.500 − 1.100 = 0.400 atm = 2x, and x = 0.200 atm.

P_G = 0.600 − 0.200 = 0.400 atm; P_H = 0.900 − 0.400 = 0.500 atm; P_GH₂ = 0.200 atm.

(b) Kp = P_GH₂ / (P_G × P_H²) = 0.200 ÷ (0.400 × 0.500²) = 0.200 ÷ 0.100 = 2.00.

(c) No. K depends only on temperature, so the new pressures still give Kp = 2.00.

Common slip: not squaring P_H (gives 1.00).

If you missed this: read the ICE table in the Topic 7.4 study guide.

Question 5 (multiple choice · 7.5)

For X(g) ⇌ Y(g), K = 5 × 10¹² at a fixed temperature. A flask starts with only Y. Which describes the mixture at equilibrium?

(A) Almost all X, because pure Y can only react in reverse.
(B) Almost all Y, with a tiny but non-zero amount of X.
(C) Only Y, with no X at all.
(D) Equal amounts of X and Y.

Answer and explanation

Answer: (B). At equilibrium [X]/[Y] = 1/K = 2 × 10⁻¹³, whichever side you start from.

(A) The starting side sets the direction, not the final position.
(C) Even a huge K never makes a concentration exactly zero.
(D) Equal amounts would need K = 1.

If you missed this: read the Topic 7.5 study guide.

Question 6 (multiple choice · 7.6)

Two invented reactions at the same temperature:

(1) X₂(g) + O₂(g) ⇌ 2XO(g), K₁ = 4.0 × 10⁻³
(2) 2XO(g) + O₂(g) ⇌ 2XO₂(g), K₂ = 2.5 × 10⁵

What is K for XO₂(g) ⇌ ½X₂(g) + O₂(g)?

(A) 1.0 × 10⁻³
(B) 5.0 × 10⁻⁴
(C) 3.2 × 10⁻²
(D) 32

Answer and explanation

Answer: (C). Adding (1) and (2) cancels 2XO: X₂ + 2O₂ ⇌ 2XO₂, K = K₁K₂ = 1.0 × 10³. Reverse (invert) and halve (square root): K = (1 ÷ 1.0 × 10³)^½ = 3.2 × 10⁻².

(A) reverses but forgets to halve.
(B) divides by 2 instead of taking the square root.
(D) takes the square root but forgets to reverse.

If you missed this: read the Topic 7.6 study guide.

Question 7 (multiple choice · 7.7)

For A(aq) ⇌ B(aq), K = 4.0. A solution starts with [A] = 0.40 M and [B] = 0.10 M. What are the equilibrium concentrations?

(A) [A] = 0.10 M, [B] = 0.40 M
(B) [A] = 0.080 M, [B] = 0.32 M
(C) [A] = 0.40 M, [B] = 0.10 M; it is already at equilibrium
(D) [A] = 0.25 M, [B] = 0.25 M

Answer and explanation

Answer: (A). Q = 0.10 ÷ 0.40 = 0.25 < K: forward. (0.10 + x) ÷ (0.40 − x) = 4.0 gives x = 0.30 M. Check: 0.40 ÷ 0.10 = 4.0.

(B) leaves out the B already present.
(C) uses [A]/[B], the expression upside down.
(D) assumes equal amounts (that needs K = 1).

If you missed this: read the Topic 7.7 study guide.

Question 8 (short answer · 7.7)

For the fictional reaction R₂(g) ⇌ 2R(g), Kp = 2.0 × 10⁻⁶. A flask holds P_R₂ = 0.50 atm and P_R = 0.010 atm.

(a) Calculate Qp and state the direction of the net reaction. (b) Calculate P_R at equilibrium. State any approximation you make.

Worked answer

(a) Qp = (0.010)² ÷ 0.50 = 2.0 × 10⁻⁴. Q > K, so the net reaction goes in reverse: R atoms combine to form R₂.

(b) K is tiny, so nearly all R recombines: P_R₂ ≈ 0.50 + 0.010 ÷ 2 = 0.505 atm (a 1% change, so the approximation is valid).

P_R² = 2.0 × 10⁻⁶ × 0.505, so P_R ≈ 1.0 × 10⁻³ atm.

If you missed this: read Worked example 2 in the Topic 7.7 study guide.

Question 9 (multiple choice · 7.8)

For A₂(g) + B₂(g) ⇌ 2AB(g), an equilibrium box holds 1 A₂, 1 B₂ and 4 AB. A second box (same volume and temperature) holds 1 A₂ and 4 B₂ at equilibrium. How many AB does it hold?

(A) 2
(B) 4
(C) 8
(D) 16

Answer and explanation

Answer: (C). Two particles each side, so counts can be used. Box 1: K = 4² ÷ (1 × 1) = 16. Box 2: (AB)² = 16 × 1 × 4 = 64, so AB = 8.

(A) gives Q = 1, far below K.
(B) copies box 1, ignoring the extra B₂.
(D) forgets to square AB in both boxes (K = 4, then AB = 4 × 1 × 4 = 16).

If you missed this: read the Topic 7.8 study guide.

Question 10 (multiple choice · 7.9)

In water, yellow chromate ions and orange dichromate ions are in equilibrium:

2CrO₄²⁻(aq) + 2H⁺(aq) ⇌ Cr₂O₇²⁻(aq) + H₂O(l)

A few drops of NaOH solution are added to an orange sample. What is observed, and why?

(A) More yellow: OH⁻ removes H⁺, so the equilibrium shifts left.
(B) More orange: OH⁻ forms water, a product.
(C) No change: neither Na⁺ nor OH⁻ is in the equation.
(D) More yellow: OH⁻ makes K smaller.

Answer and explanation

Answer: (A). OH⁻ removes H⁺ as water. Removing a reactant shifts the system left, giving more yellow chromate.

(B) Water is the solvent, so it is not a stress.
(C) OH⁻ is not in the equation, but it removes H⁺.
(D) Only temperature changes K.

If you missed this: read the Topic 7.9 study guide.

Question 11 (multiple choice · 7.10)

For A(g) ⇌ 2B(g), Kp = 0.50. At equilibrium P_A = 2.0 atm and P_B = 1.0 atm. Half of the B is removed at constant volume and temperature. Which is correct just after the change?

(A) Qp = ½Kp; the net reaction goes forward.
(B) Qp = ¼Kp; the net reaction goes forward.
(C) Qp = ¼Kp; the net reaction goes in reverse.
(D) Qp = 2Kp; the net reaction goes in reverse.

Answer and explanation

Answer: (B). Qp = 0.50² ÷ 2.0 = 0.125 = ¼Kp. Q < K, so more B forms.

(A) forgets that P_B is squared.
(C) reads Q < K backwards.
(D) treats removing a product as raising Q.

If you missed this: read the Topic 7.10 study guide.

Question 12 (multiple choice · 7.11)

The fictional salt MA₃ dissolves as MA₃(s) ⇌ M³⁺(aq) + 3A⁻(aq). Its molar solubility is 1.0 × 10⁻³ M. What is K_sp?

(A) 1.0 × 10⁻¹²
(B) 3.0 × 10⁻¹²
(C) 2.7 × 10⁻¹¹
(D) 3.0 × 10⁻⁶

Answer and explanation

Answer: (C). [M³⁺] = s and [A⁻] = 3s. K_sp = s(3s)³ = 27s⁴ = 27 × (1.0 × 10⁻³)⁴ = 2.7 × 10⁻¹¹.

(A) uses s × s³.
(B) uses 3s but forgets to cube the 3.
(D) uses 3s but forgets the power of 3.

If you missed this: read the Topic 7.11 study guide.

Question 13 (short answer · 7.12)

The fictional 1:1 salt MX has K_sp = 4.0 × 10⁻¹⁰.

(a) Calculate its molar solubility in pure water. (b) What concentration of X⁻ (from soluble NaX) would cut the molar solubility to 1.00% of its value in pure water? (c) Would the same concentration of M⁺ (from soluble MNO₃) have the same effect? Explain.

Worked answer

(a) K_sp = s², so s = √(4.0 × 10⁻¹⁰) = 2.0 × 10⁻⁵ M.

(b) Target s = 2.0 × 10⁻⁷ M = [M⁺]. [X⁻] = 4.0 × 10⁻¹⁰ ÷ 2.0 × 10⁻⁷ = 2.0 × 10⁻³ M. (The salt’s own X⁻ is negligible.)

(c) Yes. For a 1:1 salt, K_sp = [M⁺][X⁻] treats both ions alike, so [M⁺] = 2.0 × 10⁻³ M gives s = 2.0 × 10⁻⁷ M again.

If you missed this: read the Topic 7.12 study guide.

Your next step

Tick each question you missed (any wrong part of a short answer counts).

Topic	Question(s)	If you missed it
7.1 Introduction to Equilibrium	1	Study guide
7.2 Direction of Reversible Reactions	2	Study guide
7.3 Q and K	3	Study guide
7.4 Calculating the Equilibrium Constant	4	Study guide
7.5 Magnitude of the Equilibrium Constant	5	Study guide
7.6 Properties of the Equilibrium Constant	6	Study guide
7.7 Calculating Equilibrium Concentrations	7, 8	Study guide
7.8 Representations of Equilibrium	9	Study guide
7.9 Le Châtelier’s Principle	10	Study guide
7.10 Q and Le Châtelier’s Principle	11	Study guide
7.11 Solubility Equilibria	12	Study guide
7.12 Common-Ion Effect	13	Study guide

How to use your result

Look at topics, not the total. Count a guess as a miss.
Fix gaps in order, starting with the earliest topic you missed: 7.1 to 7.3 are the base, 7.4 to 7.8 the calculations, 7.9 to 7.12 the applications.
Then practise each topic’s questions, then try the Unit 7 mixed review.