Thermochemistry: Unit Diagnostic (Chemistry Unit 6)

What this diagnostic is for

Use this page to find which Unit 6 topics to revisit. There is one question per topic (two for 6.4 and 6.9). 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. Water and dilute solutions: c = 4.18 J g⁻¹ °C⁻¹; water ΔH_fus = 6.01 kJ mol⁻¹. Molar masses: H 1.008, O 16.00, Mg 24.31 g mol⁻¹. Other data are given in the questions. Substances named with letters are fictional.

Question 1 (multiple choice · 6.1)

For which process is the change endothermic for the substance named in brackets?

(A) Molten wax setting in a cold mould (the wax)
(B) A cup of tea cooling on a desk (the tea)
(C) A metal spoon warming up in a bowl of hot soup (the spoon)
(D) Steam condensing on a bathroom mirror (the steam)

Answer and explanation

Answer: (C). The spoon gains energy from the soup. Heating is endothermic for the substance heated.

(A) Freezing forms attractions, releasing energy: exothermic for the wax.
(B) The tea loses energy to the air and desk. Cooling is exothermic for the tea.
(D) Condensing is exothermic for the steam; it warms the mirror.

If you missed this: read “Three kinds of process” in the Topic 6.1 study guide.

Question 2 (multiple choice · 6.2)

An energy profile for a reaction shows the reactants at 40 kJ mol⁻¹, the top of the hump (the transition state) at 150 kJ mol⁻¹ and the products at 95 kJ mol⁻¹. Which statement is correct?

(A) The reaction is exothermic, with ΔH = −55 kJ mol⁻¹.
(B) The reaction is endothermic, with ΔH = +110 kJ mol⁻¹.
(C) The reaction is endothermic, with ΔH = +95 kJ mol⁻¹.
(D) The reaction is endothermic, with ΔH = +55 kJ mol⁻¹.

Answer and explanation

Answer: (D). ΔH depends only on the end levels: 95 − 40 = +55 kJ mol⁻¹. The products are higher: endothermic.

(A) subtracts the wrong way round (reactants − products).
(B) measures to the top of the hump: that is the activation energy.
(C) reads the product level as if the reactants were at zero.

If you missed this: read “Diagrams with a hump” in the Topic 6.2 study guide.

Question 3 (multiple choice · 6.3)

A 2.00 kg iron block at 30 °C touches a 10.0 g copper bead at 60 °C on an insulated tray. Which statement is correct?

(A) Net energy flows from block to bead, because the block has more total energy.
(B) No net energy flows, because the block’s larger mass balances the bead’s higher temperature.
(C) Net energy flows from bead to block, because the bead’s atoms have a greater average kinetic energy.
(D) Net energy flows from bead to block until both have the same total energy.

Answer and explanation

Answer: (C). The bead is warmer, so its atoms have the greater average kinetic energy. In collisions at the surface, energy passes on average from bead to block.

(A) The direction is set by temperature, not by total energy.
(B) Size does not cancel a temperature difference.
(D) Transfer stops at equal temperatures; the block still has far more total energy.

If you missed this: read “How collisions move energy” in the Topic 6.3 study guide.

Question 4 (multiple choice · 6.4)

An 80.0 g sample of a metal absorbs 0.898 kJ of energy and warms from 21.0 °C to 46.0 °C. What is its specific heat capacity?

(A) 0.244 J g⁻¹ °C⁻¹
(B) 0.449 J g⁻¹ °C⁻¹
(C) 4.49 × 10⁻⁴ J g⁻¹ °C⁻¹
(D) 2.23 J g⁻¹ °C⁻¹

Answer and explanation

Answer: (B). c = q / (mΔT) = 898 J ÷ (80.0 g × 25.0 °C) = 0.449 J g⁻¹ °C⁻¹ (iron has this value).

(A) uses the final temperature, 46.0 °C, as ΔT.
(C) leaves q in kilojoules but writes the unit as joules.
(D) inverts the equation: (80.0 × 25.0) ÷ 898.

If you missed this: read “Specific heat capacity and q = mcΔT” in the Topic 6.4 study guide.

Question 5 (short answer · 6.4)

A student dissolves 3.20 g of a fictional salt LQ (M = 64.0 g mol⁻¹) in 60.0 g of water in a calorimeter. The heat capacity of the calorimeter itself is 25.0 J °C⁻¹. The temperature falls from 22.6 °C to 18.4 °C.

(a) Is dissolving LQ endothermic or exothermic? Justify your answer from the direction of energy flow. (b) Calculate the energy absorbed or released by the dissolving process. Include the calorimeter. (c) Calculate the enthalpy change of solution of LQ in kJ mol⁻¹, with its sign, to an appropriate precision.

Worked answer

(a) Endothermic. The temperature fell, so the solution and calorimeter lost energy to the dissolving process.

(b)

ΔT = 18.4 − 22.6 = −4.2 °C. Mass of solution = 60.0 + 3.20 = 63.20 g.
Solution: q = 63.20 × 4.18 × (−4.2) = −1109.5 J.
Calorimeter: q = CΔT = 25.0 × (−4.2) = −105.0 J.
Energy is conserved: q(dissolving) = −(−1109.5 − 105.0) = +1214.5 J (about +1.2 kJ, absorbed).

(c) n(LQ) = 3.20 ÷ 64.0 = 0.0500 mol. ΔH = +1.2145 kJ ÷ 0.0500 mol = 24.29 kJ mol⁻¹, so +24 kJ mol⁻¹ (2 significant figures, set by ΔT = 4.2 °C).

Common slips: leaving out the calorimeter (+22 kJ mol⁻¹); a negative sign because the temperature fell.

If you missed this: read Worked example 3 in the Topic 6.4 study guide.

Question 6 (multiple choice · 6.5)

What is the energy change when 27.0 g of liquid water at 0 °C freezes completely?

(A) 9.01 kJ absorbed
(B) 9.01 kJ released
(C) 61.0 kJ released
(D) 162 kJ released

Answer and explanation

Answer: (B). n = 27.0 ÷ 18.016 = 1.499 mol. Freezing is the reverse of melting: q = 1.499 × (−6.01) = −9.01 kJ, so 9.01 kJ is released.

(A) has the wrong direction: freezing forms attractions, so it is exothermic.
(C) uses ΔH_vap (40.7 kJ mol⁻¹); freezing needs ΔH_fus.
(D) multiplies grams by ΔH_fus, skipping the conversion to moles.

If you missed this: read “Opposite changes, opposite signs” in the Topic 6.5 study guide.

Question 7 (multiple choice · 6.6)

Magnesium burns in oxygen: 2Mg(s) + O₂(g) → 2MgO(s) ΔH = −1203 kJ mol⁻¹. How much heat is released when 6.08 g of magnesium burns completely?

(A) 49.5 kJ
(B) 150 kJ
(C) 301 kJ
(D) 7.31 × 10³ kJ

Answer and explanation

Answer: (B). n(Mg) = 6.08 ÷ 24.31 = 0.2501 mol. The equation has 2 Mg, so moles of reaction = 0.1251 mol. q = 0.1251 × (−1203) = −150 kJ: 150 kJ is released.

(A) is the energy per gram (1203 ÷ 24.31), not for 6.08 g.
(C) forgets to divide by the coefficient 2.
(D) multiplies the mass in grams by ΔH.

If you missed this: read Worked example 2 in the Topic 6.6 study guide.

Question 8 (multiple choice · 6.7)

Estimate ΔH for C₂H₄(g) + H₂O(g) → C₂H₅OH(g), using these average bond enthalpies in kJ mol⁻¹: C–H 413, C–C 347, C=C 614, C–O 358, O–H 467. Ethanol is CH₃–CH₂–O–H.

(A) −37 kJ mol⁻¹
(B) +37 kJ mol⁻¹
(C) +43 kJ mol⁻¹
(D) −504 kJ mol⁻¹

Answer and explanation

Answer: (A). Cancel the bonds that survive (four C–H and one O–H). Broken: C=C + O–H = 614 + 467 = 1081 kJ. Formed: C–C + C–H + C–O = 347 + 413 + 358 = 1118 kJ. ΔH ≈ broken − formed = 1081 − 1118 = −37 kJ mol⁻¹.

(B) uses formed − broken, which reverses the sign.
(C) counts the C=C bond as two C–C bonds (694 kJ). A double bond has its own value.
(D) forgets that one O–H bond in water must break.

If you missed this: read “Counting bonds without mistakes” in the Topic 6.7 study guide.

Question 9 (multiple choice · 6.8)

Use ΔH°f values in kJ mol⁻¹, NH₃(g) −46.1, H₂O(l) −285.8, H₂O(g) −241.8, to calculate ΔH° for:

4NH₃(g) + 3O₂(g) → 2N₂(g) + 6H₂O(l)

(A) −1266.4 kJ mol⁻¹
(B) +1530.4 kJ mol⁻¹
(C) −1530.4 kJ mol⁻¹
(D) −239.7 kJ mol⁻¹

Answer and explanation

Answer: (C). N₂ and O₂ are elements in their standard states (0). ΔH° = 6(−285.8) − 4(−46.1) = (−1714.8) − (−184.4) = −1530.4 kJ mol⁻¹.

(A) uses H₂O(g), but the equation forms liquid water.
(B) calculates reactants − products.
(D) ignores the coefficients: −285.8 − (−46.1).

If you missed this: read Worked example 2 in the Topic 6.8 study guide.

Question 10 (multiple choice · 6.9)

Two equations for the fictional elements P and R are known:

(1) 2P(s) + 3R₂(g) → 2PR₃(g) ΔH₁
(2) PR₃(g) + R₂(g) → PR₅(g) ΔH₂

Which expression gives ΔH for P(s) + 5/2 R₂(g) → PR₅(g)?

(A) ½ΔH₁ + ΔH₂
(B) ΔH₁ + ΔH₂
(C) ½ΔH₁ − ΔH₂
(D) ΔH₁ + 2ΔH₂

Answer and explanation

Answer: (A). The target has 1 P(s): ½ × (1) gives P(s) + 3/2 R₂(g) → PR₃(g). Add (2): PR₃ cancels and R₂ totals 5/2. So ΔH = ½ΔH₁ + ΔH₂.

(B) uses all of (1), which needs 2 P(s).
(C) reverses (2), which would put PR₅ on the left.
(D) scales the whole sum to 2 mol of P, not the 1 mol in the target.

If you missed this: read “A strategy for building the target equation” in the Topic 6.9 study guide.

Question 11 (short answer · 6.9)

A fictional element Q forms two compounds with a fictional gas Z₂:

(1) 2Q(s) + Z₂(g) → 2QZ(s) ΔH₁ = −430 kJ mol⁻¹
(2) QZ(s) + Z₂(g) → QZ₃(g) ΔH₂ = −150 kJ mol⁻¹

(a) Show how to combine the equations to give Q(s) + 3/2 Z₂(g) → QZ₃(g), and calculate its ΔH. (b) Give ΔH for 2QZ₃(g) → 2Q(s) + 3Z₂(g). (c) A student measures the target reaction directly and asks whether the answer could differ from (a). Use conservation of energy to explain why it cannot (apart from experimental error).

Worked answer

(a) The target needs 1 Q, so halve (1): Q(s) + ½Z₂(g) → QZ(s), ΔH = −215 kJ mol⁻¹. Add (2) as written: QZ(s) cancels and Z₂ totals ½ + 1 = 3/2.

ΔH = −215 + (−150) = −365 kJ mol⁻¹.

(b) This is the target reversed (sign flips) and doubled: −2 × (−365) = +730 kJ mol⁻¹.

(c) Both routes have the same start and end. Energy is conserved, so the two steps together must transfer the same energy as the one-step route. Otherwise, running one route forwards and the other backwards would create energy.

Common slips: −580 kJ mol⁻¹ (not halving (1)); +365 kJ mol⁻¹ in (b) (not doubling).

If you missed this: read “Why the steps must add up” in the Topic 6.9 study guide.

Your next step

Tick each question you got wrong; a short-answer question counts as missed if any part was wrong.

Topic	Question(s)	If you missed it, read
6.1 Endothermic and Exothermic Processes	1	Topic 6.1 study guide
6.2 Energy Diagrams	2	Topic 6.2 study guide
6.3 Heat Transfer and Thermal Equilibrium	3	Topic 6.3 study guide
6.4 Heat Capacity and Calorimetry	4, 5	Topic 6.4 study guide
6.5 Energy of Phase Changes	6	Topic 6.5 study guide
6.6 Introduction to Enthalpy of Reaction	7	Topic 6.6 study guide
6.7 Bond Enthalpies	8	Topic 6.7 study guide
6.8 Enthalpy of Formation	9	Topic 6.8 study guide
6.9 Hess’s Law	10, 11	Topic 6.9 study guide

How to use your result

Look at topics, not the total. The table tells you what to read.
Read every explanation. If you guessed, count it as missed.
Fix gaps in order. Topics 6.1 to 6.3 give the direction of energy transfer, 6.4 to 6.6 put numbers on it, and 6.7 to 6.9 find ΔH without an experiment. Start with the earliest gap.
Watch for sign errors. Right size, wrong sign? Revisit Topics 6.1 and 6.4: q(process) = −q(solution).
Then practise. After each study guide, try that topic’s practice questions, then the Unit 6 mixed review.