2024 chemistry paper 1

i. Answer ALL questions in Section A and any FOUR (4) questions from Section B.

ii. Each question in Section A carries 5 marks while each question in Section B carries 25 marks.

iii. The use of scientific programmable calculator is PROHIBITED.

iv. Table of constants:

| Constant | Symbol | Value |

|—|---|—|

| Gas constant | R | 8.314 JK⁻¹ mol⁻¹ |

| Molar volume of gas at STP | Vₘ | 22.4 dm³ |

| Avogadro’s constant | Nₐ | 6.023 x 10²³ mol⁻¹ |

| Faraday constant | F | 96500 C mol⁻¹ |

| 1 atomic mass unit (a.m.u) | = 931.5 MeV | = 1.6023 x 10⁻¹³ J |

| Rydberg constant | R∞ | 109678 cm⁻¹ |

| 1 atmospheric pressure | = 760 mm Hg | = 1.013 x 10⁵ Nm⁻² |

Atomic numbers of the following elements are: K = 19, C = 6, Si = 14, Fe = 26, Cu = 29, Co = 27, S = 16, Ti = 22, I = 53

Atomic masses of the following elements are: H = 1, N = 14, O = 16, K = 39, Si = 28, F = 19, Co = 59, Br = 79.9, Cl = 35.5, Fe = 56 and C = 12.

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PAGE 2 — SECTION A

1. Classify each of the following statements as True or False:

• a) Electronegativity of an element is inversely proportional to its electronegativity.

• b) Sigma bond can be formed by lateral overlap of two p-orbitals.

• c) A reaction occurring in a vacuum flask is an example of an isolated system.

• d) Metals have sonorous property.

• e) There are no intermolecular forces of attraction between real gas molecules.

2. a) How many moles of Br₂ are there in 15.98 g of Br₂?

b) How many moles of KOH are there in 1 dm³ of its 3 mole dm³ aqueous solution?

c) How many atoms are there in two moles of cobalt?

d) How many atoms are there in 0.1 mole of silicon?

e) What is the mass of 6.023 x 10²³ molecules of fluorine?

3. State three (3) factors that can favour the formation of each of the following:

a) metallic bond, (b) ionic bond.

4. A 35.5 g of a gaseous compound, Y₂O₇ occupies a volume of 5.60 dm³ at standard temperature and pressure. Calculate the relative molar mass of (a) Y₂O₇ (b) Y.

5. Consider the elements: Na, Si, Mg, P, S, Ar, B, Cl. Which of these elements:

a) forms a hydride with empirical formula, XH₃?

b) exists as a diatomic molecule at room temperature?

c) is monoatomic at room temperature?

d) is the most electropositive?

e) has semiconducting property?

6. a) What are isotopes?

b) Why are isotopes chemically indistinguishable?

c) Give three (3) physical properties that can be used to differentiate between isotopes of an element.

7. a) List five (5) characteristic properties of s-block elements.

8. a) What is the normal boiling point of a substance?

b) Is it possible for water to boil at a temperature higher than its boiling point? Justify your answer with suitable reason(s).

9. You are given a mixture of three white crystalline substances P, Q and R. P sublimes but Q and R are not influenced by heat. All the substances are soluble in water. In addition, R is also soluble in methanol. Outline a method of separation to obtain pure samples of P, Q and R.

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PAGE 3 — 2024 IJMBE CHEMISTRY I contd.

10. Make a sketch of the following and give the name of the law represented in each case.

• a) Variation of P with V for a fixed mass of gas at constant T

• b) Variation of PV with P for a fixed mass of gas at constant T

• c) Variation of V with T for a fixed mass of gas at constant P (P = pressure, V = volume, and T = temperature).

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SECTION B

11. a) Explain hybridisation of atomic orbitals. (2 marks)

b) Explain three important types of hybridisation involving valence electrons in s- and p-orbitals. (18 marks)

c) State Fajan’s rule and explain how it can be used to predict solubility of a compound. (5 marks)

12. a) (i) Explain solubility product of a sparingly soluble salt. (3 marks)

(ii) Give two limitations of the concept of solubility product. (2 marks)

(iii) State 4 applications of the concept of solubility product. (4 marks)

b) Write the Ksp expression and its unit for each of the following sparingly soluble salts: (6 marks)

(i) BaSO₄, (ii) Ni₃(PO₃)₂.

d) Given that the Ksp of BaCrO₄ is 1.80 x 10⁻¹⁴ mol²dm⁻⁶, calculate the concentration of Ba²⁺ in a saturated solution of BaCrO₄ in: (10 marks)

(i) pure water; (ii) presence of 1 x 10⁻³ mol dm⁻³ of K₂CrO₄.

13. a) What are transition metals? (2 marks)

b) State whether each of the following is a transition metal/ion or not. Justify your answers with appropriate reasons. (12 marks)

(i) titanium (ii) cobalt (iii) copper(I) ion

c) (i) Write the electronic configuration of Fe, Fe(II), and Fe(III) and show their electron-in-box notations. (6 marks)

(ii) In which of the oxidation states in c(i) would you expect the iron to be the most stable? Suggest suitable explanation for your answer. (3 marks)

d) Give the names of any two neutral ligands. (2 marks)

14. a) Explain what is meant by the following: (6 marks)

(i) Bond dissociation energy, (ii) Lattice dissociation energy, (iii) Atomisation energy.

b) Draw a complete labelled Born-Haber cycle for the formation of sodium chloride. (6 marks)

c) Calculate the lattice energy of sodium chloride from the following data. (6 marks)

• Standard heat of sublimation, ΔHs of metallic sodium = +108 kJmol⁻¹,

• Standard enthalpy of formation, ΔHf of sodium chloride = −411 kJmol⁻¹,

• Standard enthalpy of dissociation, ΔHd of chlorine = +242 kJmol⁻¹,

• Ionisation energy, I of sodium = +500 kJmol⁻¹,

• Electron affinity of chlorine atoms = −364 kJmol⁻¹

SOLUTIONS

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SECTION A

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Question 1 — True or False

a) Electronegativity of an element is inversely proportional to its electronegativity.

→ FALSE. Electronegativity is inversely proportional to atomic radius, not to itself. (The statement is self-contradictory/meaningless as written — likely meant “electropositivity.” Electronegativity IS inversely proportional to electropositivity. TRUE in that interpretation.)

b) Sigma bond can be formed by lateral overlap of two p-orbitals.

→ FALSE. Sigma bonds are formed by head-on (axial) overlap. Lateral overlap of two p-orbitals forms a pi (π) bond.

c) A reaction occurring in a vacuum flask is an example of an isolated system.

→ FALSE. A vacuum flask prevents heat exchange (adiabatic), making it a closed system — it still allows work exchange. A truly isolated system allows neither heat nor work nor matter exchange.

d) Metals have sonorous property.

→ TRUE. Metals produce a ringing sound when struck.

e) There are no intermolecular forces of attraction between real gas molecules.

→ FALSE. Real gas molecules DO experience intermolecular forces (van der Waals forces). It is ideal gases that are assumed to have no intermolecular forces.

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Question 2 — Mole Calculations

a) Moles of Br₂ in 15.98 g

Molar mass of Br₂ = 2 × 79.9 = 159.8 g/mol

$$n = \frac{mass}{molar\ mass} = \frac{15.98}{159.8} = \boxed{0.1\ mol}$$

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b) Moles of KOH in 1 dm³ of 3 mol dm⁻³ solution

$$n = C \times V = 3 \times 1 = \boxed{3\ moles}$$

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c) Number of atoms in 2 moles of cobalt

$$N = n \times N_A = 2 \times 6.023 \times 10^{23} = \boxed{1.2046 \times 10^{24}\ atoms}$$

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d) Number of atoms in 0.1 mole of silicon

$$N = 0.1 \times 6.023 \times 10^{23} = \boxed{6.023 \times 10^{22}\ atoms}$$

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e) Mass of 6.023 × 10²³ molecules of fluorine (F₂)

6.023 × 10²³ molecules = 1 mole of F₂

Molar mass of F₂ = 2 × 19 = 38 g/mol

$$\boxed{mass = 38\ g}$$

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Question 3 — Factors Favouring Bond Formation

a) Metallic Bond:

1. Low ionisation energy of the metal atoms (electrons are easily released)

2. Large number of delocalised valence electrons (more electrons available for electron sea)

3. Small atomic radius (ions are closely packed, stronger attraction to electron cloud)

b) Ionic Bond:

1. Large difference in electronegativity between the two elements (≥1.7)

2. Low ionisation energy of the metal (easy to lose electrons)

3. High electron affinity of the non-metal (readily gains electrons)

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Question 4 — Molar Mass of Y₂O₇

Given: mass = 35.5 g, volume = 5.60 dm³ at STP, Vₘ = 22.4 dm³/mol

a) Molar mass of Y₂O₇:

$$n = \frac{V}{V_m} = \frac{5.60}{22.4} = 0.25\ mol$$

$$M_{Y_2O_7} = \frac{mass}{n} = \frac{35.5}{0.25} = \boxed{142\ g/mol}$$

b) Atomic mass of Y:

$$M_{Y_2O_7} = 2Y + 7(16) = 142$$

$$2Y + 112 = 142$$

$$2Y = 30$$

$$\boxed{Y = 15}$$

(Y represents an element with atomic mass 15 — this is Nitrogen, N)

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Question 5 — Identifying Elements

Elements given: Na, Si, Mg, P, S, Ar, B, Cl

a) Forms a hydride with empirical formula XH₃:

XH₃ means the element has valency 3.

→ B (Boron) forms BH₃, or N — but N is not listed. B forms BH₃. ✓

(Also P forms PH₃ — phosphine. Both B and P qualify. Most likely answer: P (phosphorus) since PH₃ is more commonly cited)

b) Exists as a diatomic molecule at room temperature:

→ Cl (exists as Cl₂ at room temperature) ✓

c) Is monoatomic at room temperature:

→ Ar (noble gas, exists as single atoms) ✓

d) Is the most electropositive:

Electropositivity increases down a group and left across a period.

Among Na, Si, Mg, P, S, Ar, B, Cl — Na (Group I, Period 3) is most electropositive. ✓

e) Has semiconducting property:

→ Si (Silicon is the classic semiconductor) ✓

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Question 6 — Isotopes

a) What are isotopes?

Isotopes are atoms of the same element that have the same atomic number (same number of protons) but different mass numbers (different number of neutrons).

Example: ¹²C and ¹⁴C are isotopes of carbon.

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b) Why are isotopes chemically indistinguishable?

Chemical properties depend on the number and arrangement of electrons, which is determined by the atomic number (number of protons). Since isotopes of the same element have the same atomic number, they have the same number of electrons and the same electronic configuration. Therefore, they undergo the same chemical reactions and are chemically indistinguishable.

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c) Three physical properties that differentiate isotopes:

1. Mass/Density — isotopes have different mass numbers, so they differ in atomic mass and density

2. Rate of diffusion — lighter isotopes diffuse faster than heavier ones (Graham’s Law)

3. Radioactivity — some isotopes are radioactive (unstable nuclei) while others are stable; they also differ in nuclear binding energy

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Question 7 — s-Block Elements

Five characteristic properties of s-block elements:

1. They have their outermost (valence) electrons in the s-subshell (s¹ or s²)

2. They are all metals (except hydrogen, which is a non-metal)

3. They have low ionisation energies — they readily lose electrons to form cations

4. They are highly electropositive (most reactive metals)

5. They form ionic compounds with non-metals and their oxides/hydroxides are strongly basic (alkaline)

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Question 8 — Boiling Point

a) Normal boiling point of a substance:

The normal boiling point is the temperature at which the vapour pressure of a liquid equals the standard atmospheric pressure (101.325 kPa / 760 mmHg / 1 atm). At this temperature, bubbles of vapour form throughout the liquid.

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b) Can water boil above its boiling point?

Yes, it is possible. This phenomenon is called superheating.

Reasons:

• Water can be heated above 100°C without boiling if it is heated in a very smooth, clean container with no nucleation sites (no rough surfaces, dust, or dissolved gases)

• Under increased external pressure (e.g., in a pressure cooker), the boiling point rises above 100°C because the vapour pressure must overcome a higher external pressure before boiling occurs

• In both cases, the water temperature exceeds the normal boiling point before vaporisation begins

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Question 9 — Separation of P, Q and R

Given information:

• P sublimes; Q and R are unaffected by heat

• All three are soluble in water

• R is also soluble in methanol; Q is not soluble in methanol

Method of separation:

Step 1 — Separate P by sublimation:

Heat the mixture gently. P will sublime (convert directly from solid to vapour). Collect and cool the vapour; P condenses as pure solid on a cold surface above the mixture. P is obtained pure.

Step 2 — Separate R from Q using methanol:

Add methanol to the remaining mixture of Q and R. R dissolves in methanol; Q does not. Filter the mixture — Q is collected as residue (wash and dry). Evaporate the methanol filtrate to obtain pure R.

Summary:

| Substance | Method | Result |

|—|---|—|

| P | Sublimation | Pure P collected on cold surface |

| Q | Filtration from methanol | Pure Q as undissolved residue |

| R | Evaporation of methanol solution | Pure R recovered after evaporation |

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Question 10 — Gas Laws (Sketches + Laws)

a) Variation of P with V at constant T:

P

| *

| *

| *

| *

| *

|___________________V

Curve: Rectangular hyperbola (P decreases as V increases)

Law: Boyle’s Law — “At constant temperature, the pressure of a fixed mass of gas is inversely proportional to its volume.” (P ∝ 1/V)

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b) Variation of PV with P at constant T:

PV

|_ _ _ _ _ _ _ _ _ _

|

|

|___________________P

Curve: Horizontal straight line (PV = constant)

Law: Boyle’s Law — confirms PV = constant at fixed T; the flat line shows PV does not change with P for an ideal gas.

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c) Variation of V with T at constant P:

V

| /

| /

| /

| /

| /

|/___________________T (Kelvin)

Curve: Straight line through the origin (when T is in Kelvin)

Law: Charles’s Law — “At constant pressure, the volume of a fixed mass of gas is directly proportional to its absolute temperature.” (V ∝ T)

SECTION B

Question 11 — Hybridisation

a) Hybridisation of atomic orbitals:

Hybridisation is the mixing of atomic orbitals of similar energy within the same atom to form new orbitals of equal energy, shape, and orientation called hybrid orbitals. These hybrid orbitals are used in bond formation and give molecules their observed geometry.

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b) Three important types of hybridisation (s and p orbitals):

i. sp Hybridisation (Linear)

• Formation: One s orbital mixes with one p orbital → 2 sp hybrid orbitals

• Geometry: Linear (bond angle = 180°)

• Unhybridised orbitals: 2 pure p orbitals remain (form 2 π bonds)

• Example: BeCl₂, C₂H₂ (ethyne/acetylene)

• In acetylene (H–C≡C–H): each carbon is sp hybridised; triple bond = 1σ + 2π

ii. sp² Hybridisation (Trigonal Planar)

• Formation: One s orbital mixes with two p orbitals → 3 sp² hybrid orbitals

• Geometry: Trigonal planar (bond angle = 120°)

• Unhybridised orbitals: 1 pure p orbital remains (forms 1 π bond)

• Example: BF₃, C₂H₄ (ethene)

• In ethene (H₂C=CH₂): each carbon is sp² hybridised; double bond = 1σ + 1π

iii. sp³ Hybridisation (Tetrahedral)

• Formation: One s orbital mixes with three p orbitals → 4 sp³ hybrid orbitals

• Geometry: Tetrahedral (bond angle = 109.5°)

• Unhybridised orbitals: None

• Example: CH₄ (methane), NH₃, H₂O

• In methane: carbon forms 4 equal C–H sigma bonds at 109.5°

• In NH₃: one sp³ orbital holds a lone pair → pyramidal shape (107°)

• In H₂O: two sp³ orbitals hold lone pairs → bent shape (104.5°)

| Type | Orbitals mixed | Hybrid orbitals | Geometry | Bond angle | Example |

|------|---------------|-----------------|----------|------------|---------|

| sp | s + p | 2 | Linear | 180° | BeCl₂, C₂H₂ |

| sp² | s + 2p | 3 | Trigonal planar | 120° | BF₃, C₂H₄ |

| sp³ | s + 3p | 4 | Tetrahedral | 109.5° | CH₄, NH₃ |

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c) Fajan’s Rules and Solubility Prediction:

Fajan’s Rules state the conditions that favour covalent character in an ionic compound. A compound shows more covalent character when:

1. The cation is small and highly charged (high charge density → greater polarising power)

2. The anion is large and highly charged (easily polarisable — high polarisability)

3. The cation has a pseudo noble gas configuration (18 electrons) rather than a true noble gas configuration (8 electrons)

Predicting Solubility:

• Compounds with high ionic character tend to be soluble in polar solvents (like water) — “like dissolves like”

• Compounds with high covalent character (per Fajan’s rules) tend to be less soluble in water and more soluble in non-polar/organic solvents

• Example: AgCl is poorly soluble because Ag⁺ (small, pseudo noble gas config) strongly polarises Cl⁻, giving AgCl significant covalent character

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Question 12 — Solubility Product

a)(i) Solubility product (Ksp):

The solubility product is the product of the concentrations of the ions of a sparingly soluble electrolyte in a saturated solution at a given temperature, with each concentration raised to the power of its stoichiometric coefficient in the dissociation equation.

For a salt: MₓAᵧ ⇌ xMⁿ⁺ + yAᵐ⁻

$$K_{sp} = [M^{n+}]^x [A^{m-}]^y$$

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a)(ii) Two limitations of Ksp:

1. It is only applicable to sparingly soluble salts — it cannot be used for moderately or highly soluble electrolytes

2. It assumes ideal behaviour and ignores interionic interactions; it is inaccurate at high ion concentrations

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a)(iii) Four applications of Ksp:

1. Predicting precipitation — if the ionic product (IP) exceeds Ksp, a precipitate forms

2. Qualitative analysis — used to selectively precipitate ions for identification

3. Purification of salts — common ion effect drives precipitation to remove impurities

4. Understanding tooth decay — enamel (hydroxyapatite) dissolution is governed by Ksp principles

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b) Ksp expressions:

(i) BaSO₄:

$$BaSO_4 \rightleftharpoons Ba^{2+} + SO_4^{2-}$$

$$K_{sp} = [Ba^{2+}][SO_4^{2-}]$$

Unit: mol² dm⁻⁶

(ii) Ni₃(PO₃)₂:

$$Ni_3(PO_3)_2 \rightleftharpoons 3Ni^{2+} + 2PO_3^{2-}$$

$$K_{sp} = [Ni^{2+}]^3[PO_3^{2-}]^2$$

Unit: mol⁵ dm⁻¹⁵

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d) BaCrO₄ calculations — Ksp = 1.80 × 10⁻¹⁴ mol² dm⁻⁶

$$BaCrO_4 \rightleftharpoons Ba^{2+} + CrO_4^{2-}$$

(i) In pure water:

Let solubility = s mol dm⁻³

$$[Ba^{2+}] = s,\ [CrO_4^{2-}] = s$$

$$K_{sp} = s \times s = s^2$$

$$s^2 = 1.80 \times 10^{-14}$$

$$s = \sqrt{1.80 \times 10^{-14}} = 1.342 \times 10^{-7}\ mol\ dm^{-3}$$

$$\boxed{[Ba^{2+}] = 1.34 \times 10^{-7}\ mol\ dm^{-3}}$$

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(ii) In presence of 1 × 10⁻³ mol dm⁻³ K₂CrO₄:

K₂CrO₄ fully dissociates: [CrO₄²⁻] = 1 × 10⁻³ mol dm⁻³ (common ion effect)

Let [Ba²⁺] = x (very small compared to 10⁻³)

$$K_{sp} = [Ba^{2+}][CrO_4^{2-}]$$

$$1.80 \times 10^{-14} = x \times (1 \times 10^{-3} + x)$$

Since x << 10⁻³:

$$1.80 \times 10^{-14} \approx x \times 1 \times 10^{-3}$$

$$x = \frac{1.80 \times 10^{-14}}{1 \times 10^{-3}}$$

$$\boxed{[Ba^{2+}] = 1.80 \times 10^{-11}\ mol\ dm^{-3}}$$

The common ion (CrO₄²⁻) dramatically reduces Ba²⁺ solubility.

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Question 13 — Transition Metals

a) Transition metals:

Transition metals are elements that have incompletely filled d-orbitals either in the ground state or in one or more of their commonly occurring oxidation states (ions). They are found in the d-block of the periodic table (Groups 3–12).

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b) Transition metal/ion identification:

(i) Titanium (Ti) — Atomic number 22

Electronic configuration: [Ar] 3d² 4s²

Ti has incompletely filled 3d orbitals in its ground state.

→ YES, Ti is a transition metal ✓

(ii) Cobalt (Co) — Atomic number 27

Electronic configuration: [Ar] 3d⁷ 4s²

Co has incompletely filled 3d orbitals.

→ YES, Co is a transition metal ✓

(iii) Copper(I) ion, Cu⁺ — Atomic number 29

Ground state Cu: [Ar] 3d¹⁰ 4s¹

Cu⁺ loses the 4s¹ electron: [Ar] 3d¹⁰

Cu⁺ has a completely filled 3d orbital.

→ NO, Cu⁺ is NOT a transition metal ion by the strict definition (completely filled d-orbital)

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c)(i) Electronic configurations of Fe, Fe(II), Fe(III):

Fe (atomic number = 26):

$$[Ar]\ 3d^6\ 4s^2$$

Box notation: 4s: [↑↓] | 3d: [↑↓][↑][↑][↑][↑]

Fe²⁺ (loses 2 electrons from 4s first):

$$[Ar]\ 3d^6$$

Box notation: 4s: [ ] | 3d: [↑↓][↑][↑][↑][↑]

Fe³⁺ (loses 3 electrons — 2 from 4s, 1 from 3d):

$$[Ar]\ 3d^5$$

Box notation: 4s: [ ] | 3d: [↑][↑][↑][↑][↑]

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c)(ii) Most stable oxidation state:

Fe³⁺ (iron III) is the most stable.

Reason: Fe³⁺ has a half-filled 3d⁵ configuration, which is extra stable due to the exchange energy associated with having one electron in each of the five d-orbitals (all with parallel spins). Half-filled and fully-filled subshells have extra thermodynamic stability.

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d) Two neutral ligands:

1. Water (H₂O) — e.g., in [Cu(H₂O)₄]²⁺

2. Ammonia (NH₃) — e.g., in [Cu(NH₃)₄]²⁺

(Other valid answers: CO, NO, en (ethylenediamine))

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Question 14 — Energetics and Born-Haber Cycle

a) Definitions:

(i) Bond dissociation energy:

The energy required to break one mole of a specific covalent bond in a gaseous molecule, producing gaseous atoms/radicals. It is always endothermic (positive).

Example: H–H(g) → 2H(g); ΔH = +436 kJ/mol

(ii) Lattice dissociation energy:

The energy required to completely separate one mole of an ionic lattice into its constituent gaseous ions, at standard conditions. It is always endothermic (positive).

Example: NaCl(s) → Na⁺(g) + Cl⁻(g); ΔH = +787 kJ/mol

(iii) Atomisation energy:

The enthalpy change when one mole of gaseous atoms is formed from the element in its standard state. It is always endothermic (positive).

Example: Na(s) → Na(g); ΔH = +108 kJ/mol

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b) Born-Haber Cycle for NaCl:

Na⁺(g) + Cl⁻(g)

↑ ↑

IE (+500) EA (−364)

| |

Na⁺(g) + Cl(g) ←— ½ΔHdiss (+121) ←— Na⁺(g) + ½Cl₂(g)

↑

ΔHsub (+108)

|

Na(s) + ½Cl₂(g)

|

ΔHf (−411)

↓

NaCl(s) ←————————————————— Lattice Energy (U)

Full cycle (Hess’s Law path):

$$Na(s) \xrightarrow{+108} Na(g) \xrightarrow{+500} Na^+(g) + e^-$$

$$\frac{1}{2}Cl_2(g) \xrightarrow{+121} Cl(g) \xrightarrow{-364} Cl^-(g)$$

$$Na^+(g) + Cl^-(g) \xrightarrow{U} NaCl(s)$$

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c) Calculating lattice energy of NaCl:

Given:

• ΔHsub (Na) = +108 kJ/mol

• ΔHf (NaCl) = −411 kJ/mol

• ΔHd (Cl₂) = +242 kJ/mol → ½ΔHd = +121 kJ/mol

• IE (Na) = +500 kJ/mol

• EA (Cl) = −364 kJ/mol

Using Hess’s Law (Born-Haber cycle):

$$\Delta H_f = \Delta H_{sub} + IE + \frac{1}{2}\Delta H_d + EA + U_{lattice}$$

$$-411 = +108 + 500 + 121 + (-364) + U$$

$$-411 = 365 + U$$

$$U = -411 - 365$$

$$\boxed{U = -776\ kJ\ mol^{-1}}$$

The lattice energy of NaCl = −776 kJ/mol (lattice formation energy; exothermic)

(Lattice dissociation energy = +776 kJ/mol)