CHM 004: ORGANIC CHEMISTRY
7. (a) Differentiate between the following sets of terms:
(i) Thermoplastics and Thermosets
(ii) Natural and Synthetic polymers
(iii) Addition and condensation polymers
[3 Marks]
(b)
(i) List the characteristic reactions Alkyl Halides undergo.
(ii) Give ONE equation each for the reactions listed in (b)(i) above.
(iii) Write the equation to show the reaction of methyl Iodide with aqueous potassium hydroxide.
# SECTION B: COMPLETE SOLUTIONS
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## CHM 001: GENERAL CHEMISTRY
### Question 1
**(a) Definitions:**
**(i) Molality (m):** The number of moles of solute dissolved in 1 kilogram of solvent.
**Unit:** mol/kg (molal)
**(ii) Molarity (M):** The number of moles of solute dissolved in 1 litre (1 dm³) of solution.
**Unit:** mol/dm³ (or mol/L)
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**(b) Frequency and wavelength of photon emitted (n = 5 → n = 2):**
Using the Rydberg formula:
$$\frac{1}{\lambda} = R_H\left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right)$$
Wait — using energy:
$$\Delta E = R_H\left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right)$$
$$\Delta E = 2.18 \times 10^{-18}\left(\frac{1}{2^2} - \frac{1}{5^2}\right)$$
$$= 2.18 \times 10^{-18}\left(\frac{1}{4} - \frac{1}{25}\right)$$
$$= 2.18 \times 10^{-18}\left(\frac{25 - 4}{100}\right)$$
$$= 2.18 \times 10^{-18} \times \frac{21}{100}$$
$$= 2.18 \times 10^{-18} \times 0.21 = 4.578 \times 10^{-19} \text{ J}$$
**Frequency:**
$$\nu = \frac{\Delta E}{h} = \frac{4.578 \times 10^{-19}}{6.63 \times 10^{-34}} = 6.906 \times 10^{14} \text{ Hz}$$
**Wavelength:**
$$\lambda = \frac{c}{\nu} = \frac{3.00 \times 10^8}{6.906 \times 10^{14}} = 4.344 \times 10^{-7} \text{ m} = 434.4 \text{ nm}$$
*(This is the violet line of the Balmer series.)*
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**(c) Reaction predictions:**
**(i) Na₂SO₄(aq) + Sr(NO₃)₂(aq):**
Sr²⁺ + SO₄²⁻ → SrSO₄ (insoluble — reaction OCCURS)
- **Molecular:** Na₂SO₄(aq) + Sr(NO₃)₂(aq) → SrSO₄(s) + 2NaNO₃(aq)
- **Total ionic:** 2Na⁺(aq) + SO₄²⁻(aq) + Sr²⁺(aq) + 2NO₃⁻(aq) → SrSO₄(s) + 2Na⁺(aq) + 2NO₃⁻(aq)
- **Net ionic:** Sr²⁺(aq) + SO₄²⁻(aq) → SrSO₄(s)
**(ii) NH₄ClO₄(aq) + NaBr(aq):**
All possible products (NaClO₄ and NH₄Br) are soluble. **No reaction occurs.**
- **Molecular:** NH₄ClO₄(aq) + NaBr(aq) → No reaction
- **Net ionic:** No net ionic equation (no precipitate, gas, or weak electrolyte formed)
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**(d) Oxidation number of Zinc:**
**(i) Zn(NH₃)₄(SO₄):**
NH₃ = 0, SO₄²⁻ = –2
∴ Zn + (–2) = 0 → **Zn = +2**
**(ii) [Zn(NH₃)₄Cl]²⁺:**
NH₃ = 0, Cl = –1, overall charge = +2
∴ Zn + 0 + (–1) = +2 → **Zn = +3**
---
### Question 2
**(a) Limiting reagent and yield:**
Equation: 2FeBr₂(aq) + 3Na₂S(aq) → Fe₂S₃(s) + 6NaBr(aq)
Molar masses:
- FeBr₂ = 55.8 + 2(79.9) = **215.6 g/mol**
- Na₂S = 2(23) + 32 = **78 g/mol**
- Fe₂S₃ = 2(55.8) + 3(32) = **207.6 g/mol**
Moles of FeBr₂ = 7.0/215.6 = **0.03247 mol**
Moles of Na₂S = 3.2/78 = **0.04103 mol**
From stoichiometry: 2 mol FeBr₂ requires 3 mol Na₂S
∴ 0.03247 mol FeBr₂ requires = (3/2) × 0.03247 = **0.04871 mol Na₂S**
Available Na₂S = 0.04103 mol < 0.04871 mol required
**(i) Na₂S is the limiting reagent.**
**(ii) Moles of Fe₂S₃ formed:**
From stoichiometry: 3 mol Na₂S → 1 mol Fe₂S₃
∴ 0.04103 mol Na₂S → 0.04103/3 = 0.01368 mol Fe₂S₃
Mass of Fe₂S₃ = 0.01368 × 207.6 = **2.84 g**
**(iii) Percentage yield:**
$$\% \text{ yield} = \frac{\text{actual yield}}{\text{theoretical yield}} \times 100 = \frac{1.92}{2.84} \times 100 = \mathbf{67.6\%}$$
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**(b) Precipitation Reaction:**
A precipitation reaction is a chemical reaction in which two soluble ionic solutions are mixed and an insoluble solid product (precipitate) is formed that separates from the solution.
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**(c)(i) Accuracy vs Precision:**
- **Accuracy** refers to how close a measured value is to the true or accepted value.
- **Precision** refers to how close repeated measurements are to each other, regardless of whether they are close to the true value.
**(c)(ii) Balance the redox reaction in basic medium:**
$$H_2O_2 + MnO_4^- \rightarrow O_2 + MnO_2$$
**Step 1 — Separate half-reactions:**
*Oxidation:* H₂O₂ → O₂
*Reduction:* MnO₄⁻ → MnO₂
**Oxidation half-reaction:**
H₂O₂ → O₂ + 2H⁺ + 2e⁻
In basic medium, add 2OH⁻ to both sides:
H₂O₂ + 2OH⁻ → O₂ + 2H₂O + 2e⁻
**Reduction half-reaction:**
MnO₄⁻ → MnO₂
MnO₄⁻ + 2H₂O + 3e⁻ → MnO₂ + 4OH⁻
**Balance electrons** (LCM of 2 and 3 = 6):
Multiply oxidation × 3: 3H₂O₂ + 6OH⁻ → 3O₂ + 6H₂O + 6e⁻
Multiply reduction × 2: 2MnO₄⁻ + 4H₂O + 6e⁻ → 2MnO₂ + 8OH⁻
**Add and simplify:**
3H₂O₂ + 6OH⁻ + 2MnO₄⁻ + 4H₂O → 3O₂ + 6H₂O + 2MnO₂ + 8OH⁻
Cancel: 4H₂O from both sides, 6OH⁻ from both sides:
$$\boxed{3H_2O_2 + 2MnO_4^- \rightarrow 3O_2 + 2MnO_2 + 2OH^- + 2H_2O}$$
---
## CHM 002: PHYSICAL CHEMISTRY
### Question 3
**(a)(i) Half-life of a radioactive element:**
The half-life is the time required for half of the atoms in a given sample of a radioactive element to decay (disintegrate).
**(a)(ii) THREE parameters for nuclear stability:**
1. Neutron-to-proton ratio (n/p ratio)
2. Binding energy per nucleon
3. Magic numbers (specific numbers of protons or neutrons: 2, 8, 20, 28, 50, 82, 126)
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**(b) Half-life of cobalt-60 = 6.0 years**
**(i) Decay constant (λ):**
Convert half-life to seconds:
t½ = 6.0 × 365 × 24 × 3600 = 6.0 × 3.156 × 10⁷ = **1.894 × 10⁸ s**
$$\lambda = \frac{0.693}{t_{1/2}} = \frac{0.693}{1.894 \times 10^8} = \mathbf{3.659 \times 10^{-9} \text{ s}^{-1}}$$
**(ii) Activity of 1.0 mg sample:**
Molar mass of Co-60 = 60 g/mol
Mass = 1.0 mg = 1.0 × 10⁻³ g
$$N = \frac{m}{M} \times N_A = \frac{1.0 \times 10^{-3}}{60} \times 6.022 \times 10^{23} = 1.004 \times 10^{19} \text{ atoms}$$
$$A = \lambda N = 3.659 \times 10^{-9} \times 1.004 \times 10^{19} = \mathbf{3.67 \times 10^{10} \text{ disintegrations/s (Bq)}}$$
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**(c) Thermodynamic definitions:**
**(i) Entropy (S):** A thermodynamic state function that measures the degree of disorder or randomness in a system. Unit: J/mol·K
**(ii) Gibbs' Free Energy (G):** The thermodynamic potential that measures the maximum reversible work obtainable from a system at constant temperature and pressure. Defined as: G = H – TS. It determines spontaneity of a reaction (ΔG < 0 = spontaneous).
**(iii) Enthalpy (H):** The total heat content of a system at constant pressure. Defined as H = U + PV, where U = internal energy. ΔH < 0 = exothermic; ΔH > 0 = endothermic.
---
### Question 4
**(a) Integrated rate laws:**
**(i) Zeroth-order reaction:**
$$[A] = [A]_0 - kt$$
Unit of k: **mol dm⁻³ s⁻¹**
**(ii) First-order reaction:**
$$\ln[A] = \ln[A]_0 - kt \quad \text{or} \quad [A] = [A]_0 e^{-kt}$$
Unit of k: **s⁻¹**
---
**(b) Bond energy calculation for C–Cl bond:**
Reaction: C₂H₄ + Cl₂ → C₂H₄Cl₂, ΔH = –129 kJ/mol
**Bonds broken (reactants):**
- C₂H₄: one C=C (+612) + four C–H (+415 × 4 = +1660)
- Cl₂: one Cl–Cl (+243)
- Total bonds broken = 612 + 1660 + 243 = **+2515 kJ**
**Bonds formed (products):**
C₂H₄Cl₂: one C–C (+330) + four C–H (+415 × 4 = +1660) + two C–Cl (2x)
- Total bonds formed = –[330 + 1660 + 2x] = –(1990 + 2x)
**Using:** ΔH = Bonds broken – Bonds formed
$$-129 = 2515 - (1990 + 2x)$$
$$-129 = 2515 - 1990 - 2x$$
$$-129 = 525 - 2x$$
$$2x = 525 + 129 = 654$$
$$x = \frac{654}{2} = \mathbf{+327 \text{ kJmol}^{-1}}$$
**The bond energy of C–Cl = 327 kJmol⁻¹**
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**(c) Activation energy calculation:**
Given: k₁ = 4.52 dm³mol⁻¹s⁻¹ at T₁ = 298 K; k₂ = 12.94 dm³mol⁻¹s⁻¹ at T₂ = 400 K
Using the Arrhenius equation:
$$\ln\frac{k_2}{k_1} = \frac{E_a}{R}\left(\frac{1}{T_1} - \frac{1}{T_2}\right)$$
$$\ln\frac{12.94}{4.52} = \frac{E_a}{8.314}\left(\frac{1}{298} - \frac{1}{400}\right)$$
$$\ln(2.863) = \frac{E_a}{8.314}\left(0.003356 - 0.002500\right)$$
$$1.052 = \frac{E_a}{8.314} \times 8.557 \times 10^{-4}$$
$$E_a = \frac{1.052 \times 8.314}{8.557 \times 10^{-4}} = \frac{8.746}{8.557 \times 10^{-4}}$$
$$\boxed{E_a = 1.022 \times 10^4 \text{ J/mol} = 10.22 \text{ kJ/mol}}$$
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**(d) Positive deviation from Raoult's law:**
Positive deviation occurs when the intermolecular forces between solute–solvent molecules are **weaker** than the forces between pure solvent–solvent or solute–solute molecules. This causes molecules to escape more easily, raising vapour pressure above the ideal value.
Example: Ethanol–water mixture.
---
## CHM 003: INORGANIC CHEMISTRY
### Question 5
**(a) Reasons:**
**(i) Electronegativity increases left to right:**
Moving across a period, the nuclear charge increases while the number of electron shells remains the same. This increases the effective nuclear charge, pulling bonding electrons closer to the nucleus and increasing electronegativity.
**(ii) Ionisation enthalpy decreases down a group:**
Going down a group, atomic size increases and additional electron shells are added, increasing shielding effect. The outermost electrons are further from the nucleus and more shielded, so less energy is needed to remove them.
**(iii) Helium (1s²) is placed in group 18 (p-block):**
Although He has no p-electrons, it is placed in group 18 because it has a completely filled outermost shell (like other noble gases) and exhibits similar chemical inertness. Its properties (zero reactivity) align with group 18 elements.
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**(b) Accounts:**
**(i) Second ionisation energy of alkali metals > alkaline earth metals:**
After removing one electron, alkali metals attain the stable noble gas configuration (fully filled shells), making it extremely difficult to remove a second electron. Alkaline earth metals after losing one electron still have one electron in the outer shell, which is easier to remove.
**(ii) Alkali metal solutions in liquid ammonia are conducting:**
When alkali metals dissolve in liquid ammonia, they release free electrons (solvated electrons) and metal cations:
M → M⁺(am) + e⁻(am)
These mobile solvated electrons act as charge carriers, making the solution electrically conducting.
**(iii) HF is the weakest halogen acid despite F being most electronegative:**
The H–F bond has the highest bond dissociation energy among hydrogen halides due to the small size of fluorine and strong bond formed. This makes ionisation of HF in water very difficult, giving it a relatively high pKa (≈3.2) and making it a weak acid compared to HCl, HBr, and HI.
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**(c) Reactions of Calcium:**
**(i) With Nitrogen:**
$$3Ca + N_2 \rightarrow Ca_3N_2$$
(Calcium nitride is formed)
**(ii) With dilute H₂SO₄:**
$$Ca + H_2SO_4(dil) \rightarrow CaSO_4 + H_2\uparrow$$
Note: The reaction soon stops because CaSO₄ is sparingly soluble and forms a protective layer on the calcium surface.
---
**(b) Order of increasing electrical conductivity of complexes:**
Conductivity depends on the number of ions produced in solution:
| Complex | Ions produced |
|---|---|
| [Co(NH₃)₃Cl₃] | 0 ions (non-electrolyte) |
| [Co(NH₃)₄Cl]Cl₂ | 3 ions (1:2 electrolyte) |
| [Co(NH₃)₅Cl]Cl₂ | 3 ions (1:2 electrolyte) |
| [Co(NH₃)₆]Cl₃ | 4 ions (1:3 electrolyte) |
**Increasing conductivity:**
$$[Co(NH_3)_3Cl_3] < [Co(NH_3)_4Cl]Cl_2 \approx [Co(NH_3)_5Cl]Cl_2 < [Co(NH_3)_6]Cl_3$$
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**(c) Hydrogen compounds with W(9), X(11), Y(12), Z(17):**
- W = F (at. no. 9), X = Na (at. no. 11), Y = Mg (at. no. 12), Z = Cl (at. no. 17)
**(i) Balanced equations:**
- H₂ + F₂ → **2HF**
- 2Na + H₂ → **2NaH**
- Mg + H₂ → **MgH₂**
- H₂ + Cl₂ → **2HCl**
**(ii) Type of hydride:**
- HF — **Covalent (molecular) hydride**
- NaH — **Ionic (saline) hydride**
- MgH₂ — **Ionic hydride** (with some covalent character)
- HCl — **Covalent (molecular) hydride**
---
**(d) Formulas of Fe³⁺ complexes:**
**(i)** One OH⁻, two NH₃, three Cl⁻:
Total ligands = 1 + 2 + 3 = 6 (octahedral)
$$[Fe(OH)(NH_3)_2Cl_3]^{-1} \text{ or } [Fe(OH)(NH_3)_2Cl_3]$$
**(ii)** Three CN⁻, three NH₃:
$$[Fe(CN)_3(NH_3)_3]$$
---
## CHM 004: ORGANIC CHEMISTRY
### Question 7
**(a) Differentiation:**
**(i) Thermoplastics vs Thermosets:**
| Thermoplastics | Thermosets |
|---|---|
| Soften on heating and can be remoulded | Permanently harden on heating; cannot be remoulded |
| Linear or branched polymer chains | Highly cross-linked 3D network |
| Example: Polyethylene, PVC | Example: Bakelite, epoxy resin |
**(ii) Natural vs Synthetic polymers:**
| Natural Polymers | Synthetic Polymers |
|---|---|
| Occur naturally in living organisms | Manufactured by humans in laboratories/industries |
| Example: Starch, cellulose, proteins, rubber | Example: Nylon, polyethylene, PVC |
**(iii) Addition vs Condensation polymers:**
| Addition Polymers | Condensation Polymers |
|---|---|
| Formed by successive addition of monomer units with no by-product | Formed by reaction between monomers with elimination of small molecules (H₂O, HCl) |
| Monomers must contain C=C double bonds | Monomers must have two functional groups |
| Example: Polyethylene, polystyrene | Example: Nylon, polyester |
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**(b) Alkyl Halides:**
**(i) Characteristic reactions:**
1. Nucleophilic Substitution (SN1 and SN2)
2. Elimination reaction (dehydrohalogenation)
3. Formation of Grignard reagents (reaction with Mg)
4. Reduction reaction
**(ii) ONE equation each:**
1. **Nucleophilic Substitution:**
$$CH_3I + NaOH \rightarrow CH_3OH + NaI$$
2. **Elimination:**
$$CH_3CH_2Br \xrightarrow{KOH/alc} CH_2=CH_2 + HBr$$
3. **Grignard reagent:**
$$CH_3Br + Mg \xrightarrow{dry\ ether} CH_3MgBr$$
4. **Reduction:**
$$CH_3Cl + 2[H] \xrightarrow{LiAlH_4} CH_4 + HCl$$
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**(iii) Reaction of methyl iodide (CH₃I) with aqueous KOH:**
This is a nucleophilic substitution (SN2) reaction:
$$CH_3I + KOH(aq) \rightarrow CH_3OH + KI$$
The hydroxide ion (OH⁻) acts as the nucleophile and displaces the iodide ion (I⁻), producing **methanol (CH₃OH)** and potassium iodide (KI).