Grade 12 — Term 2
3 study notes · 18 practice questions · Aligned with NIE syllabus
Counted in the Idasara Academy app, which you open with a free account. The free notes, revision, practice and library pages for each lesson are linked below.
This term in the Idasara Academy app has 3 lessons (NIE Units 4, 5, 6). For each lesson, this page gives you the points you must know and one exam-style question with its answer. None of it needs an account.
Lessons in This Term
Gaseous State of Matter
What you must know
- Gas collected over water: P_dry_gas = P_total - P_H2O(vapor pressure at that temperature).
- van der Waals equation: (P + an²/V²)(V - nb) = nRT, where 'a' corrects for intermolecular attractions and 'b' corrects for finite molecular volume.
- Under what conditions does a real gas come closest to ideal behaviour?: Real gases approach ideal gas behavior under conditions of low pressure and high temperature.
Try one question
A rigid closed vessel contains 4.0 g of methane (CH₄, molar mass = 16.0 g mol⁻¹) and 28.0 g of nitrogen (N₂, molar mass = 28.0 g mol⁻¹) at 300 K. If the total pressure in the vessel is 1.50 atm, what is the partial pressure of methane?
- A. 0.25 atm
- B. 0.30 atm
- C. 0.75 atm
- D. 1.20 atm
Show the answer
Answer: B. 0.30 atm. According to Dalton's Law of Partial Pressures, P_i = x_i × P_total. Moles of CH₄ = 4.0 / 16.0 = 0.25 mol. Moles of N₂ = 28.0 / 28.0 = 1.00 mol. Total moles = 0.25 + 1.00 = 1.25 mol. Mole fraction of CH₄ = 0.25 / 1.25 = 0.20. Therefore, P(CH₄) = 0.20 × 1.50 atm = 0.30 atm.
A model question written by Idasara on the exam pattern. It is not from a past paper.
Energetics
What you must know
- Gibbs free energy equation: ΔG = ΔH - TΔS. A process is spontaneous at constant T and P when ΔG < 0.
- When ΔH < 0 and ΔS > 0: reaction is spontaneous at all temperatures (ΔG is always negative).
- Calorimetry formula: q = mcΔT (where m is solution mass in g, c is specific heat capacity 4.18 J g⁻¹ K⁻¹ for aqueous solutions).
Try one question
Given the standard enthalpies of formation (ΔH_f°): C₂H₆(g) = -84.7 kJ mol⁻¹, CO₂(g) = -393.5 kJ mol⁻¹, and H₂O(l) = -285.8 kJ mol⁻¹, calculate the standard enthalpy of combustion (ΔH_c°) of ethane gas.
- A. -3119.4 kJ mol⁻¹
- B. -1559.7 kJ mol⁻¹
- C. -1475.0 kJ mol⁻¹
- D. -1644.4 kJ mol⁻¹
Show the answer
Answer: B. -1559.7 kJ mol⁻¹. The combustion equation is C₂H₆(g) + 3.5O₂(g) → 2CO₂(g) + 3H₂O(l). Using Hess's law: ΔH_c° = [2 × ΔH_f°(CO₂) + 3 × ΔH_f°(H₂O)] - [ΔH_f°(C₂H₆) + 3.5 × ΔH_f°(O₂)]. Since ΔH_f°(O₂) = 0: ΔH_c° = [2(-393.5) + 3(-285.8)] - [-84.7] = [-787.0 - 857.4] + 84.7 = -1644.4 + 84.7 = -1559.7 kJ mol⁻¹.
A model question written by Idasara on the exam pattern. It is not from a past paper.
Chemistry of s, p, and d Block Elements
What you must know
- Coordination complexes: ligands donate lone pairs into vacant metal hybrid orbitals (e.g. [Fe(CN)₆]⁴⁻, coordination number 6, octahedral).
- Origin of complex color: d-orbital splitting by ligands results in d-d electronic transitions absorbing specific visible wavelengths.
- Disproportionation reaction: element is simultaneously oxidized and reduced (e.g. Cl₂ + 2OH⁻(cold dilute) → Cl⁻ + ClO⁻ + H₂O).
Try one question
Which statement correctly explains why aqueous solutions of [Ti(H₂O)₆]³⁺ are violet in color, whereas aqueous solutions of [Sc(H₂O)₆]³⁺ are colorless?
- A. Sc³⁺ is paramagnetic, while Ti³⁺ is diamagnetic in aqueous solution
- B. Ti³⁺ has a 3d¹ configuration allowing d-d electronic transitions in visible light, while Sc³⁺ has a 3d⁰ configuration with no d-electrons
- C. Sc³⁺ undergoes charge transfer transitions in the infrared region, whereas Ti³⁺ does not
- D. Ti³⁺ forms a tetrahedral complex, whereas Sc³⁺ forms an octahedral complex
Show the answer
Answer: B. Ti³⁺ has a 3d¹ configuration allowing d-d electronic transitions in visible light, while Sc³⁺ has a 3d⁰ configuration with no d-electrons. Titanium (atomic number 22) has ground state configuration [Ar] 3d² 4s². For Ti³⁺, the configuration is [Ar] 3d¹. In an octahedral crystal field, the d-orbitals split into t₂g and e_g levels. Absorption of visible light promotes the single 3d electron from t₂g to e_g (d-d transition), resulting in a violet transmitted color. Scandium (atomic number 21) has configuration [Ar] 3d¹ 4s²; Sc³⁺ is [Ar] 3d⁰, possessing no d-electrons, so d-d transitions are impossible and the complex is colorless.
A model question written by Idasara on the exam pattern. It is not from a past paper.
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