IB DP Chemistry Structure 1.5 Ideal gases IB Style Question Bank SL Paper 2

Detailed Solution:

Step 1: Convert temperature
T = 85 °C = 85 + 273 = 358 K

Step 2: Use ideal gas law
PV = nRT, where n = m/M. Rearrange: M = mRT / PV
– P = 101 kPa = 101 × 10³ Pa
– V = 400 cm³ = 400 × 10⁻⁶ m³ = 0.0004 m³
– m = 1.17 g
– R = 8.31 J mol⁻¹ K⁻¹ (section 2)
– T = 358 K

Step 3: Calculate
M = (1.17 × 8.31 × 358) / (101 × 10³ × 0.0004) = 3479.886 / 40.4 ≈ 86.2 g mol⁻¹

Conclusion: The molar mass is 86.2 g mol⁻¹.

Detailed Solution:

Step 1: Empirical formula mass
C₃H₇: (3 × 12.01) + (7 × 1.01) = 36.03 + 7.07 = 43.1 g mol⁻¹

Step 2: Ratio
Molar mass = 86.2 g mol⁻¹ (from a(i)). Ratio = 86.2 / 43.1 ≈ 2

Step 3: Molecular formula
(C₃H₇) × 2 = C₆H₁₄

Conclusion: The molecular formula is C₆H₁₄.

Detailed Solution:

Step 1: Identify isomers
C₅H₁₂ isomers: n-pentane, isopentane (2-methylbutane), neopentane (2,2-dimethylpropane).

Step 2: Compare boiling points
Neopentane (C(CH₃)₄) is the most branched, with the smallest surface area.

Step 3: Explain
Smaller surface area reduces London dispersion forces, lowering the boiling point due to less intermolecular attraction.

Conclusion: Neopentane (C(CH₃)₄) has the lowest boiling point due to weakest London forces.

Detailed Solution:

Step 1: Ethanal production
Distillation removes ethanal immediately, preventing further oxidation.

Step 2: Ethanoic acid production
Reflux allows prolonged reaction with excess oxidant, fully oxidizing ethanol to ethanoic acid.

Conclusion: Distillation for ethanal, reflux for ethanoic acid.

Detailed Solution:

Step 1: Ethanol (C₂H₅OH)
For C₁: Bonds to 1 O (+1), 3 H (–1 each), 1 C (0) → +1 – 3 + 0 = –2.

Step 2: Ethanal (CH₃CHO)
For C₁ (CHO): Bonds to 1 O (+2, double bond), 1 H (–1), 1 C (0) → +2 – 1 + 0 = –1.

Conclusion: Ethanol: –2, Ethanal: –1.

Detailed Solution:

Step 1: Write oxidation
Ethanol (CH₃CH₂OH) oxidizes to ethanal (CH₃CHO), losing 2 H atoms.

Step 2: Balance charge
Left: neutral. Right: CH₃CHO (neutral) + 2H⁺ (+2) + 2e⁻ (–2) = 0.

Conclusion: CH₃CH₂OH → CH₃CHO + 2H⁺ + 2e⁻.

Detailed Solution:

Step 1: Balance electrons
Oxidation: CH₃CH₂OH → CH₃CHO + 2H⁺ + 2e⁻ (×3 to match 6e⁻)
Reduction: Cr₂O₇²⁻ + 14H⁺ + 6e⁻ → 2Cr³⁺ + 7H₂O

Step 2: Combine
3CH₃CH₂OH + Cr₂O₇²⁻ + 14H⁺ + 6e⁻ → 3CH₃CHO + 6H⁺ + 6e⁻ + 2Cr³⁺ + 7H₂O

Step 3: Simplify
Net H⁺: 14 – 6 = 8H⁺. Result: 3CH₃CH₂OH + Cr₂O₇²⁻ + 8H⁺ → 3CH₃CHO + 2Cr³⁺ + 7H₂O.

Conclusion: The redox equation is balanced as shown.

Detailed Solution:

Step 1: Characteristic 1
At equilibrium, the forward and reverse reaction rates are equal, so no net change occurs.

Step 2: Characteristic 2
Concentrations of all species remain constant over time in a closed system.

Conclusion: These define a dynamic equilibrium.

Detailed Solution:

Step 1: Alternative pathway
A catalyst offers a different reaction mechanism with lower energy barriers.

Step 2: Activation energy
Lower Eₐ means more molecules have sufficient energy to react, increasing rate.

Conclusion: Catalysts accelerate reactions by lowering Eₐ.

Detailed Solution:

Step 1: No shift
A catalyst does not alter the equilibrium position.

Step 2: Rate effect
It lowers activation energy for both forward and reverse reactions equally, speeding up attainment of equilibrium.

Conclusion: No change in equilibrium position.

Detailed Solution:

Step 1: Exothermic reaction
Forward reaction releases heat, so increasing temperature favors the reverse reaction (Le Chatelier’s principle).

Step 2: Effect on Kₓ
Kₓ = [products]/[reactants]. Reverse shift reduces product concentrations, decreasing Kₓ.

Conclusion: Kₓ decreases with increasing temperature.