Stoichiometry
Calculate the number of moles of NaCl in a 117 g sample (Mᵣ = 58.5).
n = m/Mᵣ = 117 g ÷ 58.5 g mol⁻¹
n = 2.0 mol
Give the condensed electron configuration for sulfur (Z = 16).
[Ne] 3s² 3p⁴
According to VSEPR, give the molecular geometry of BI₃.
Trigonal Planar
According to collision theory, list the two simultaneous conditions a molecular collision must satisfy for a reaction to occur.
The colliding particles must possess energy ≥ activation energy and have the correct orientation.
The sign of ΔS° for dissolving ammonium nitrate in water at 298 K.
Positive
State the volume, in dm³, occupied by 1.50 mol of an ideal gas at STP (273 K, 1 × 10⁵ Pa).
Use molar volume at STP: 22.7 dm³ mol⁻¹ (IB data booklet value for 1 × 10⁵ Pa).
V = n × Vₘ = 1.50 mol × 22.7 dm³ mol⁻¹ = 34 dm³ (2 sig. fig.).
Magnesium occurs as ²⁴Mg (78.99 %), ²⁵Mg (10.00 %) and ²⁶Mg (11.01 %). Calculate its relative atomic mass to two decimal places.
Aᵣ = (24 × 0.7899) + (25 × 0.1000) + (26 × 0.1101) = 24.32
...
In the reaction rate = k[A]², what happens to the rate if the concentration of A is halved?
Rate falls to one-quarter
Direction in which equilibrium shifts when pressure is increased for N₂O₄ ⇌ 2 NO₂.
Left (toward N₂O₄)
Combusting an unknown hydrocarbon produces 4.40 g CO₂ and 1.80 g H₂O. Determine its empirical formula.
moles C = 4.40 g ÷ 44.0 g mol⁻¹ = 0.100 mol
moles H = (1.80 g ÷ 18.0 g mol⁻¹) × 2 = 0.200 mol
Divide by the smallest: C 1 H 2 → empirical formula CH₂
Name the Period-2 element whose first-ionization energy is lower than that of the element immediately before it, and state the underlying reason in one phrase.
Oxygen; Spin Pair Repulsion in the paired 2p orbital
Identify the hybridization and approximate bond angle around the carbon atom in HCN.
sp, 180 °
The overall reaction is NO₂ + F₂ → 2 NO₂F, and its experimental rate law is rate = k [NO₂][F₂].
A proposed mechanism is:
(1) NO₂ + F₂ → NO₂F + F
(2) F + NO₂ → NO₂F
Which elementary step is the rate-determining step?
Step (1) NO₂ + F₂ → NO₂F + F
Calculate ΔH° for the reaction ½ N₂ + ½ O₂ → NO, given bond enthalpies
N≡N = 945, O=O = 498, N=O = 631.
Bonds broken: ½ N≡N + ½ O=O
ΔH₍break₎ = 0.5(945) + 0.5(498) = 721 kJ
Bonds formed: 1 N=O → ΔH₍form₎ = 631 kJ
ΔH° = Σ broken – Σ formed = 721 – 631 = +91 kJ mol⁻¹
25.0 cm³ of 0.200 mol dm⁻³ H₂SO₄ neutralises NaOH. Calculate the volume (cm³) of 0.150 mol dm⁻³ NaOH required. (H₂SO₄ + 2 NaOH → Na₂SO₄ + 2 H₂O)
H₂SO₄ + 2 NaOH → Na₂SO₄ + 2 H₂O
n(H₂SO₄) = 0.0250 dm³ × 0.200 mol dm⁻³ = 0.00500 mol
n(NaOH) required = 2 × 0.00500 = 0.0100 mol
V(NaOH) = n ÷ c = 0.0100 mol ÷ 0.150 mol dm⁻³ = 0.0667 dm³ = 66.7 cm³
The first four successive ionization energies of a period three element are 578, 1817, 2745, and 11 578 kJ mol⁻¹. Identify the element and state what the large jump after the third value reveals about its electron configuration.
Al; the huge increase after the third IE shows that three outer-shell electrons have been removed and the next electron comes from a lower (core) shell.
Explain why the C—O bond in methanoate ion (HCOO⁻) is shorter than the C—O bond in methanol (CH₃OH).
Delocalized π resonance gives a bond order of 1 ½, increasing bond strength and shortening the bond
A straight-line Arrhenius plot of ln k versus 1/T (K⁻¹) has a gradient of –5.40 × 10³ K. Using the Arrhenius equation determine the activation energy Eₐ in kJ mol⁻¹ (two significant figures).
Slope m = –Eₐ/R → Eₐ = –m R
Eₐ = (5.40 × 10³ K)(8.31 J mol⁻¹ K⁻¹) = 4.49 × 10⁴ J mol⁻¹
= 45 kJ mol⁻¹ (2 sig. fig.)
For a reaction where ΔH = +35 kJ mol⁻¹ and ΔS = +150 J mol⁻¹ K⁻¹, the minimum temperature (K) at which it becomes spontaneous (two sig. fig.).
Convert ΔS to kJ: 150 J K⁻¹ = 0.150 kJ K⁻¹
T = ΔH/ΔS = 35 kJ ÷ 0.150 kJ K⁻¹ = 233 K → 2.3 × 10² K (2 sig. fig.)
2 Al + 3 Cl₂ → 2 AlCl₃. Starting with 15.0 g Al and 40.0 g Cl₂, 35.0 g AlCl₃ is isolated. Find (i) the limiting reagent, (ii) the theoretical AlCl₃ yield (g), and (iii) the % yield.
n(Al) = 15.0 g ÷ 27.0 g mol⁻¹ = 0.556 mol
n(Cl₂) = 40.0 g ÷ 70.9 g mol⁻¹ = 0.564 mol
Stoichiometric ratios need: 0.556 mol Al needs 0.834 mol Cl₂ → Cl₂ limits.
Theoretical n(AlCl₃) = (2/3) × n(Cl₂) = 0.376 mol
m(theor.) = 0.376 mol × 133 g mol⁻¹ ≈ 50.1 g
% yield = (35.0 g ÷ 50.1 g) × 100 = 69.8 %
The frequency that corresponds to the ionization limit of a ground-state H atom is 3.29 × 10¹⁵ Hz. Using E = h*f and Nₐ = 6.02 × 10²³ mol⁻¹, calculate the ionization energy of hydrogen in kJ mol⁻¹ (three significant figures).
E (per atom) = h f = (6.626 × 10⁻³⁴ J s)(3.29 × 10¹⁵ s⁻¹) = 2.18 × 10⁻¹⁸ J
E (per mol) = E × Nₐ = 2.18 × 10⁻¹⁸ J × 6.02 × 10²³ mol⁻¹
= 1.31 × 10⁶ J mol⁻¹ = 1.31 × 10³ kJ mol⁻¹
Rank these compounds from highest to lowest lattice enthalpy, and justify briefly: MgO, KBr, and CaF₂.
MgO > CaF₂ > KBr; higher charges and smaller ionic radii raise electrostatic attraction, so 2+ /2− ions in MgO give the greatest lattice energy, followed by CaF₂ (2+/1−, small F⁻), then KBr (1+/1−, larger ions)
Experiments on A + B → products give these initial-rate data:
1. [A]=0.100 M, [B]=0.100 M, rate=2.0×10⁻⁴ M s⁻¹; 2. [A]=0.200 M, [B]=0.100 M, rate=4.0×10⁻⁴ M s⁻¹; 3. [A]=0.100 M, [B]=0.200 M, rate=8.0×10⁻⁴ M s⁻¹.
Deduce the rate law and overall order of reaction.
rate = k[A]¹[B]² ; overall third order
Value of the equilibrium constant Kc for the Haber process at 450 °C if ΔG° = +19 kJ mol⁻¹ (R = 8.31 J mol⁻¹ K⁻¹, give two sig. fig.).
T = 450 + 273 = 723 K
ΔG° = –RT ln K ⟹ ln K = –ΔG°/(RT)
ln K = –19 000 J ÷ (8.31 J mol⁻¹ K⁻¹ × 723 K) ≈ –3.16
K = e^(–3.16) ≈ 4.2 × 10⁻²