Flashcards · H432 · Topic 5

Physical chemistry and transition elements flashcards

54 cards on Topic 5, subtopics 5.1.1 to 5.3.2. Pick a subtopic or study them all.

Study mode needs JavaScript. Every card is listed below, so you can still cover the answers and test yourself.

All 54 cards

For a quick read-through or printing. Cover the right-hand side to test yourself.

5.1.1 How fast?

Half-life, and how it shows first order: The time for the concentration of a reactant to fall to half its value. Constant successive half-lives mean first order.
Rate constant from the half-life of a first-order reaction with t½ = 46 s: k = ln 2 ÷ t½ = 0.693 ÷ 46 = 1.51 × 10⁻² s⁻¹
[P] trebles and the rate trebles; [Q] doubles and the rate quadruples. Rate equation?: Rate = k[P][Q]² (first order in P, second order in Q).
Units of k for overall order 2 and 3 (OCR form): Order 2: dm³ mol⁻¹ s⁻¹. Order 3: dm⁶ mol⁻² s⁻¹.
Clock reaction: what is measured and why is it valid?: The time t for a fixed small amount of product to form; initial rate ∝ 1/t. Valid only while the reaction has not proceeded too far, so the rate stays nearly constant.
Arrhenius plot: gradient −8.42 × 10³ K, R = 8.314 J mol⁻¹ K⁻¹. Ea?: Ea = −gradient × R = 8.42 × 10³ × 8.314 = 7.00 × 10⁴ J mol⁻¹ = 70.0 kJ mol⁻¹
2NO₂ + F₂ → 2NO₂F has rate = k[NO₂][F₂]. Suggest the rate-determining step.: NO₂ + F₂ → NO₂F + F (slow), then NO₂ + F → NO₂F (fast). One NO₂ and one F₂ react in the slow step.

5.1.2 How far?

Mole fraction: x(A) = moles of A ÷ total moles of gas present. It has no units, and the values for one mixture sum to 1.
Partial pressure: x(A) multiplied by the total pressure. Summing every gas’s partial pressure gives back the total pressure.
Kp for CaO(s) + SO₂(g) ⇌ CaSO₃(s): Kp = 1 ÷ p(SO₂), units kPa⁻¹. Pure solids are left out of heterogeneous expressions.
J(g) + 3K(g) ⇌ 2L(g) in 3.00 dm³: 0.30 mol J, 0.90 mol K, 0.60 mol L. Kc?: [J] = 0.100, [K] = 0.300, [L] = 0.200 mol dm⁻³. Kc = 0.200² ÷ (0.100 × 0.300³) = 14.8 dm⁶ mol⁻²
Kc = 4.6 at 400 K and 1.9 at 500 K. Is the forward reaction exothermic or endothermic?: Exothermic: K falls as temperature rises.
In terms of K, why does adding a reactant shift an equilibrium right?: K is fixed at that temperature. Adding reactant makes the quotient smaller than K, so the reaction goes forwards until the quotient equals K again.

5.1.3 Acids, bases and buffers

Conjugate acid–base pair: Two species that differ by one H⁺, e.g. HNO₂ and NO₂⁻.
pH of a strong base solution: method: Find [OH⁻] (double it for a base like Sr(OH)₂), then [H⁺] = Kw ÷ [OH⁻] with Kw = 1.00 × 10⁻¹⁴ mol² dm⁻⁶, then pH = −log[H⁺].
The two approximations for the pH of a weak acid: [H⁺] = [A⁻] (H⁺ from water is negligible), and [HA] at equilibrium ≈ [HA] undissociated. Then [H⁺] = √(Ka × [HA]).
When does the weak-acid approximation break down?: For "stronger" weak acids (large Ka), especially in dilute solution: a significant fraction dissociates, so the calculated [H⁺] is too high and the pH too low.
Buffer solution, and the two ways to make an acid buffer: A system that minimises pH changes on addition of small amounts of an acid or a base. Made from a weak acid plus its salt, or excess weak acid plus a strong alkali.
How does the H₂CO₃/HCO₃⁻ buffer control blood pH?: It holds plasma between pH 7.35 and 7.45. Extra H⁺ reacts with HCO₃⁻ to form H₂CO₃; extra OH⁻ removes H⁺ and H₂CO₃ dissociates to replace it.
30.0 cm³ of 0.180 mol dm⁻³ HZ (Ka 7.2 × 10⁻⁵) + 27.0 cm³ of 0.100 mol dm⁻³ NaOH. pH?: n(HZ) = 0.00540, n(OH⁻) = 0.00270 mol: half-neutralised, so [HZ] = [Z⁻] and pH = pKa = −log(7.2 × 10⁻⁵) = 4.14
End point of a titration in terms of the indicator: The point where the indicator changes colour, when [HIn] = [In⁻]. The indicator’s range must lie within the steep section of the curve.

5.2.1 Lattice enthalpy

Lattice enthalpy, ΔLEH (OCR definition): The enthalpy change when 1 mol of an ionic lattice is formed from its gaseous ions under standard conditions, e.g. Li⁺(g) + I⁻(g) → LiI(s). Always exothermic.
First ionisation energy (as a Born–Haber term): The energy required to remove 1 mol of electrons from 1 mol of gaseous atoms, giving 1 mol of gaseous 1+ ions.
Enthalpy change of formation: The enthalpy change when 1 mol of a compound forms, under standard conditions, from its constituent elements, each in its standard state.
Lattice enthalpy of KF: ΔfH −568, ΔatH(K) +89, IE(K) +419, ΔatH(F) +79, EA(F) −328 kJ mol⁻¹: Sum of steps = 89 + 419 + 79 − 328 = +259. −568 = 259 + ΔLEH, so ΔLEH = −827 kJ mol⁻¹.
Enthalpy change of solution from lattice and hydration enthalpies (OCR sign convention): ΔsolH = −ΔLEH + ΣΔhydH, counting every ion. E.g. CsBr: +640 + (−276) + (−336) = +28 kJ mol⁻¹.
Effect of ionic charge and radius on lattice and hydration enthalpies: Smaller ions or larger charges give stronger attraction, so both become more exothermic. E.g. Mg²⁺ compounds are far more exothermic than Na⁺ ones.
Atomisation equations for bromine and iodine: ½Br₂(l) → Br(g) ½I₂(s) → I(g) Per mole of atoms formed, from the element in its standard state.

5.2.2 Enthalpy and entropy

Entropy, S: A measure of the dispersal of energy in a system; the more disordered, the greater the entropy. Units J mol⁻¹ K⁻¹. Elements have non-zero S values.
Predicting the sign of ΔS: Positive when solids become liquids or gases, or when the number of gas molecules increases. Negative when the number of gas molecules decreases.
ZnCO₃ → ZnO + CO₂: ΔH = +68.8 kJ mol⁻¹, ΔS = +175.1 J mol⁻¹ K⁻¹. Feasible at 298 K?: ΔG = 68.8 − 298 × 0.1751 = +16.6 kJ mol⁻¹: not feasible. It becomes feasible above T = 68.8 ÷ 0.1751 = 393 K.
Feasibility condition: A process is feasible when ΔG = ΔH − TΔS is negative (T in K, ΔS converted to kJ).
Feasible versus fast: Thermodynamics versus kinetics.
Why might a reaction with negative ΔG not be observed?: A large activation energy makes it too slow, e.g. methane in air until it is ignited.

5.2.3 Redox and electrode potentials

Standard hydrogen electrode: H₂(g) at 100 kPa over platinum, in 1 mol dm⁻³ H⁺(aq), at 298 K. E⦵ = 0.00 V by definition.
Standard cell potential: E⦵cell = E⦵(positive electrode) − E⦵(negative electrode). Reduction takes place at the more positive half-cell.
Half-equation: dichromate(VI) to chromium(III) in acid: Cr₂O₇²⁻ + 14H⁺ + 6e⁻ → 2Cr³⁺ + 7H₂O
Manganate(VII) titration of Fe²⁺: conditions, end point and ratio: Excess dilute H₂SO₄ (not HCl, as Cl⁻ is oxidised); no indicator needed; end point is the first permanent pale pink. MnO₄⁻ : Fe²⁺ = 1 : 5.
Iodine–thiosulfate titration: indicator, end point and ratio: Add starch when the iodine colour is pale yellow; end point is when the blue-black colour just disappears. I₂ : S₂O₃²⁻ = 1 : 2 (I₂ + 2S₂O₃²⁻ → 2I⁻ + S₄O₆²⁻).
Why might a reaction predicted feasible by E⦵ values not happen?: The rate may be too slow (high activation energy), and real concentrations differ from standard, shifting each half-cell equilibrium.
Fuel cell: A cell that uses energy from a fuel reacting with oxygen to create a voltage, with reactants supplied continuously. The fuel is oxidised at the negative electrode.

5.3.1 Transition elements

Transition element: A d-block element that forms an ion with an incomplete d-sub-shell. In Period 4, Ti to Cu; Sc (Sc³⁺ 3d⁰) and Zn (Zn²⁺ 3d¹⁰) are not.
Electron configuration of Mn²⁺ (Mn is [Ar]3d⁵4s²): [Ar]3d⁵ (1s² 2s² 2p⁶ 3s² 3p⁶ 3d⁵). The 4s electrons are removed first.
Shapes of [Cu(H₂O)₆]²⁺, [CuCl₄]²⁻ and Pt(NH₃)₂Cl₂: [Cu(H₂O)₆]²⁺: octahedral, 90°. [CuCl₄]²⁻: tetrahedral, 109.5°. Pt(NH₃)₂Cl₂: square planar, 90°.
How does cis-platin work as an anticancer drug?: Cl⁻ ligands are replaced in the cell and Pt binds to nitrogen atoms on DNA bases (often two adjacent guanines). The distorted DNA cannot be copied, so the cell cannot divide.
[Cu(H₂O)₆]²⁺ with concentrated HCl: [Cu(H₂O)₆]²⁺ + 4Cl⁻ ⇌ [CuCl₄]²⁻ + 6H₂O. Blue → yellow (green in between); octahedral → tetrahedral.
Why is CO toxic, in terms of haemoglobin?: Haemoglobin has octahedral Fe²⁺ with O₂ on the sixth site. CO substitutes for O₂, binds far more strongly and is not easily released, so less oxygen is carried.
Cu²⁺ with iodide ions: observations and equation: A white precipitate of CuI in a brown solution (I₂). 2Cu²⁺ + 4I⁻ → 2CuI + I₂
Which isomerism does [Ni(en)₃]²⁺ show?: Optical isomerism: it has no plane of symmetry, so it forms non-superimposable mirror images.

5.3.2 Qualitative analysis

Cu²⁺, Fe²⁺ and Fe³⁺ with a little NaOH(aq) or NH₃(aq): Cu²⁺: pale blue precipitate. Fe²⁺: green precipitate, turning orange-brown at the surface on standing. Fe³⁺: orange-brown precipitate.
Mn²⁺ and Cr³⁺ with a little NaOH(aq): Mn²⁺: light brown precipitate that darkens on standing in air. Cr³⁺: grey-green precipitate.
Which hydroxides dissolve in excess NaOH(aq), and which in excess NH₃(aq)?: Excess NaOH: only Cr(OH)₃ (dark green [Cr(OH)₆]³⁻). Excess NH₃: Cr(OH)₃ (purple [Cr(NH₃)₆]³⁺) and Cu(OH)₂ (deep blue solution [Cu(NH₃)₄(H₂O)₂]²⁺).
Fe²⁺ and Cr³⁺ solutions can both look green. How do you tell them apart?: Add excess NaOH(aq): Cr(OH)₃ dissolves to a dark green solution; Fe(OH)₂ does not dissolve.
Why does aqueous ammonia precipitate metal hydroxides?: NH₃ + H₂O ⇌ NH₄⁺ + OH⁻ provides hydroxide ions.