Answer
Flashcards · IB Diploma Programme Chemistry · Topic 5
Reactivity 2 – How much, how fast and how far? flashcards
25 cards on Topic 5, subtopics 2.1 to 2.3. 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.
Choose your cards
Session complete
Worth another look:
All 25 cards
For a quick read-through or printing. Cover the right-hand side to test yourself.
2.1 How much? The amount of chemical change
- Limiting reactant
- The reactant that is used up first. It caps the amount of product.
- Theoretical yield
- The amount of product calculated from the limiting reactant, assuming complete reaction and no losses.
- Atom economy
- Molar mass of the desired product as a percentage of the total molar mass of all reactants in the balanced equation; a green chemistry measure of efficiency.
- 4Li + O₂ → 2Li₂O with 0.50 mol Li and 0.15 mol O₂. Which is limiting, and how much Li₂O forms?
- Divide by coefficients: Li 0.50 ÷ 4 = 0.125; O₂ 0.15 ÷ 1 = 0.15. Lithium is limiting. Li₂O = 0.50 × 2/4 = 0.25 mol; 0.025 mol O₂ is left over.
- 2C₃H₆(g) + 9O₂(g) → 6CO₂(g) + 6H₂O(l). Volumes for 10 cm³ of propene?
- Oxygen needed: 10 × 9/2 = 45 cm³. CO₂ made: 10 × 6/2 = 30 cm³. Liquid water takes up no gas volume.
- Percentage yield vs atom economy
- Yield depends on how the reaction is carried out (experimental ÷ theoretical × 100). Atom economy is fixed by the equation, so better technique cannot change it.
2.2 How fast? The rate of chemical change
- Rate of reaction, and the instantaneous rate
- Rate: how fast the concentration of a named reactant or product changes per unit time (mol dm⁻³ s⁻¹). Instantaneous rate: gradient of the tangent to the curve at that moment.
- The two conditions for a successful collision
- Energy at least equal to the activation energy, Ea, and the proper orientation (collision geometry).
- Which factors change collision frequency only, and which change the fraction with E ≥ Ea?
- Concentration, pressure, surface area: frequency only. Temperature: the fraction with E ≥ Ea (main effect), and frequency slightly. Catalyst: the fraction, because Ea is lower.
- Maxwell–Boltzmann: temperature vs catalyst
- Temperature changes the curve; a catalyst moves the Ea line.
- Kinetic vs thermodynamic stability
- A high Ea makes a reaction slow; a positive or negative ΔH says nothing about speed. A high Ea lets a fuel be stored in air without igniting.
- An endothermic reaction has Ea = 140 kJ mol⁻¹ and ΔH = +35 kJ mol⁻¹. Ea of the reverse reaction?
- Ea(reverse) = 140 − 35 = 105 kJ mol⁻¹.
- How are the orders in a rate equation found?
- Only by experiment, never from the coefficients of the balanced equation. rate = k[A]ᵐ[B]ⁿ; overall order = m + n. (HL only)
- Mechanism: L + M → LM (slow); LM + M → LM₂ (fast). Rate equation?
- rate = k[L][M]. Only one M is in the rate-determining (slow) step, so it is first order in M even though the overall equation has 2M. LM is an intermediate. (HL only)
- Intermediate vs transition state
- Intermediate: made in one step and used in a later one; a trough on the profile; can be detected. Transition state: the highest-energy arrangement in one step; a peak; cannot be isolated. (HL only)
- Units of k for zero, first, second and third order overall
- Zero: mol dm⁻³ s⁻¹. First: s⁻¹. Second: mol⁻¹ dm³ s⁻¹. Third: mol⁻² dm⁶ s⁻¹. (HL only)
- Arrhenius plot: what is plotted, and how are Ea and A found?
- ln k against 1/T (K). Gradient = −Ea/R, so Ea = −gradient × 8.31 J mol⁻¹. The y-intercept is ln A. (HL only)
2.3 How far? The extent of chemical change
- Features of a dynamic equilibrium
- Closed system; forward and backward rates equal and not zero; concentrations constant (not necessarily equal); observable properties constant; it can be reached from either side.
- Equilibrium constant expression for 2N₂O(g) + 3O₂(g) ⇌ 4NO₂(g)
- K = [NO₂]⁴ ÷ ([N₂O]²[O₂]³). Products on top; coefficients as powers.
- K for the forward reaction is 0.25. K for the reverse at the same temperature?
- K(reverse) = 1 ÷ K(forward) = 4.0.
- Effect of heating on K for an exothermic forward reaction
- K gets smaller and there is less product. Only temperature changes K; concentration and pressure move the position but leave K the same.
- Effect of a catalyst on an equilibrium
- No shift in position and no change in K: it increases the forward and backward rates by the same factor, so equilibrium is reached faster.
- Q compared with K: which way does the reaction go?
- Q < K: forward, making more products. Q = K: at equilibrium. Q > K: backward. Compare only at the same temperature. (HL only)
- 0.50 mol SO₂Cl₂ in 2.0 dm³ gives 0.10 mol Cl₂ at equilibrium (SO₂Cl₂ ⇌ SO₂ + Cl₂). K?
- Equilibrium: SO₂Cl₂ 0.40 mol, SO₂ 0.10, Cl₂ 0.10. Concentrations 0.20, 0.050, 0.050 mol dm⁻³. K = (0.050 × 0.050) ÷ 0.20 = 0.0125. (HL only)
- ΔG⦵ = +10.0 kJ mol⁻¹ at 400 K. K?
- lnK = −10 000 ÷ (8.31 × 400) = −3.01, so K = 0.0494. Positive ΔG⦵ and K < 1: reactants favoured. (HL only)