Reactivity 2: How much, how fast and how far?IB Chemistry SL: Topic test

20 questions, 54 marks

IB Chemistry SL

Reactivity 2: How much, how fast and how far? topic test

Total 54 marks

Name

Class

Date

1
A student adds 4.03 g of magnesium oxide, MgO, to 40.0 cm³ of 2.00 mol dm⁻³ sulfuric acid. They react completely to form magnesium sulfate solution: MgO(s) + H₂SO₄(aq) → MgSO₄(aq) + H₂O(l). After evaporating and drying the solid, the student collects 8.19 g of magnesium sulfate. Molar masses: MgO 40.3 g mol⁻¹, H₂SO₄ 98.1 g mol⁻¹, MgSO₄ 120.4 g mol⁻¹.
(a)
What amount, in mol, of sulfuric acid is available?
[1 mark]
A0.0200 mol
B0.0800 mol
C0.100 mol
D0.200 mol
(b)
Which reactant is the limiting reactant?
[1 mark]
AMgO, because it has the smaller molar mass
BMgO, because fewer moles are present
CH₂SO₄, because it is used up first
DH₂SO₄, because it has the larger molar mass
(c)
Calculate the percentage yield of magnesium sulfate.
[2 marks]
Total for question 1: 4 marks
2
Ethene reacts with hydrogen over a nickel catalyst: C₂H₄(g) + H₂(g) → C₂H₆(g), ΔH = −137 kJ mol⁻¹. Without a catalyst, this reaction has an activation energy of about 180 kJ mol⁻¹ and proceeds immeasurably slowly at 25 °C, even though it is exothermic. With a nickel catalyst present, the activation energy falls to about 40 kJ mol⁻¹ and the reaction is fast at 25 °C.
(a)
Why is the uncatalysed reaction immeasurably slow at 25 °C even though it releases energy?
[1 mark]
AThe reaction is endothermic, so few collisions have enough energy
BOnly a very small fraction of collisions have kinetic energy ≥ 180 kJ mol⁻¹, the activation energy
CThe particles are too far apart to collide at room temperature
DThe products are more stable than the reactants, so the reaction cannot occur
(b)
Which statement correctly explains why the nickel catalyst increases the rate?
[1 mark]
AIt increases the average kinetic energy of the particles at 25 °C
BIt increases the frequency of collisions between particles
CIt provides an alternative pathway with lower activation energy, so a larger fraction of collisions are successful
DIt shifts the position of equilibrium to favour the products
(c)
Using the Maxwell–Boltzmann energy distribution at 25 °C, explain why lowering the activation energy from 180 kJ mol⁻¹ to 40 kJ mol⁻¹ increases the reaction rate.
[2 marks]
Total for question 2: 4 marks
3
Hydrogen and iodine vapour are mixed in a sealed, rigid container at a constant temperature of 700 K and allowed to reach equilibrium: H₂(g) + I₂(g) ⇌ 2HI(g). At equilibrium the container holds 0.100 mol H₂, 0.100 mol I₂ and 1.56 mol HI in a fixed volume of 2.00 dm³.
(a)
Deduce the equilibrium constant expression for this reaction, and calculate its value at 700 K, including units if appropriate.
[3 marks]
(b)
The temperature is then raised to 900 K and a new equilibrium is reached at which Kc = 65. Explain, using Le Châtelier's principle and the change in Kc, what has happened to the position of equilibrium and whether the forward reaction is exothermic or endothermic.
[4 marks]
Total for question 3: 7 marks
4
In industry, hydrogen is manufactured by the water-gas shift reaction: CO(g) + H₂O(g) ⇌ CO₂(g) + H₂(g), ΔH = −41 kJ mol⁻¹. The forward reaction is catalysed by an iron oxide catalyst. Engineers must choose an operating temperature and pressure, and must decide whether to remove CO₂ continuously from the reactor.
(a)
Using collision theory and the idea of activation energy, explain how the iron oxide catalyst increases the rate of the forward reaction, and explain why increasing the temperature also increases the rate, even though the reaction is exothermic.
[6 marks]
(b)
The engineers are considering removing CO₂ continuously from the reactor as it forms. Using Le Châtelier's principle and the equilibrium law, explain the effect this would have on the amount of hydrogen produced, and explain why this approach does not require a higher operating temperature to be effective, given that the forward reaction is exothermic.
[6 marks]
Total for question 4: 12 marks
5
A student adds 3.35 g of iron filings to 100.0 cm³ of 0.500 mol dm⁻³ copper(II) sulfate solution: Fe(s) + CuSO₄(aq) → FeSO₄(aq) + Cu(s). The student collects and dries the copper formed, obtaining 2.86 g. Molar masses: Fe 55.8 g mol⁻¹, CuSO₄ 159.6 g mol⁻¹, Cu 63.5 g mol⁻¹.
(a)
What amount, in mol, of CuSO₄ is available?
[1 mark]
A0.0500 mol
B0.0300 mol
C0.0600 mol
D0.100 mol
(b)
Which reactant is the limiting reactant?
[1 mark]
AFe, because it is in excess
BFe, because it has the smaller molar mass
CCu, because it is the product
DCuSO₄, because fewer moles are available
(c)
Calculate the percentage yield of copper obtained.
[2 marks]
Total for question 5: 4 marks
6
A student measures the initial rate of reaction between excess dilute hydrochloric acid and a fixed 0.500 g strip of magnesium ribbon by timing how long it takes to collect 20.0 cm³ of hydrogen gas. Experiment 1 is carried out at 20 °C; the gas is collected in 40 s. Experiment 2 uses an identical strip of magnesium and the same acid, but at 50 °C; the gas is collected in only 9 s.
(a)
What is the ratio of the initial rate of reaction in Experiment 2 to Experiment 1?
[1 mark]
A2.22
B0.225
C5.56
D4.44
(b)
Which explanation accounts for the large increase in rate for only a 30 °C rise in temperature?
[1 mark]
AA small rise in temperature causes a disproportionately large increase in the fraction of particles with energy ≥ the activation energy, because of the shape of the Maxwell–Boltzmann distribution
BA 30 °C rise roughly doubles the average kinetic energy of the particles, doubling the rate
CThe activation energy of the reaction decreases as temperature rises
DThe concentration of the hydrochloric acid increases as temperature rises
(c)
Explain why, at a fixed temperature, increasing the concentration of the hydrochloric acid increases the rate of this reaction.
[2 marks]
Total for question 6: 4 marks
7
Carbon monoxide reacts reversibly with chlorine in a sealed, rigid 5.00 dm³ container at a fixed temperature to form phosgene: CO(g) + Cl₂(g) ⇌ COCl₂(g), ΔH = −108 kJ mol⁻¹. At equilibrium the container holds 0.0200 mol CO, 0.0300 mol Cl₂ and 0.480 mol COCl₂.
(a)
Deduce the equilibrium constant expression for this reaction and calculate its value, including units, at this temperature.
[3 marks]
(b)
The temperature of the container is then increased. State and explain, using Le Châtelier's principle, what happens to the value of Kc and to the equilibrium amount of phosgene, given that the forward reaction is exothermic. State what this implies for the effect of raising temperature on Kc for a reaction whose forward reaction is instead endothermic.
[4 marks]
Total for question 7: 7 marks
8
In an industrial batch process, phosphorus pentachloride is partially decomposed in a sealed reactor: PCl₅(g) ⇌ PCl₃(g) + Cl₂(g), ΔH = +92 kJ mol⁻¹. The desired product is PCl₃, which is separated and purified; the theoretical yield is calculated from the initial amount of PCl₅ used, assuming complete decomposition. Engineers can increase the fractional decomposition of PCl₅ (and so the percentage yield of PCl₃) by using a high operating temperature and/or a low pressure, since this equilibrium is not catalysed.
(a)
Explain, using Le Châtelier's principle and the equilibrium law, why increasing the temperature and lowering the pressure both increase the equilibrium amount of PCl₃ formed from a fixed initial amount of PCl₅, and hence the percentage yield of PCl₃.
[6 marks]
(b)
A rival process uses a catalyst that speeds up the attainment of equilibrium but does not change Kc. Explain why this catalyst cannot, on its own, increase the percentage yield of PCl₃ beyond that from part (a), and evaluate whether operating at a very high temperature is nonetheless the best strategy for maximising the percentage yield, given that the reactor materials degrade rapidly above 550 K and heating the reactor is the largest energy cost of the process.
[6 marks]
Total for question 8: 12 marks

End of questions