NCEA Level 2 Chemistry 91166 Chemical reactivity: past papers, practice and marking
- NCEA Level 2
- Chemistry
- AS91166
- 4 credits
- External exam
About 91166
AS91166, Demonstrate understanding of chemical reactivity. Why reactions speed up and slow down, what a catalyst really does, how an equilibrium answers back when you change it, Kc, and strong and weak acids, bases and salts with their pH and conductivity: the 4-credit external, in the six question types it keeps reusing.
What 91166 covers
- Reaction ratesConcentration, surface area and temperature explained by collision theory: collisions with enough energy (at least the activation energy) and the right orientation, frequency of successful collisions per unit time, reading gas-volume and mass-loss data section by section
- CatalystsAn alternative pathway with a lower activation energy, a greater proportion of collisions successful, not used up, does not supply energy, does not change the equilibrium position or Kc, sketching the faster line
- EquilibriumForward and reverse rates equal, concentrations constant but not equal; the response to adding or removing a species (including by neutralisation or precipitation), pressure (fewer moles of gas) and temperature; industrial compromise between rate and yield
- KcProducts over reactants with powers from the coefficients, the equation from a given Kc, the meaning of a large or small Kc, calculating Kc and an unknown concentration, comparing a quotient with Kc, only temperature changes Kc
- Acids and basesBrønsted–Lowry proton transfer, conjugate pairs, amphiprotic species, strong (complete, single arrow) and weak (partial, equilibrium arrow) acids and bases, salt dissociation then the reaction of the ion with water, pH and conductivity explained by the ions present
- pHUsing pH = −log[H₃O⁺], [H₃O⁺] = 10^−pH, Kw = [H₃O⁺][OH⁻] = 1.00 × 10⁻¹⁴, strong acids and strong bases only, two OH⁻ per formula unit, dilution, units and significant figures
91166 past papers
Open a 91166 past paper beside your own Ako page, then get it marked, part by part.
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Practice questions
Question 1: Reaction rates
0.20 g of zinc granules is added to 25.0 mL of 1.0 mol L⁻¹ sulfuric acid (in excess) at four temperatures. The time to collect 30.0 mL of hydrogen is measured. | Temperature (°C) | 15 | 25 | 35 | 45 | |---|---|---|---|---| | Time to collect 30.0 mL (s) | 150 | 74 | 38 | 19 |
- Describe the trend in the results.(up to Achieved)
- Calculate the average rate of hydrogen production at 15 °C and at 35 °C.(up to Achieved)
Check your answer
Part a: what gets the marks
- As the temperature rises the time to collect 30.0 mL falls, so the rate increases; each 10 °C roughly halves the time.
A full answer: As the temperature increases, the time decreases, so the rate increases. Each 10 °C rise roughly halves the time (doubles the rate).
Part b: what gets the marks
- The rate at 15 °C is the volume divided by the time.
- The rate at 35 °C is the volume divided by the time.
A full answer: 15 °C: 30.0 mL ÷ 150 s = 0.20 mL s⁻¹.
35 °C: 30.0 mL ÷ 38 s = 0.79 mL s⁻¹ (about four times faster).
What gets each grade
Question 2: Reaction rates
The magnesium experiment is repeated with the same strip of magnesium ribbon (0.050 g) in 50.0 mL of 0.500 mol L⁻¹ hydrochloric acid, still in excess. In the first experiment, with 1.00 mol L⁻¹ acid, the rate was 1.5 mL s⁻¹ over the first 20 s and the volume levelled off at 50 mL by 80 s.
- Compare the second experiment with the first, referring to the rate at the start, when the reaction stops, and the final volume of hydrogen.(up to Excellence)
- Calculate the amount, in mol, of magnesium and of hydrochloric acid used in the second experiment, and use your answers to show that the acid is in excess. (M(Mg) = 24.3 g mol⁻¹; Mg + 2HCl → MgCl₂ + H₂)(up to Merit)
Check your answer
Part a: what gets the marks
- With 0.500 mol L⁻¹ acid the reaction is slower at the start: the average rate over the first 20 s is less than 1.5 mL s⁻¹, so the line is less steep.
- There are fewer acid particles per unit volume, so fewer collisions per second with the magnesium surface and fewer successful collisions per second.
- The first experiment's line is steeper and levels off by 80 s; the second line rises more slowly and takes longer to level off, because the successful collisions per second are fewer all the way through.
- Both level off at the same 50 mL, because the same 0.050 g of magnesium is the limiting reactant and the acid is in excess, so the same amount of hydrogen forms.
A full answer: With 0.500 mol L⁻¹ acid there are fewer acid particles per unit volume, so there are fewer collisions per second with the magnesium and fewer successful collisions per second. The line starts less steep than the first experiment's 1.5 mL s⁻¹ and takes longer to level off: the first reaction is finished by 80 s, the second is still going. The final volume is the same, 50 mL, because the magnesium (0.050 g) is the limiting reactant in both and the acid is in excess, so the same amount of hydrogen forms. Only the time it takes changes.
Part b: what gets the marks
- The amount of magnesium is the mass divided by the molar mass.
- The amount of acid is the concentration multiplied by the volume in litres.
- The equation shows 2 mol of HCl react with 1 mol of Mg, so 0.00412 mol of HCl is needed; 0.0250 mol is far more than this, so the acid is in excess.
A full answer: n(Mg) = 0.050 ÷ 24.3 = 0.00206 mol. n(HCl) = 0.500 × 0.0500 = 0.0250 mol. The equation shows 2 mol of HCl react with 1 mol of Mg, so only 2 × 0.00206 = 0.00412 mol of HCl is needed. There is 0.0250 mol, about six times more than needed, so the acid is in excess and the magnesium is the limiting reactant.
What gets each grade
Question 3: Reaction rates
Zinc granules react with sulfuric acid about four times faster at 35 °C than at 15 °C, with the same concentration of acid and the same mass of zinc. Zn(s) + H₂SO₄(aq) → ZnSO₄(aq) + H₂(g)
- Explain why the reaction is faster at 35 °C.(up to Merit)
- The experiment is repeated at 15 °C using acid with double the concentration. Predict how the rate changes, and explain your answer in terms of collisions.(up to Merit)
- A student says, "The reaction is faster at 35 °C because the higher temperature lowers the activation energy." Explain why this statement is wrong.(up to Merit)
Check your answer
Part a: what gets the marks
- At 35 °C the particles have more kinetic energy and move faster.
- Faster-moving particles collide more often, so there are more collisions per second between H₃O⁺ ions and zinc atoms.
- A greater proportion of the collisions have energy equal to or greater than the activation energy, so more successful collisions per second and a faster rate.
A full answer: At 35 °C the particles have more kinetic energy and move faster. They collide more often (more collisions per second), and a greater proportion of the collisions have energy equal to or greater than the activation energy. Both effects give more successful collisions per second, so the rate is higher.
Part b: what gets the marks
- The rate of reaction increases, because the acid is more concentrated.
- There are more H₃O⁺ ions in each unit volume of the more concentrated acid, so they collide with the zinc atoms more frequently (more collisions per second).
- The temperature is unchanged, so the same proportion of collisions have at least the activation energy, and more collisions per second gives more successful collisions per second.
A full answer: The rate increases. In the more concentrated acid there are more H₃O⁺ ions per unit volume, so they hit the zinc more frequently (more collisions per second). The temperature is the same, so the same proportion of collisions have energy equal to or greater than the activation energy. More collisions per second therefore means more successful collisions per second, and a faster rate.
Part c: what gets the marks
- Temperature does not change the activation energy of the reaction; only a catalyst provides an alternative pathway with a lower activation energy.
- At 35 °C the particles have more kinetic energy, so a greater proportion of the collisions have energy equal to or greater than the same activation energy.
- A greater proportion of successful collisions, together with more frequent collisions, gives more successful collisions per second and a faster rate.
A full answer: The activation energy stays the same: only a catalyst lowers it, by providing an alternative pathway. At 35 °C the particles have more kinetic energy, so a greater proportion of collisions have energy equal to or greater than the same activation energy. Together with more frequent collisions, this gives more successful collisions per second and a faster rate.