Questions on collision theory, activation energy and the rate of reaction

Question 1

Explain, using collision theory, the results of three experiments in which 1.0 g of calcium carbonate reacts with 50 cm³ of hydrochloric acid, an excess, at 20 °C. A: chips in 1.0 mol dm⁻³ acid. B: powder in 1.0 mol dm⁻³ acid. C: chips in 2.0 mol dm⁻³ acid. B and C both have a greater initial rate than A, and all three give the same final volume of carbon dioxide.

Accepted Answer

Indicative content

a reaction happens only when particles collide with at least the activation energy and in the correct orientation
in B the powder has a larger surface area, so more calcium carbonate particles are exposed to the acid
in C there are more hydrogen ions in each unit volume of the acid than in A
in both B and C collisions are more frequent than in A, so there are more successful collisions per unit time
the temperature is the same in all three, so the fraction of collisions with at least the activation energy is unchanged
calcium carbonate is the limiting reactant and the same mass is used each time, so the same amount of carbon dioxide forms in all three

Levels

Level 3 (5–6 marks) all three results are explained through collision theory: the condition for a successful collision, the larger surface area in B and the higher concentration in C both giving more frequent collisions, the unchanged fraction with the activation energy at the same temperature, and the same final volume from the same limiting mass of carbonate
Level 2 (3–4 marks) the faster rates of B and C are put down to more frequent collisions from surface area and concentration; the condition for a successful collision, the unchanged temperature, or why the final volume is the same is left out
Level 1 (1–2 marks) powder or stronger acid is said to react faster, typically because the particles have more energy, and the same final volume is not explained

Question 2

Calculate the initial rate of reaction, and the rate 60 s after mixing, from the volumes of hydrogen in the table, collected when magnesium ribbon reacted with excess dilute hydrochloric acid. Plot the results and draw tangents to the curve.

Accepted Answer

M1 the points plotted with a smooth curve through them from the origin, levelling off near 75 cm³ M1 a tangent ruled at t = 0 and its gradient found as rise divided by run between two widely separated points A1 initial rate = 2.5 cm³ s⁻¹; accept 2.2 to 2.8 cm³ s⁻¹ M1 a second tangent ruled to touch the curve at t = 60 s, with its gradient found the same way A1 rate at 60 s = 0.34 cm³ s⁻¹ to two significant figures; accept 0.28 to 0.40 cm³ s⁻¹

Question 3

State the two conditions that a collision between two reactant particles must meet for it to lead to a reaction, and explain why most collisions between gas molecules at room temperature do not lead to a reaction.

Accepted Answer

B1 the particles must collide with energy at least equal to the activation energy; reject 'enough energy' without reference to the activation energy B1 the particles must collide in the correct orientation, with the reacting parts facing each other B1 at room temperature most colliding particles have less energy than the activation energy, so the bonds that must break are not broken B1 many collisions are in the wrong orientation, so the reacting parts of the particles do not meet and they bounce apart unchanged

Question 4

Calculate the initial rate at which the concentration of the acid falls when marble chips react with 25.0 cm³ of hydrochloric acid, given that carbon dioxide is produced at an initial rate of 1.8 cm³ s⁻¹ at room temperature and pressure. CaCO₃(s) + 2HCl(aq) → CaCl₂(aq) + H₂O(l) + CO₂(g). The molar volume is 24.0 dm³ mol⁻¹.

Accepted Answer

M1 rate of production of CO₂ = 1.8 ÷ 24 000 = 7.5 × 10⁻⁵ mol s⁻¹ M1 rate of use of HCl = 2 × 7.5 × 10⁻⁵ = 1.5 × 10⁻⁴ mol s⁻¹, from the 2 : 1 ratio in the equation M1 divide by the volume of acid in dm³: 1.5 × 10⁻⁴ ÷ 0.0250 A1 6.0 × 10⁻³ mol dm⁻³ s⁻¹, to two significant figures; allow error carried forward from a missing factor of two, 3.0 × 10⁻³, for the last mark only

Question 5

Calculate the mean rate, as the rate at which the concentration of the acid falls, over the first 25 s of the reaction of marble chips with hydrochloric acid in an open flask on a balance, using the results in the table. CaCO₃(s) + 2HCl(aq) → CaCl₂(aq) + H₂O(l) + CO₂(g).

Accepted Answer

M1 amount of CO₂ = 0.22 ÷ 44.0 = 5.0 × 10⁻³ mol, so the mean rate of production is 5.0 × 10⁻³ ÷ 25 = 2.0 × 10⁻⁴ mol s⁻¹ M1 rate of use of HCl = 2 × 2.0 × 10⁻⁴ = 4.0 × 10⁻⁴ mol s⁻¹, from the 2 : 1 ratio in the equation M1 divide by the volume of acid in dm³: 4.0 × 10⁻⁴ ÷ 0.0250 A1 1.6 × 10⁻² mol dm⁻³ s⁻¹, to two significant figures

Question 6

Explain why halving the volume of a mixture of two reacting gases, at constant temperature, increases the rate of the reaction, but pumping argon into the same mixture in a sealed vessel of fixed volume raises the total pressure without changing the rate.

Accepted Answer

B1 halving the volume doubles the pressure, so there are twice as many reactant molecules in each unit volume B1 collisions between reactant molecules are more frequent, so there are more successful collisions per unit time B1 argon does not react, and at fixed volume the number of reactant molecules in each unit volume is unchanged, so their collisions are no more frequent B1 the temperature is unchanged in both cases, so the same fraction of collisions have at least the activation energy; only the collision frequency decides the change

Question 7

Give the meaning of the term activation energy and of the term rate of reaction, and give the units of a rate of reaction.

Accepted Answer

B1 activation energy is the minimum energy that colliding particles must have for a reaction to take place; the word minimum is needed B1 rate of reaction is the change in the concentration of a reactant or a product per unit time B1 the units are mol dm⁻³ s⁻¹

Question 8

Explain why increasing the concentration of hydrochloric acid increases the rate of its reaction with magnesium ribbon at the same temperature.

Accepted Answer

B1 there are more hydrogen ions in the same volume of solution B1 so collisions between the ions and the magnesium surface are more frequent B1 the same fraction of collisions have at least the activation energy, so there are more successful collisions per unit time; reject 'the particles collide harder'

Question 9

Explain why powdered marble reacts faster with dilute hydrochloric acid than the same mass of marble in large chips.

Accepted Answer

B1 the powder has a larger surface area, so more calcium carbonate particles are exposed to the acid B1 so collisions between acid particles and the solid are more frequent B1 the energy of each collision is unchanged, so the larger number of collisions per unit time gives more successful collisions per unit time

Question 10

Explain why the rate of reaction between excess marble chips and dilute hydrochloric acid decreases as the reaction proceeds.

Accepted Answer

B1 the hydrochloric acid is used up as it reacts, so its concentration falls B1 there are fewer acid particles in each unit volume, so collisions with the marble are less frequent B1 there are fewer successful collisions per unit time, and the reaction stops when the acid, the limiting reactant, has been used up

Question 11

Calculate the activation energy of the reverse reaction, with and without the catalyst, from the enthalpy profile.

Accepted Answer

M1 the reverse reaction climbs from the products, which are 45 kJ mol⁻¹ below the reactants, to the same peak, so Ea(reverse) = Ea(forward) + 45 A1 without the catalyst, Ea(reverse) = 120 + 45 = 165 kJ mol⁻¹ A1 with the catalyst, Ea(reverse) = 80 + 45 = 125 kJ mol⁻¹; allow error carried forward from a wrong method, once

Question 12

Describe how a graph of the volume of gas produced against time is used to find the initial rate of a reaction.

Accepted Answer

B1 a smooth curve is drawn through the plotted points B1 a tangent is ruled that touches the curve at t = 0, extended across at least half of the grid B1 the gradient of the tangent, its rise divided by its run between two well-separated points, is the initial rate in cm³ s⁻¹

Question 13

Suggest why a sack of flour burns only slowly in a fire, but flour dust spread through the air of a mill can explode when a spark reaches it.

Accepted Answer

B1 dust has a far larger surface area than flour packed in a sack, and every particle is surrounded by air B1 collisions between oxygen molecules and flour particles are therefore far more frequent, so the rate of combustion is very high B1 the heat released by the exothermic reaction spreads the reaction to neighbouring particles faster than it is lost, so the whole cloud burns at once

Question 14

Compare the initial rate with the mean rate over the whole reaction for a reaction in which the tangent at t = 0 has a gradient of 2.0 cm³ s⁻¹ and the reaction finishes after 120 s, having produced 90 cm³ of gas.

Accepted Answer

B1 the mean rate is 90 ÷ 120 = 0.75 cm³ s⁻¹ B1 the initial rate is greater, about 2.7 times the mean rate B1 the rate is highest at the start, when the reactant concentration is highest, and falls as reactants are used up, so the mean includes the slow later stages

Question 15

Evaluate the claim that a reaction with ΔH = −300 kJ mol⁻¹ must be faster than a reaction with ΔH = −50 kJ mol⁻¹ under the same conditions.

Accepted Answer

B1 the claim is not valid: the rate depends on the activation energy, not on the enthalpy change B1 the fraction of collisions with at least the activation energy decides how many succeed, and ΔH does not fix the height of the barrier B1 methane burning has ΔH = −890 kJ mol⁻¹ yet does not react with air at room temperature until lit, so a very exothermic reaction can be very slow

Question 16

Suggest why a reaction between two large molecules, each with a single reactive group at one end, is slower than a reaction between two small molecules with the same activation energy, at the same concentrations and temperature.

Accepted Answer

B1 for the large molecules only collisions in which the two reactive groups meet can succeed, so a smaller fraction of collisions have the correct orientation B1 the activation energy is the same, so the fraction of collisions with enough energy is the same, and the lower rate comes from orientation

Measurement	Value
Volume of hydrochloric acid	25.0 cm³
Time interval	25 s
Mass lost in that time	0.22 g
Mr of CO₂	44.0