Edexcel IAL Chemistry Topic 9 Introduction to kinetics and equilibria

Move between rate data and particle explanations, annotate Maxwell–Boltzmann or profile diagrams precisely and distinguish equal rates from equal concentrations.

Exam points

Calculate rate from time or graph gradients and explain changes with collision theory.
Draw energy distributions and reaction profiles with Eₐ, ΔH and catalyst effects.
Explain dynamic equilibrium through equal forward and reverse rates in a closed system.

Covered syllabus points

9.1—Understand, in terms of the collision theory, the effect of changes in concentration, temperature, pressure and surface area 9.2—Reactions take place only when collisions have sufficient energy, known as the activation energy 9.3—The rate of a reaction from: i the time taken for a reaction 9.4—Qualitatively, in terms of the Maxwell-Boltzmann distribution of molecular energies, how changes in temperature affect 9.5—The role of catalysts in providing alternative reaction routes of lower activation energy 9.6—Draw the reaction profiles for uncatalysed and catalysed reactions 9.7—The use of catalysts in industry to make processes more sustainable by using less energy and/or higher atom economy 9.8—Interpret the action of a catalyst in terms of a qualitative understanding of the Maxwell-Boltzmann distribution 9.9—Many reactions are readily reversible and that they can reach a state of dynamic equilibrium in which: i the rate of 9.10—Predict and justify the qualitative effects of changes of temperature, pressure and concentration on the position 9.11—Evaluate data to explain the necessity, for many industrial processes, to reach a compromise between the yield and the rate

[Maximum number: 11]

Ammonium nitrate, $\mathrm{NH}_{4} \mathrm{NO}_{3}$ , is used in the manufacture of fertilisers and explosives. It is produced on a large scale using only methane, water and air.
The process has four stages.

(a)

The first two reactions in Stage 1 involve the production of hydrogen.

At temperature $T_{1}$ , methane reacts with excess steam to give hydrogen.

\mathrm{CH}_{4}(\mathrm{~g})+\mathrm{H}_{2} \mathrm{O}(\mathrm{~g}) \rightleftharpoons 3 \mathrm{H}_{2}(\mathrm{~g})+\mathrm{CO}(\mathrm{~g}) \quad \Delta H=+206 \mathrm{~kJ} \mathrm{~mol}^{-1}

At a different temperature, $T_{2}$ , the carbon monoxide reacts with more steam.

\mathrm{CO}(\mathrm{~g})+\mathrm{H}_{2} \mathrm{O}(\mathrm{~g}) \rightleftharpoons \mathrm{H}_{2}(\mathrm{~g})+\mathrm{CO}_{2}(\mathrm{~g}) \quad \Delta H=-42 \mathrm{~kJ} \mathrm{~mol}^{-1}

[ 2 ]

(i)

Give the reason why excess steam is used in the first reaction.

[ 1 ]

An answer that makes reference to one of the following points:

- shifts position of equilibrium to the right

OR

increases the (equilibrium) yield (of $\mathrm{H}_{2}$ )

Ignore to increase rate (of forward reaction) Ignore cheaper to have steam in excess Ignore to react with most of the $\mathrm{CH}_{4}$

Allow to increase yield (of CO / products)

Do not award so all of the $\mathrm{CH}_{4}$ reacts / so reaction goes to completion

Do not award to increase the moles of gas/pressure

(ii)

Predict which of $T_{1}$ and $T_{2}$ is the higher temperature.

Justify your answer.

[ 1 ]

An answer that makes reference to the following points:

- $T_{1}$ (is higher)

and

(first reaction is) endothermic

Accept reverse argument

Allow positive enthalpy change for endothermic Allow (first reaction) absorbs (heat) energy for endothermic

Ignore just +206 for endothermic

Ignore correct reference to effect of temperature on equilibrium yields

Do not award absorbs more energy to break (reactant) bonds

(b)

In Stage 2, the hydrogen from Stage 1 reacts with nitrogen (from the air) to produce ammonia. The conditions for this reaction are:
- a temperature of 700 K
- a pressure in the range $100-200 \mathrm{~atm}$
- an iron catalyst

\mathrm{N}_{2}(\mathrm{~g})+3 \mathrm{H}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NH}_{3}(\mathrm{~g}) \quad \Delta H=-92 \mathrm{~kJ} \mathrm{~mol}^{-1}

Give one advantage and one disadvantage of using a pressure of 200 atm, compared to a pressure of 100 atm , in Stage 2.

[ 2 ]

An answer that makes reference to the following points:

- advantage of using high pressure

- disadvantage of using high pressure

Examples of advantage: shifts position of equilibrium to right / products

OR increases (equilibrium) yield (of $\mathrm{NH}_{3}$ )

OR increases rate

OR increases occupation of catalyst active sites

Ignore any reference to collisions

(c)

The reaction in Stage 2 has an activation energy, $E_{\text {cat }}=+70 \mathrm{~kJ} \mathrm{~mol}^{-1}$ .

The uncatalysed reaction between $\mathrm{N}_{2}$ and $\mathrm{H}_{2}$ has an activation energy, $E_{\mathrm{a}}=+290 \mathrm{~kJ} \mathrm{~mol}^{-1}$ .

[ 4 ]

(i)

Complete the profile for the catalysed and uncatalysed reactions.

Label the activation energies and the enthalpy change of reaction, $\Delta H$ .
Your diagram must match the scale shown for the production of $\mathrm{NH}_{3}$ .

Figure for Question (i) — Edexcel A-Level Chemistry AS

[ 3 ]

An answer that makes reference to the following
points:

- $\Delta H$ labelled

and

arrow pointing downwards

(1)

- labelled reaction profiles for uncatalysed and catalysed reactions

(1)

- correct scale for activation energies

(1)

Example of labelled reaction profile:
$\mathrm{H} / \mathrm{kJ} \mathrm{mol}^{-1}$
reaction progress
Allow arrows to start/end within one small square of correct
placement and penalise incorrect placement once only
Allow -92 / 'enthalpy change' for $\Delta H$
Do not award double headed arrow
Allow any form of unambiguous labelling, eg values
Allow double headed arrows
Do not award downward arrows
Do not award $E_{\text {cat }}>E_{\text {a }}$
Accept accuracy of ± one small square
Ignore scale shown on y-axis

Figure for Answer to Question (i) — Edexcel A-Level Chemistry AS

(ii)

Suggest why the use of the catalyst makes Stage 2 more sustainable.

[ 1 ]

(d)

In Stage 3, nitrogen monoxide, NO, is produced in the reaction between $\mathrm{NH}_{3}$ (from Stage 2) and $\mathrm{O}_{2}$ (from the air). The conditions used are a temperature of 1100 K in the presence of a platinum-rhodium catalyst.

4 \mathrm{NH}_{3}(\mathrm{~g})+5 \mathrm{O}_{2}(\mathrm{~g}) \rightleftharpoons 4 \mathrm{NO}(\mathrm{~g})+6 \mathrm{H}_{2} \mathrm{O}(\mathrm{~g}) \quad \Delta H=-905 \mathrm{~kJ} \mathrm{~mol}^{-1}

[ 1 ]

(i)

Give one reason why a high temperature is needed in this reaction.

[ 1 ]

An answer that makes reference to one of the following points:

- increase rate

OR

rate is slow at low temperature

OR

catalyst does not work at low temperature

OR

so more reactants/collisions have $E \geq E_{\mathrm{a}}$

OR

to break O=O / N-H bonds

Do not award to increase yield

Do not award to shift position of equilibrium (to left / right)

Do not award reverse reaction is endothermic

Allow to increase the number of successful collisions Ignore to increase collision frequency

Allow catalyst more efficient at high temperature

Allow to activate the catalyst

Accept (to reach) high activation energy

Allow to break bonds in oxygen/ammonia/reactants

(e)

The NO from the first reaction in Stage 3 is cooled and then converted to nitrogen dioxide, $\mathrm{NO}_{2}$ , by reaction with more $\mathrm{O}_{2}$ .

2 \mathrm{NO}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NO}_{2}(\mathrm{~g})

Nitric acid, $\mathrm{HNO}_{3}(\mathrm{aq})$ , is produced by the addition of water.

3 \mathrm{NO}_{2}(\mathrm{~g})+\mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \rightarrow 2 \mathrm{HNO}_{3}(\mathrm{aq})+\mathrm{NO}(\mathrm{~g})

Explain how adding water in the second reaction affects the yield of $\mathrm{NO}_{2}$ in the first reaction.

[ 2 ]

An explanation that makes reference to the following points:

- $\mathrm{NO}_{2}$ removed (in second reaction)

- shifting position of equilibrium (in first reaction) to right and increasing the yield (of $\mathrm{NO}_{2}$ )

(1)

Allow (as) NO formed (in second reaction)

Ignore $\mathrm{HNO}_{3}$ is formed (in second reaction) Ignore reaction is irreversible Ignore $\mathrm{NO}_{2}$ dissolves

[Maximum number: 2]

A Maxwell-Boltzmann distribution curve for the particles present in a reaction mixture is shown. Number of particles with energy, E

Figure for Question Topic 9: Introduction to Kinetics and Equilibria question 2 — Edexcel A-Level Chemistry AS

(a)

Which statement is correct?

position X represents the mean energy of the particles

activation energy, $E_{\mathrm{a}}$ , is the mean energy of the particles that react

the area under the curve to the right of the activation energy, $E_{\mathrm{a}}$ , represents the number of particles with enough energy to react

adding a catalyst moves the activation energy, $E_{\mathrm{a}}$ , to the right

[ 1 ]

(b)

What happens to the distribution curve when the temperature of the gas is decreased?
□ A
□ B
□ C
□ D

Table for Question (b) — Edexcel A-Level Chemistry AS

[ 1 ]

[Maximum number: 7]

Rhubarb stems contain ethanedioic acid. Ethanedioic acid reacts with acidified potassium manganate(VII), $\mathrm{KMnO}_{4}$ , and decolorises the solution.

(a)

A student carried out an experiment using rhubarb juice as a source of ethanedioic acid.

The student used different volumes of rhubarb juice and a constant volume of manganate(VII) as shown. An indication of rate was calculated using $1 \div \mathrm{t}$ .

[ 3 ]

(i)

Complete the table.

Table for Question (i) — Edexcel A-Level Chemistry AS

[ 1 ]

807

υ.υσο-й

0.40

(1)

- boxes completed to 3SF

261

0.00383

3.83

, 40

ก กค.A.T

$1{ }^{m}$

(ii)

Explain why a rate of reaction is greater at higher concentrations.

[ 2 ]

(b)

The Maxwell-Boltzmann distribution for the particles in a reaction is shown.

[ 4 ]

(i)

Draw a curve to represent the distribution at a higher temperature.

[ 1 ]

a line that has a similar shape with peak that is
lower and to the right

Example of a drawing

Ignore shading

Do not award if the line cut the curve more once

Do not award if the line plateaus above the height of the $E_{\mathrm{a}}$ line Do not award if line does not start at the origin Ignore changes to the $E_{\mathrm{a}}$ line

(ii)

Explain why increasing the temperature increases the rate of reaction.

[ 3 ]

All question bank results loaded

Covered syllabus points

Topic 9: Introduction to Kinetics and Equilibria question 1

Question (a)

Question (i)

Question (ii)

Question (b)

Question (c)

Question (i)

Question (ii)

Question (d)

Question (i)

Question (e)

Topic 9: Introduction to Kinetics and Equilibria question 2

Question (a)

Question (b)

Topic 9: Introduction to Kinetics and Equilibria question 3

Question (a)

Question (i)

Question (ii)

Question (b)

Question (i)

Question (ii)