Unit 2 Equilibria, energetics and elements (F325) Rates ...UNIT+2+Module+1+Rat… · 1 | P a g e Unit 2 Equilibria, energetics and elements (F325) Rates, equilibrium and pH Topics

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Unit 2 Equilibria, energetics and elements (F325) Rates, equilibrium and pH

Topics covered in this module:

1. Rate of reaction

2. Measuring reaction rates

3. Orders and the rate equation

4. Half-lives

5. Orders from rate-concentration graphs

6. Initial rates and the rate constant

7. Rate-determining step

8. The equilibrium constant, KC

9. Calculations using KC

10. The equilibrium position and KC

11. The equilibrium constant, KC, and the rate constant, k

12. The road to acids

13. The role of H+ in reactions of acids

14. Conjugate acid-base pairs

15. What is pH?

16. Strong and weak acids

17. Calculating pH for strong and weak acids

18. The ionisation of water

19. pH values of bases

20. Buffer solutions

21. pH values of buffer solutions

22. Neutralisation – titration curves

23. Neutralisation – enthalpy changes

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1. Rate of reaction

Unit 2 – Equilibria, energetics and elements

Rates, equilibrium and pH Done

Explain and use the term rate of reaction.

Deduce the rate of a reaction from a concentration-time graph.

Key areas to concentrate on…

During your AS course you learned about rates:

how quickly a reaction reaches equilibrium based upon collision theory

how and why factors affect the rate in terms of particles

how to draw Boltzmann distribution curves

how these diagrams can be used to explain how a catalyst or a temperature change alters the rate of a reaction.

Now at A2, numbers become involved:

rate equations need to be worked out

orders determined

rate constants (with correct units) need to be calculated.

This can be done either by looking at:

concentration-time graphs and understanding the importance of a half-life

initial rate-concentration graphs

the importance of a rate-determining step needs to be understood.

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2. Measuring reaction rates



Plot a concentration-time graph from experimental results.

Deduce the rate of a reaction from a concentration-time graph.









orders determined






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3. Orders and the rate equation



Explain and use the terms order and rate constant.

Deduce a rate equation from orders.









orders determined






When you write down units, you should put positive indices first. Units are then written alphabetically. Don’t worry if you put negative indices first. In exams, you will still be awarded a mark provided the powers are correct.

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Past paper questions

Nitrogen dioxide is one of the major pollutants in air, formed by reaction of nitrogen monoxide with oxygen.

2NO(g) + O2(g) 2NO2(g)

(a) What is meant by the rate of reaction?

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[1]

(b) A series of experiments was carried out to investigate the kinetics of this reaction. The results are shown in the table below.

Experiment [O2]

/ mol dm–3

[NO]

/ mol dm–3

initial rate

/ mol dm–3 s–1

1 0.00100 0.00100 7.10

2 0.00400 0.00100 28.4

3 0.00400 0.00300 256

(i) For each reactant, deduce the order of reaction. Show your reasoning.

O2(g) .................................................................................................

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NO(g) ................................................................................................

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[4]

(ii) Deduce the rate equation for this reaction.

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[1]

[Turn over]

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(iii) Calculate the rate constant, k, for this reaction. State the units for k.

k = ................................... units ...................................

[2]

[Total 8 marks]

Examiner’s comments

(a) Explaining the rate of reaction caused candidates some difficulty and a number resorted to descriptions involving the tome at which a reaction takes place.

(b) (i) In general this was very well answered and there is no doubt that this topic is one that most candidates are able to master.

(ii) Again this very well answered. Very few candidates lost marks here and even those few who had failed to gain full marks in part (i) still gave a correct expression based on their previous answer.

(iii) This was undoubtedly a little more challenging than the preceding two parts but many candidates were able to calculate a correct value although fewer were able to work out the appropriate units.

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4. Half-lives



Explain and use the terms order and half-life.

Deduce the half-life of a first-order reaction from a concentration-time graph.

State that the half-life of a first order reaction is independent of the concentration.









orders determined






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5. Orders from rate-concentration graphs



Deduce the order (0, 1 or 2) with respect to a reactant from a rate-concentration graph.

Determine, using the initial rates method, the order (0, 1 or 2) with respect to a reactant.









orders determined






Remember, you do not need to include a reactant that is zero order in a rate equation. Any value to the power ‘0’ is equal to ‘1’.

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6. Initial rates and the rate constant



Deduce, from orders, a rate equation of the form rate = k[A]m[B]n.

Calculate the rate constant, k, from a rate equation.

Explain qualitatively the effect of temperature change on a rate constant and the rate of a reaction.









orders determined






Remember – the higher the temperature, the larger the value of the rate constant, k.

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One cause of low-level smog is the reaction of ozone, O3, with ethene, C2H4.

The smog contains methanal, HCHO(g).

The equation for methanal production is shown below.

O3(g) + C2H4(g) 2HCHO(g) + ½ O2(g)

The rate of the reaction was investigated, using a series of different concentrations of either C2H4(g) or O3(g), by measuring the initial rate of

formation of HCHO(g).

The results are shown below.

experiment [O3(g)]

/ 10–7mol dm–3

[C2H4(g)]

/ 10–8 mol dm–3

initial rate

/ 10–12 mol dm–3 s–1

1 0.5 1.0 1.0

2 2.0 1.0 4.0

3 4.0 2.0 16.0

(i) Analyse and interpret the results to deduce the order of reaction of each reactant and the rate equation.

Explain your reasoning.

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[Turn over]

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(ii) Calculate the value of the rate constant and state the units.

rate constant =.............................. units…….

[3]

(iii) Using the equation above, deduce the initial rate of formation of O2(g) in

experiment 1.

Explain your reasoning.

answer = ................................. mol dm–3 s–1

[1]

(iv) The experiment was repeated at a higher temperature.

How would the new conditions affect the rate of the reaction and the value of the rate constant?

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[1]

[Total 10 marks]

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Nitrogen monoxide reacts with hydrogen at 500 °C as in the equation below.

2NO(g) + 2H2(g) → N2(g) + 2H2O(g)

A series of experiments was carried out to investigate the kinetics of this reaction. The results are shown in the table below.

experiment [NO]

/ mol dm–3

[H2]

/ mol dm–3

initial rate

/ mol dm–3 s–1

1 0.10 0.20 2.6

2 0.10 0.50 6.5

3 0.30 0.50 58.5

In this question, one mark is available for the quality of spelling, punctuation and grammar.

(i) For each reactant, deduce the order of reaction. Show your reasoning.

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Quality of Written Communication [1]

(ii) Deduce the rate equation for this reaction.

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[1]

[Turn over]

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(iii) Calculate the rate constant, k, for this reaction. State the units for k.

k = .................................................... units ................................................

[3]

[Total 9 marks]


(i) Candidates are mostly very comfortable with problems involving the initial rates method and this part proved to be a relatively easy starter to the paper. Surprisingly, the second order factor of 3 for [NO] seemed to pose less difficulties than the first order factor of 2.5 for [H2], with some candidates instead vaguely stating ‘just over 2’.

Almost all candidates expressed themselves clearly and were awarded the mark for quality of written communication.

(ii) Most candidates used their results from (i) to construct a rate equation.

(iii) This calculation caused few problems to most candidates. When errors did occur, these were usually in the form of an incorrectly rearranged equation or with incorrect units.

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One cause of low-level smog is the reaction of ozone, O3, with ethene. The

smog contains methanal, CH2O(g), and the equation for its production is

shown below.

O3(g) + C2H4(g) 2CH2O(g) + O2(g) equation 1

(a) The rate of the reaction doubles when the initial concentration of either O3(g) or C2H4(g) is doubled.

(i) What is the order of reaction with respect to

O3 .............................

C2H4? .......................

[1]

(ii) What is the overall order of the reaction?

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[1]

(iii) Write the rate equation for this reaction.

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[1]

(b) For an initial concentration of ozone of 0.50 × 10–7 mol dm–3 and one of

ethene of 1.0 × 10–8 mol dm–3, the initial rate of methanal formation was

1.0 × 10–12 mol dm–3 s–1.

(i) How could the initial rate of methanal formation be measured from a concentration/time graph?

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[2]

(ii) Calculate the value of the rate constant and state the units.

rate constant =.......................... units....................................

[3]

[Turn over]

2

1

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(iii) The initial rate of methanal formation is different from that of oxygen formation in equation 1.

Explain why.

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[1]

(iv) The experiment was repeated but at a higher temperature. What would be the effect of this change on the rate and the rate constant of the reaction?

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[2]

[Total 11 marks]


Most candidates scored all three marks in this part with comparative ease.

In (i), the majority of candidates correctly stated that the gradient or tangent was required but it was surprising that many ignored the emphasised reference to the ‘initial’ rate. Weak candidates often responded in terms of half-life.

In (ii), many candidates correctly calculated the correct answer of 2000 dm3 mol–1 s–1. However, as in 1(c), it was disappointing that many candidates were unable to rearrange the rate equation and it was common to see the inverse value of 0.0005 in the final answer.

In (iii), most correctly responded in terms of the stoichiometry of methanal and oxygen.

Part (iv) was usually answered correctly although some candidates were reluctant to state that the reaction rate and the rate constant would both increase.

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The decomposition of dinitrogen pentoxide, N2O5, at 45 °C was investigated.

The reaction that takes place is shown below.

2N2O5 → 4NO2 + O2

In an experiment, N2O5 with a concentration of 0.60 mol dm–3 was

decomposed at 45 °C.

At this temperature, the reaction has a constant half-life of 1200 s.

(i) How can you tell that this reaction is first order with respect to N2O5?

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[1]

(ii) Write down an expression for the rate equation of this decomposition.

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[1]

[Turn over]

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(iii) Complete the graph below to show how the [N2O5] changes over the first

3600 s of the reaction.

[2]

[Turn over]

00

400

800

1200

160

02000

2400

28

00

3200

3600

0.1

0

0.2

0

0.3

0

0.4

0

0.5

0

0.6

0

time/s

[N2O5] / mol dm–3

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(iv) The rate of this reaction can be determined from the graph.

Show on the graph how the rate can be measured after 1200 s.

[1]

(v) The rate can also be calculated from the rate equation. The rate constant

for this reaction is 6.2 × 10–4 s–1.

Calculate the initial rate of this reaction. State the units.

rate =………….…..……. units………..….………..

[2]

[Total 7 marks]


(i) Very few candidates failed to be guided by the final statement of the question towards the correct answer.

(ii) This part was also extremely well answered but a handful of candidates suggesting that the reaction was second order, presumable by incorrectly using the stoichiometry in the balanced equation.

(iii) The graph was nearly always plotted correctly.

(iv) Nearly all candidates correctly drew a tangent to their curve at 1200 s.

(v) Again this part was well answered. It was noteworthy that of the candidates who had proposed a second order reaction in part (ii), many gave incorrect units here, presumably confusing rate with rate constant.

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7. Rate-determining step



Propose a rate equation that is consistent with the rate-determining step.

Propose the steps in a reaction mechanism from the rate equation and the balanced equation for the overall reaction.









orders determined






You are not expected to remember specific examples, e.g. NO2 + CO.

You are expected to be able to interpret data to identify a rate-determining step and to suggest possible steps in the mechanism for the reaction.

The overall balanced equation does not tell you anything about the reaction mechanism. For that we have to carry out rate experiments. Read this repeatedly until it is ingrained into your head! This is the whole crux of being able to predict a reaction mechanism for a multi-step reaction using a rate equation.

It is very unlikely that more than two molecules will be involved in each step of a reaction. Remember that a collision may lead to a reaction – the chances of more than two molecules colliding simultaneously are very slim! So try to restrict each step to either one molecule (which may decompose) or two molecules (which may react on collision). The overall equation is the sum of the equations from each step in the reaction mechanism.

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Nitrogen monoxide, NO, is involved in formation of ozone at low levels and the breakdown of ozone at high levels.

(i) In the lower atmosphere, NO is produced by combustion in car engines. Ozone is then formed following the series of reactions shown below.

NO(g) + 1/2O2(g) → NO2(g)

NO2(g) → NO(g) + O(g)

O2(g) + O(g) → O3(g)

Write the overall equation for this reaction sequence.

Identify the catalyst and justify your answer.

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[3]

(ii) In the upper atmosphere, NO removes O3 by the following reaction

mechanism.

NO(g) + O3(g) → NO2(g) + O2(g) slow

O(g) + NO2(g) → NO(g) + O2(g) fast

Suggest the rate equation for this process. Explain your reasoning.

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[2]

[Total 5 marks]


(i) Candidates found this part more difficult than anticipated and there was a reluctance to cancel species occurring on both sides of the equation. Some candidates changed ½O2(g) to O(g). Most candidates who identified NO as the catalyst were able to

explain its action as being used and then regenerated.

(ii) The majority were able to show a rate equation using the species in the slow step.

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Nitrogen dioxide reacts with carbon monoxide emitted from car exhausts in the following reaction.

NO2 + CO NO + CO2

The rate equation for this reaction is rate = k[NO2]2.

This is a multi-step reaction. The first step is the rate-determining step.

(i) What is meant by the rate-determining step?

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[1]

(ii) Suggest a two-step reaction mechanism for this reaction that is consistent with the kinetic data and the overall reaction.

[2]

[Total 3 marks]


(i) Only the very weakest candidate failed to describe correctly the rate-determining step as the slow stage in a multi-stage reaction.

(ii) This was a very challenging question and there were only a handful of candidates who achieved full marks here. It was a little depressing to see how many candidates suggested steps that were chemical nonsense and involved steps that were neither balanced nor added to give the overall reaction.

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In this question, one mark is available for the quality of use and organisation of scientific terms.

Propanone reacts with iodine in the presence of dilute hydrochloric acid.

A student carried out an investigation into the kinetics of this reaction.

He measured how the concentration of propanone changes with time. He also investigated how different concentrations of iodine and hydrochloric acid affect the initial rate of the reaction.

The graph and results are shown below.

[CH3COCH3]

/ mol dm–3

[I2]

/ mol dm–3

[H+]

/ mol dm–3

initial rate

/ mol dm–3 s–1

1.5 ×10–3 0.0300 0.0200 2.1 ×10–9

1.5 ×10–3 0.0300 0.0400 4.2 ×10–9

1.5 ×10–3 0.0600 0.0400 4.2 ×10–9

[Turn over]

00

[CH3 3COCH ]

time

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The overall equation for the reaction is given below.

CH3COCH3 + I2 → CH3COCH2I + HI

This is a multi-step reaction.

• What conclusions can be drawn about the kinetics of this reaction from the student’s investigation? Justify your reasoning.

• Calculate the rate constant for this reaction, including units.

• Suggest the equations for a possible two-step mechanism for this reaction. Label the rate-determining step and explain your reasoning.

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Quality of Written Communication [1]

[Total 14 marks]

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This question tested candidates’ ability to interpret data as well as their knowledge and understanding of chemical kinetics.

Most candidates are able to interpret initial rates data with ease and the majority

deduced, and explained. that the reaction is 1st order with respect to H+ and zero order with respect to I2. Many then deduced that the reaction was also zero order with respect

to propanone because its concentration had not been varied – the graphical data was often completely ignored. These students who did consider the graph often produced strange conclusions with zero, first and second order all being seen.

Consequently, it was common to see the incorrect expression ‘rate = k[H+]’ as well as the

correct ‘rate = k[H+][CH3COCH3]’. The expected correct calculated value was

7.0 × 10–5 dm3 mol–1 s–1 but the examiners marked this calculation consequentially on the candidate’s rate equation.

For the two-step mechanism, the examiners expected candidates to show two equations which, added together, gave the overall equation. The rate-determining step required reactants which were the species in the rate equation. Many of the candidates with correct rate equations rose to this challenge and many different two-step mechanism were credited. Unfortunately, many showed a rate-determining step involving propanone and iodine, presumably because these are the reactants in the overall equation. Only the best candidates answered this part well.

The mark for Quality of Written Communication was awarded for use of technical terms.

Overall, the open nature of this question discriminated extremely well.

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A student investigated the hydration of 2-methylpropene, (CH3)2C=CH2, with

dilute aqueous acid to form 2-methylpropan-2-ol, (CH3)3COH.

The following mechanism has been proposed for this hydration.

step 1 (CH3)2C=CH2 + H+(aq) → (CH3)3C+ rate determining step

step 2 (CH3)3C+ + H2O → (CH3)3COH + H+(aq)

(i) Step 1 is the rate-determining step for this hydration.

What is meant by the term rate-determining step?

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[1]

(ii) Write a balanced equation for the overall hydration reaction.

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[1]

(iii) Suggest the role of H+(aq) in this mechanism. Explain your reason.

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[2]

(iv) Use the mechanism above to suggest the rate equation for this hydration.

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[1]

[Total 5 marks]

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(i) Nearly all candidates correctly described what is meant by the term ratedetermining step.

(ii) This part proved a little more challenging and a number of candidates tried to

involve H+ in the overall reaction.

(iii) Nearly all candidates identified H+ as a catalyst and the majority of these were able

to explain why this was the case. For this second mark, most discussed the fact that H+

was not used up rather than giving the better answer that H+ was used in the first step of the reaction and regenerated in the second step.

(iv) This part proved to be more challenging. Although many did give the correct

equation, a significant number involved the concentration of H2O rather than H+ in their

proposed rate equation.

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8. The equilibrium constant, KC



Deduce expression for the equilibrium constant, KC, for homogeneous reactions.

Determine the units for KC.


Although equilibrium at A2 is still based upon le Chatelier’s principle and how factors affect the balance of a reaction – e.g. more reactants, less products, etc., you must now be able to:

calculate values for the equilibrium constant, KC, including the units

determine the effect of temperature change upon KC.

By convention, ‘reactants’ are shown on the left-hand side of the equilibrium expression. ‘Products’ are on the right-hand side of the equilibrium expression.

A KC expression always relates to a particular equilibrium equation.

For A2 level, you will only be asked about homogeneous equilibria. These will be either gaseous, in solution or a mixture of liquids.

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9. Calculations using KC



Calculate the value of an equilibrium constant, KC, including its units.

Calculate the concentration or quantities of substances present at equilibrium.





If all values have been given to, say, three significant figures, your answer should be expressed to three significant figures.

If you have a question where KC has no units (with the same total moles on both sides of the equilibrium), then the volume cancels. If you have a problem like this, then use V to represent the volume. So don’t panic if you haven’t been given the volume in such questions.

Remember that stoichiometry is the molar relationship between the relative quantities of substances taking part in a reaction. You get the stoichiometry of the reaction from the balanced numbers in the balanced chemical equation.

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10. The equilibrium position and KC



Understand the significance of KC values.

Explain the effect of changing temperature on the value of KC.





KC only changes with temperature.

KC increases as the temperature rises if the forward reaction is endothermic

(ΔH +ve).

KC decreases as the temperature rises if the forward reaction is exothermic

(ΔH -ve).

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Syngas is a mixture of carbon monoxide and hydrogen gases, used as a feedstock for the manufacture of methanol.

A dynamic equilibrium was set up between carbon monoxide, CO, hydrogen,

H2, and methanol, CH3OH, in a 2.0 dm3 sealed vessel.

The equilibrium is shown below.

CO(g) + 2H2(g) CH3OH(g)

The number of moles of each component at equilibrium is shown below

component CO(g) H2(g) CH3OH(g)

number of moles at equilibrium 6.20 × 10–3 4.80 × 10–2 5.20 × 10–5

(a) State two features of a system that is in dynamic equilibrium.

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[2]

(b) (i) Write an expression for Kc for this equilibrium system.

[1]

(ii) Calculate Kc for this equilibrium. State the units.

Kc = ……………………... units:…………..…..

[4]

[Turn over]

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(c) The pressure was increased whilst keeping the temperature constant. The mixture was left to reach equilibrium.

The equilibrium position above shifted to the right.

(i) Explain why the equilibrium position shifted to the right.

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[1]

(ii) What is the effect, if any, on the value of Kc?

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[1]

(d) The temperature was increased whilst keeping the pressure constant. The mixture was left to reach equilibrium.

The value of Kc for the equilibrium above decreased.

(i) Explain what happened to the equilibrium position in the equilibrium.

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[1]

(ii) Deduce the sign of the enthalpy change for the forward reaction shown in the equilibrium above. Explain your reasoning.

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[1]

(e) Methanol can be used as an additive to petrol.

(i) Write an equation for the complete combustion of methanol, CH3OH.

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[1]

(ii) Suggest why methanol is added to petrol.

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[1]

[Total 13 marks]

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11. The equilibrium constant, KC, and the rate constant, k.



State that the value of KC is unaffected by changes in concentration, pressure or the presence of a catalyst.

Make predictions for shifts in equilibrium position from concentration and pressure changes.

Understand that compromise conditions rely on a balance between KC and k.





Rate constants are calculate from experimental data only.

Equilibrium constants are calculated from the overall balanced equation for a reaction.

Read these two lines repeatedly until ingrained into your head!

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When heated, phosphorus pentachloride, PCl5, dissociates.

PCl5(g) PCl3(g) + Cl2(g)

A chemist placed a mixture of the three gases into a container. The initial

concentration of each gas was the same: 0.30 mol dm–3. The container was left until equilibrium had been reached.

Under these conditions, Kc = 0.245 mol dm–3.

(a) Write an expression for Kc for this equilibrium.

[1]

(b) Use the value of Kc for this equilibrium to deduce whether the

concentration of each gas increases, decreases or stays the same as the mixture approaches equilibrium.

(i) Show your answer by placing a tick in the appropriate cells in the table below.

initial concentration

/ mol dm–3

greater than

0.30 mol dm–

3

less than

0.30 mol dm–3

equal to

0.30 mol dm–3

PCl5 0.30

PCl3 0.30

Cl2 0.30

[1]

(ii) Explain your deduction.

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(c) The chemist compressed the equilibrium mixture at constant temperature and allowed it to reach equilibrium under these new conditions.

(i) Explain what happens to the value of Kc.

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[1]

(ii) Explain what happened to the composition of the equilibrium mixture.

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[2]

(d) The chemist heated the equilibrium mixture and the equilibrium moved to the left.

(i) Explain what happens to the value of Kc.

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[1]

(ii) Explain what additional information this observation reveals about the reaction.

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[Total 9 marks]

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(a) The Kc expression was correctly shown by all but the very weakest candidates.

(b) (i) It was pleasing to see how many candidates were able to tackle this part successfully.

(ii) The best candidates were able to explain that the system would be out of equilibrium and that concentrations would adjust to generate a ratio with the correct Kc value. Many explanations, however, stated that the equilibrium would move to the left because the Kc value is less than 1. This suggests that some candidates had obtained the correct answer to part (i) using flawed logic.

(c) (i) The majority correctly stated that Kc would not change but the examiners also required some statement that Kc only changes when the temperature changes.

(ii) Most candidates correctly stated that the increase in pressure would be opposed with the equilibrium moving to the left as the side with fewer gaseous molecules.

(d) (i) Again, the majority correctly identified that Kc would decrease.

(ii) It was pleasing to see how many candidates could correctly deduce that the forward reaction must be exothermic and that the equilibrium position would move to the left to oppose the temperature increase.

37 | P a g e


The preparation of hydrogen iodide, HI(g), from hydrogen and iodine gases is a reversible reaction which reaches equilibrium at constant temperature.

H2(g) + I2(g) 2HI(g)

(a) Write the expression for Kc for this equilibrium.

[1]

(b) A student mixed together 0.30 mol H2(g) with 0.20 mol I2(g) and the

mixture was allowed to reach equilibrium. At equilibrium, 0.14 mol H2(g)

was present.

(i) Complete the table below to show the amount of each component in the equilibrium mixture.

component H2(g) I2(g) HI(g)

initial amount / mol 0.30 0.20 0

equilibrium amount / mol

[2]

(ii) Calculate Kc to an appropriate number of significant figures. State

the units, if any.

Kc = .......................................

units, if any .......................................................................................

[3]

[Turn over]

38 | P a g e

(c) The student compressed the equilibrium mixture so that its volume was reduced. The temperature was kept constant.

Comment on the value of Kc and the composition of the equilibrium

mixture under these new conditions.

....................................................................................................................

....................................................................................................................

....................................................................................................................

....................................................................................................................

[2]

[Total 8 marks]


(a) Candidates find Kc expressions easy to write and this was no exception with almost

all candidates writing the correct expression.

(b) (i) The majority could not complete the table despite being given 4 out of 6 concentrations. Predictably, the commonest mistake was an equilibrium amount of 0.16 mol HI, instead of 0.32 mol, obtained by ignoring the stoichiometry in the equation.

(ii) Calculations were marked using the candidate’s tabular data. Most candidates chose 3 significant figures as appropriate, despite all information in the question being given to 2 significant figures. Consequently the required answer of 18 was seen very rarely. The commonest candidate responses seen were 18.3 and 4.57 (obtained from 0.16 mol HI). Almost all candidates identified that this Kc value would have no units.

(c) This was well answered with almost all candidates securing both marks. The question was easier than it might have been as there is the same number of moles on either side of the equation. Consequently, it was felt that many candidates had obtained the correct answer almost accidentally.

39 | P a g e

12. The road to acids



Describe an acid as a proton donor and a base as a proton acceptor.

Understand that scientific knowledge is always evolving.


Here, there is some overlap with AS Unit F321: Module 1 - Atoms and reactions, since key points still include the definition of acids as proton donors and ionic equations of acid reactions.

The Brønsted-Lowry theory of acids and bases is used in this A2 course. You must be comfortable with this idea of acids and bases in terms of proton transfer. It underpins all the work in this section.

40 | P a g e


Chocolate mousse contains gelatine and a compound to promote fast setting of the mousse.

Compound A is such a setting agent. It has two acidic hydrogen atoms per molecule and is one of the six acids listed below.

oxalic acid HOOCCOOH

malonic acid HOOCCH2COOH

succinic acid HOOC(CH2)2COOH

glutaric acid HOOC(CH2)3COOH

adipic acid HOOC(CH2)4COOH

pimelic acid HOOC(CH2)5COOH

The student analysed a sample of compound A by titration.

The student dissolved 2.82 g of compound A in water and made the solution

up to 250 cm3 in a volumetric flask. He titrated 25.0 cm3 of this solution with

0.175 mol dm–3 NaOH.

22.05 cm3 of NaOH were required for complete neutralisation.

Use the results of the student’s analysis to identify compound A from the list above.

Show all of your working.

[Total 5 marks]

41 | P a g e


It was impressive to see how many candidates tackled this extended calculation in a series of logical steps. Of the errors that were made, only a few candidates failed to scale by a factor of ten. The most common mistake was to ignore the stoichiometry of the equation and so not to realise that the amount of A would be half the amount of the sodium hydroxide. As in question 2(d), candidates did not use the information in the question. In this case candidates are told that A contains two acidic hydrogen atoms per molecule. Even when the answer did not match the relative formula mass of any of the acids, these candidates were still unable to realise their mistake.

42 | P a g e


This question looks at different compounds used in medicine.

(a) Nitrous oxide, N2O, is the gas used as a general anaesthetic.

(i) What is the oxidation number of nitrogen in nitrous oxide?

...........................................................................................................

[1]

(ii) Suggest a ‘dot-and-cross’ diagram for nitrous oxide. Show outer electrons only. The sequence of atoms in a nitrous oxide molecule is N N O.

[1]

(b) The structure of the painkiller ibuprofen is shown below.

(i) Determine the molecular formula of ibuprofen.

[1]

[Turn over]

O

OH

43 | P a g e

(ii) Suggest a chemical that would react with a solution of ibuprofen to produce a gas. Name the gas produced and write a balanced equation for the reaction.

chemical ...........................................................................................

gas ...................................................................................................

equation

[2]

(c) Lidocaine, C13H20N2O2, is used as a local anaesthetic in dentistry.

Lidocaine is administered by syringe as a solution containing 100 mg in

5.00 cm3.

Calculate the concentration, in mol dm–3, of lidocaine in the syringe.

concentration = ......................mol dm–3

[3]

(d) Eugenol is used as a painkiller in dentistry. It is an organic compound of C, H and O.

A sample of 1.394 g of eugenol was analysed by burning in oxygen to form 3.74 g of CO2 and 0.918 g of H2O. The relative molecular mass of

eugenol was shown to be 164 using a mass spectrometer.

Calculate the molecular formula of eugenol.

[5]

[Total 13 marks]

44 | P a g e


(a) (i) Virtually all candidates stated correctly that the oxidation number of nitrogen in N2O is +1.

(ii) Candidates found this part far more challenging. The examiners credited diagrams with an octet of electrons around each atom.

(b) (i) The responses here were rather disappointing with many candidates deriving the wrong number of H atoms, usually 17 or 20.

(ii) Many candidates correctly chose a suitable chemical with associated gas. The commonest chemicals seen were Na, NaHCO3, CaCO3 and Na2CO3. Candidates responding

with Na often showed a correct equation but those choosing CaCO3 often created

problems with the formula of the salt (usually incorrectly derived from Ca+).

(c) Most candidates secured all three marks here. Most candidates derived the relative molecular mass of lidocaine as 236 and the main difficultly arose from conversion of 100 mg to 0.1 g.

Some candidates wasted a mark by excessive rounding of the calculated concentration to 0.08 g.

(d) This part discriminated extremely well. Most candidates made some headway with the problem and it was common to see the amounts of CO2 and H2O calculated as 0.085

mol and 0.051 mol respectively for two marks. The best candidates almost always obtained the correct molecular formula of C10H12O2 for the full five marks.

45 | P a g e

13. The role of H+ in reactions of acids



Illustrate the role of H+ in the reactions of acids with carbonates, bases, alkalis and metals.


Here, there is some overlap with AS Unit F321: Module 1 - Atoms and reactions, since key points still include the definition of acids as proton donors and ionic equations of acid reactions.

The H+ ion (proton) is the active ingredient in acids. H+ is responsible for typical

acid-base reactions.

Neutralisation reactions between acids and bases are very common. Make sure you learn:

acid + metal carbonate metal salt + carbon dioxide + water

acid + base salt + water

acid + alkali salt + water

(extra one) acid + ammonia ammonium salt

Each one of these involves ionic equations which you need to practice writing for you exam.

Redox reactions between acids and metals are very common. Make sure that you learn:

acid + metal metal salt + hydrogen

46 | P a g e


In its reactions, sulphuric acid, H2SO4, can behave as an acid, an oxidising

agent and as a dehydrating agent.

The displayed formula of pure sulphuric acid is shown below.

Dilute sulphuric acid takes part in the typical acid reactions, reacting with metals, carbonates and bases.

Write balanced equations for the reaction of sulphuric acid with

a metal, ...............................................................................................................

a carbonate, ........................................................................................................

a base. ................................................................................................................

[Total 3 marks]


Overall, there were some very good responses to this part with average to above candidates usually constructing at least 2 out of the three equations. Common mistakes arose when candidates chose to use sodium carbonate and hydroxide, with NaSO4 being

seen often as one of the products. Surprisingly, and perhaps sensibly for many, acid salts were seen on many occasions and this approach was credited as correct chemistry.

O

O

S

O

O

H

H

47 | P a g e

14. Conjugate acid-base pairs



Describe and use the term conjugate acid-base pairs.


Here, there is some overlap with AS Unit F321: Module 1 - Atoms and reactions, since key points still include the definition of acids as proton donors and ionic equations of acid reactions. However, at A2 you will need to be able to express conjugate acid-base pairs.

48 | P a g e

15. What is pH?



Define pH as pH = – log10[H+

(aq)].

Define [H+(aq)] = 10-pH.

Convert between pH values and [H+(aq)].


Calculations will feature heavily in your exam and could include:

Ka and pKa of weak acids

pH of strong acids, weak acids and strong bases.

In exams, it is perfectly acceptable (and a lot easier) to define pH as

pH = – log[H+(aq)] rather than trying to express this formula as a sentence.

49 | P a g e

The standard measurement for determining whether a solution is acid or alkali was

developed in the Carlsberg Laboratory.

The pH scale, the standard measurement of acidity, was developed by the head of

Carlsberg Laboratory’s Chemical Department in 1909.

Dr Søren Sørensen (1868-1939) developed the pH scale during his pioneering research into

proteins, amino acids and enzymes - the basis of today’s protein chemistry. Basically

meaning ‘the power of hydrogen’, the scale provides a simple and universal measurement

of the amount of hydrogen ions in a solution, which affects its acidity and how it reacts

chemically.

The concentration of hydrogen ions is determined by measuring the current generated in an

electrochemical cell when the ions migrate to oppositely charged electrodes. Sørensen

used a negative logarithm of the hydrogen concentration to create a scale from 0-14, where

a pH of less than 7 is an acid, 7 is neutral and higher than 7 is an alkali. So water has a pH

of 7, lemon juice 2.4 and bleach 12.5. The pH of beer is 4.5.

The applications of the pH scale have been countless, ranging from foodstuffs and

cosmetics to chemicals and pharmaceuticals. Just about every liquid has had its pH

measured at some time to determine how it will react and interact with living organisms.

http://www.carlsberggroup.com/Company/Research/Pages/pHValue.aspx

http://www.carlsberggroup.com/Company/Research/Pages/pHValue.aspx

50 | P a g e

16. Strong and weak acids



Explain qualitatively the differences between strong and weak acids.

Explain that the acid dissociation constant, Ka, shows the extent of acid dissociation.

Deduce expressions for Ka and pKa for weak acids.


Here, there is some overlap with AS Unit F321: Module 1 - Atoms and reactions, since key points still include the definition of acids as proton donors and ionic equations of acid reactions. However, at A2 you will need to be able to work out the difference between strong and weak acids.




51 | P a g e


An excess of magnesium was added to 100 cm3 of 0.0450 mol dm–3 hydrochloric acid.

The same mass of magnesium was added to 100 cm3 of 0.0450 mol dm–3 ethanoic acid.

Both reactions produced 54 cm3 of hydrogen gas, measured at room temperature and pressure, but the reaction with ethanoic acid took much longer to produce this gas volume.

Explain why the reactions produced the same volume of a gas but at different rates.

Use equations in your answer.

.............................................................................................................................

.............................................................................................................................

.............................................................................................................................

.............................................................................................................................

.............................................................................................................................

.............................................................................................................................

.............................................................................................................................

.............................................................................................................................

.............................................................................................................................

[Total 4 marks]


On a positive note, most candidates realised that the answer was linked to the different strengths of the two acids and so their different degrees of ionisation. As a result this mark was awarded the most frequently. Fewer were the candidates who were able to explain why the volume of hydrogen would be the same in each case. When it came to the equations a large number of candidates were able to give a correct equation for the reaction of magnesium with hydrochloric acid, but very few were able to do the same for the reaction with ethanoic acid. The most common error was to write CH3COOMg. This is

an extremely serious error as within two lines candidates are happy to change the valence of magnesium. More serious were the number of candidates who gave the formula of magnesium chloride as MgCl. Again it is depressing to think that candidates who can handle a reasonably sophisticated calculation are then unable to write the correct formula for magnesium chloride.

52 | P a g e


The Ka values for three acids are shown in the table below.

acid Ka / mol dm–3

ethanoic acid CH3COOH 1.70 ×10–5

phenol C6H5OH 1.28 ×10–10

sulphurous acid H2SO3 1.50 ×10–2

(a) What information is provided by Ka values?

....................................................................................................................

....................................................................................................................

[1]

(b) When sulphurous acid and ethanoic acid are mixed together, an acid-base reaction takes place.

H2SO3(aq) + CH3COOH(aq) HSO3–(aq) + CH3COOH2

+(aq)

....................... ....................... ....................... ......................

(i) In the spaces above

• label one conjugate acid-base pair as acid 1 and base 1, • label the other conjugate acid-base pair as acid 2 and base 2.

[2]

(ii) Predict and explain the acid-base reaction that would take place if ethanoic acid were mixed with phenol. Include an equation in your answer.

...........................................................................................................

...........................................................................................................

...........................................................................................................

...........................................................................................................

[2]

[Turn over]

53 | P a g e

(c) The pH value of 0.0450 mol dm–3 hydrochloric acid is different from that

of 0.0450 mol dm–3 ethanoic acid.

Calculate the pH values of these two acids. Show all your working.

[5]

[Total 10 marks]


(a) (i) This was well answered with most candidates making the link between the strength of an acid and the size of the acid dissociation constant.

(b) (i) Again the majority of candidates picked up both marks here.

(ii) The vast majority of candidates realised that one acid would protonate the other although many chose phenol as the acid rather than ethanoic acid.

(c) The quality of answers to this part was extremely pleasing. Even weaker candidates who scored poorly in the rest of the paper had clearly learned how to do these calculations and so many gained full marks here.

54 | P a g e

17. Calculating pH for strong and weak acids



Calculate pH from [H+(aq)] and [H+

(aq)] from pH for strong and weak acids.

Calculate [H+(aq)] from pH for strong and weak acids.

Calculate Ka for a weak acid.





The key to calculating the pH of a weak acid, HA, is to find [H+(aq)].

[H+(aq)] depends on [HA(aq)] and Ka.

For a weak acid HA, [H+(aq)] =√𝐾a x [HA(aq)]

When working out the square root, use brackets around the expression being square-rooted. If you don’t do this, you will only take the square root of the first value.

How accurately should you show pH values? A typical pH meter shows pH values to two decimal places. The whole number before the decimal place is the logarithmic way of showing powers of 10. In terms of significant figures, pH starts after the decimal place. So a pH of 2.66 is to two significant places only! So don’t ‘round’ pH values to one decimal place as this is only one significant figure.

You may find this explanation a little tricky to understand, so as a general rule just write pH values to two decimal places.

55 | P a g e


Hydroiodic acid, HI(aq), is a strong acid that is an aqueous solution of

hydrogen iodide. In the laboratory, hydroiodic acid can be prepared by the method below.

A mixture of 480 g of iodine and 600 cm3 of water was put into a flask. The mixture was stirred and hydrogen sulphide gas, H2S(g), was bubbled

through for several hours.

The mixture became yellow as sulphur separated out. The sulphur was filtered off and the solution was purified by fractional distillation. A fraction of

HI(aq) was collected containing 440 g of HI in a total volume of 750 cm3.

(i) Construct a balanced equation, with state symbols, for the preparation of hydroiodic acid.

...........................................................................................................

[2]

(ii) Determine the percentage yield of hydroiodic acid.

[3]

(iii) Calculate the pH of the hydroiodic acid fraction.

[2]

[Total 7 marks]

56 | P a g e


(i) Despite all species and states being provided in the experimental method provided, construction of the equation caused candidates more problems than anticipated. Many ignored state symbols or simply got them wrong. Sometimes, SO2 was

shown as a product instead of sulphur and it was not uncommon to see O2. S2 was seen a

surprising number of times. S or S8 were expected for sulphur but S8 was never seen.

Many candidates included a superfluous H2O on both sides of the equation and this

should have been cancelled out.

(ii) The calculation presented problems for many candidates. An alarming number took the Mr of I2 as 127 instead of 254. These candidates often produced a % yield of HI by

dividing the moles of I2 by the moles of HI! The weakest candidates used the masses 440

and 480 instead of moles and many used volumes. As a result, the expected answer, 91.0%, was seen less often than expected although most candidates did score at least some of the three available marks.

(iii) Although most candidates correctly calculated the molar concentration of HI, the subsequent calculated negative pH of –0.66 confused some candidates, who then decided to ignore the negative sign.

57 | P a g e


Methanoic acid, HCOOH, is a weak organic acid which occurs naturally in ants and stinging nettles.

(a) Use an equation for the dissociation of methanoic acid to show what is meant by a weak acid.

....................................................................................................................

....................................................................................................................

[1]

(b) A 1.50 × 10–2 mol dm–3 solution of HCOOH has [H+] = 1.55 × 10–3 mol

dm–3.

(i) Calculate the pH of this solution and give one reason why the pH scale is a more convenient measurement for measuring acid

concentrations than [H+].

...........................................................................................................

...........................................................................................................

...........................................................................................................

[2]

(ii) Write the expression for Ka for methanoic acid.

[1]

(iii) Calculate the values of Ka and pKa for methanoic acid.

[3]

[Turn over]

58 | P a g e

(iv) Estimate the percentage of HCOOH molecules that have dissociated in this aqueous solution of methanoic acid.

[1]

[Total 8 marks]


(a) This caused problems for very few candidates. The main problems seen were ionisation of the wrong acid, use of HA or omission of an equilibrium sign.

(b) With modern calculators, pH calculations no longer cause problems for candidates.

(i) Most candidates easily calculated the pH as 2.81 and there were many good attempts at explaining the convenience of a logarithmic scale over one with a large range of values, most with negative indices.

(ii), (iii) Most candidates could give the expression for Ka and then go on to calculate Ka as

1.60 × 10–4 mol dm–3 using the stock [H+]2/[HA] approximation. The relationship between Ka and pKa was well known and created few problems with most candidates obtaining the

correct value of 3.80.

(iv) The dissociation of the acid was a novel question but it was well answered by the better candidates with most obtaining 10% or 10.3% as the correct answer.

59 | P a g e

18. The ionisation of water



State and use the expression for the ionic product of water, Kw.

Understand the importance of Kw in controlling the concentrations of H+

(aq) and OH-(aq) in aqueous solutions.


Calculations will feature heavily in the exam and could include:



Provided you know [H+(aq)], you can always find [OH-

(aq)] using:

Kw = [H+(aq)] [OH-

(aq)]

= 1.00 x 10-14 mol2 dm-6

This value is provided in exams on the Data Sheet.

60 | P a g e

19. pH values of bases



Calculate pH from [H+(aq)] for strong bases, using Kw.

Calculate [H+(aq)] from pH for strong bases, using Kw.


Calculations will feature heavily in the exam and could include:



Remember that an alkali is a soluble base that releases hydroxide ions, OH-.

61 | P a g e


In sewage plants, biological activity can be reduced by increasing the pH of the water. This is achieved by adding small amounts of solid calcium hydroxide, Ca(OH)2, to the sewage water.

In all parts of this question, assume that measurements have been made at 25 °C.

(a) The pH of aqueous solutions is determined by Kw.

Kw has a value of 1.0 × 10–14 mol2 dm–6 at 25 °C.

(i) What name is given to Kw?

...........................................................................................................

[1]

(ii) Write the expression for Kw.

...........................................................................................................

[1]

(b) A chemist checked the concentration of aqueous calcium hydroxide,

Ca(OH)2, in the sewage water by titration with 5.00 × 10–3 mol dm–3

hydrochloric acid.

Ca(OH)2(aq) + 2HCl(aq) → CaCl2(aq) + 2H2O(l)

The chemist titrated 25.0 cm3 of the sewage water with 21.35 cm3 of HCl to reach the endpoint of the titration.

Calculate the concentration, in mol dm–3, of the calcium hydroxide in the sewage water.

concentration = .......................................... mol dm–3

[3]

[Turn over]

62 | P a g e

(c) The chemist analysed a sample of water from another part of the sewage works and he found that the calcium hydroxide concentration was 2.7 ×

10–3 mol dm–3.

When solid calcium hydroxide dissolves in water, its ions completely dissociate.

Ca(OH)2(s) → Ca2+(aq) + 2OH–(aq)

Calculate the pH of this sample.

[3]

(d) After further treatment, the water could be used for drinking. In the

drinking water produced, the OH– concentration was 100 times greater

than the H+ concentration.

What was the pH of this drinking water?

[1]

[Total 9 marks]

63 | P a g e


(a) (i) Surprisingly few candidates knew that Kw is called the ionic product of

water. For some able candidates, this was the only mark dropped in question 3.

(ii) Most candidates correctly showed the expression for Kw. The examiners accepted

expressions including either H+ or H3O+. The commonest mistake was division by [H2O].

(b) Above average candidates solved this unstructured problem with comparative ease. However, below average candidates often got completely lost with the numbers in the stem often being used at random.

(c) Candidates found this part more difficult and only the best identified that [OH–] would be twice that of Ca(OH)2. The correct answer of pH = 11.73 was seen only from the

best candidates with many candidates producing 11.43 by not using the factor of 2 above.

It was disappointing to see how many candidates just calculated the –log value of [OH–] to give an answer of 2.27, an impossible pH value for an alkaline solution.

(d) This part foxed all but the very best candidates. The answer of 8 tested real understanding of pH and could be deduced from logic rather from lengthy calculations that were often seen, usually resulting in an incorrect wrong answer.

64 | P a g e

20. Buffer solutions



Describe what is mean by a buffer solution.

State that a buffer solution can be made from a weak acid and a salt of the weak acid.

Explain the role of the conjugate acid-base pair in an acid buffer solution.


You will need to explain what is meant by the term buffer. You will need to know how to:

make buffers

explain how they work

calculate the pH of a buffer solution.

The weak acid, HA, removes most of any added alkali. The conjugate base, A-, removes most of any added acid.

Added acid and added alkali shift the buffer equilibrium in opposite directions.

Make sure that you are able to explain the role of each component in a buffer system. Also remember that the buffer cannot remove all of any added acid or alkali, but pH changes are minimised.

The mode of action of a buffer solution is often tested in exams.

65 | P a g e


Hexylresorcinol is an antiseptic used in solutions for cleansing wounds and in mouthwashes and throat lozenges.

The structure of hexylresorcinol is shown below.

Identify a compound that could be added to hexylresorcinol to make a buffer solution. Explain your answer.

[Total 2 marks]


Many candidates identified that the phenoxide would be required and the commonest responses seen were mono- or di-sodium salts, sodium or sodium hydroxide. These were all credited. Comparatively few then went on the state that this would produce a weak acid and its conjugate base.

OH

OH

CH2(CH2 4 3) CH

66 | P a g e


‘Superphosphate’ fertilisers contain calcium dihydrogenphosphate, Ca(H2PO4)2. This compound is one of the world's most important fertilisers.

When dissolved in water, Ca(H2PO4)2 dissociates forming H2PO4– ions which

are easily taken up by plants.

(a) Calcium dihydrogenphosphate, Ca(H2PO4)2, is produced by treating rock

phosphate, containing Ca3(PO4)2, with sulphuric acid, H2SO4.

Write a balanced equation for this reaction.

....................................................................................................................

[1]

(b) Aqueous H2PO4– ions can act as a weak acid.

Write an equation to represent the dissociation of the H2PO4– ion.

....................................................................................................................

[1]

(c) The H2PO4– ion can act as either an acid or a base.

(i) State the formula of the conjugate base of H2PO4–.

...........................................................................................................

[1]

(ii) State the formula of the conjugate acid of H2PO4–.

...........................................................................................................

[1]

(iii) A solution of calcium dihydrogenphosphate, Ca(H2PO4)2, in water

acts as a buffer solution.

Suggest, with the aid of equations, how this buffering action takes place.

...........................................................................................................

...........................................................................................................

...........................................................................................................

...........................................................................................................

...........................................................................................................

[3]

[Total 7 marks]

67 | P a g e


(a) Candidates found great difficulty in constructing this equation. The mistake seen on many papers was a formula of Ca2SO4 for calcium sulphate.

(b) Good candidates successfully applied their understanding of acids to show a correct equilibrium. Many however seemed to use species rather at random and it was disappointing to see the number of response with different total charges on either side of the equilibrium, with an array of impossible species.

(c) (i), (ii) Candidates produced muddled responses here, repeating the problems of part (b). Part (ii) was correct more often than part (i).

(iii) Most candidates found this part beyond them, often deciding to write instead about buffers in general. In contrast, the best candidates applied their understanding well and often secured all three available marks.

68 | P a g e

21. pH values of buffer solutions



Calculate the pH of a buffer solution from the Ka value of a weak acid and the equilibrium concentrations of the conjugate acid-base pair.

Explain the role of the carbonic acid-hydrogen carbonate ion system as a buffer in the control of blood pH.


You will need to explain what is meant by the term buffer. You will need to know how to:

make buffers

explain how they work

calculate the pH of a buffer solution.

Get plenty of practice at pH buffer calculations.

For buffer calculations, learn:

Method 1: [H+(aq)] = Ka x

[HA(aq)]

[A−(aq)] ; then pH = -log10[H

+(aq)]

or Method 2: pH = pKa + log [A−(aq)]

[HA(aq)]

69 | P a g e


Phenol, C6H5OH, is a powerful disinfectant and antiseptic.

Phenol is a weak Brønsted–Lowry acid.

C6H5OH(aq) H+(aq) + C6H5O−(aq) Ka = 1.3 × 10–10 mol dm–3

Define the following terms:

(i) A Brønsted–Lowry acid,

....................................................................................................................

[1]

(ii) A weak acid.

....................................................................................................................

[1]

When phenol is mixed with aqueous sodium hydroxide, an acid–base reaction takes place.

C6H5OH(aq) + OH–(aq) C6H5O–(aq) + H2O(l)

...................... ..................... ....................... .....................

In the available spaces,

• label one conjugate acid–base pair as acid 1 and base 1,

• label the other conjugate acid–base pair as acid 2 and base 2.

[1]

A solution of phenol in water has a concentration of 4.7 g dm–3.

(i) Write an expression for the acid dissociation constant, Ka, of phenol.

[1]

(ii) Calculate the pH of this solution of phenol.

[5]

[Turn over]

70 | P a g e

As part of an investigation, a student needed to prepare a buffer solution with a pH value of 8.71. From the Ka value of phenol, the student thought that a

mixture of phenol and sodium phenoxide could be used to prepare this buffer solution.

The student decided to use a 0.200 mol dm–3 solution of phenol, mixed with an equal volume of sodium phenoxide.

Use your knowledge of buffer solutions to determine the concentration of sodium phenoxide solution that the student would need to mix with the 0.200

mol dm–3 phenol solution.

[3]

[Total 12 marks]

71 | P a g e


A student carried out an investigation with aqueous solutions of nitric acid, sodium hydroxide, ethanoic acid and water.

Nitric acid, HNO3, is a strong Brønsted-Lowry acid.

(i) Explain what is meant by a strong acid and a Brønsted-Lowry acid.

....................................................................................................................

....................................................................................................................

....................................................................................................................

[2]

(ii) What is the conjugate base formed from HNO3?

....................................................................................................................

[1]

The student diluted 0.015 mol dm–3 nitric acid with an equal volume of water and measured the pH of the diluted acid at 25 °C.

(i) Calculate the pH of 0.015 mol dm–3 nitric acid.

[2]

(ii) Calculate the pH of the diluted acid.

[1]

The student measured the pH of a solution of sodium hydroxide as 13.54 at 25 °C.

Kw = 1.0 × 10–14 mol2 dm–6 at 25 °C.

(i) Write down an expression for the ionic product, Kw, for water.

....................................................................................................................

[1]

[Turn over]

72 | P a g e

(ii) Calculate the concentration, in mol dm–3, of this solution of sodium hydroxide.

[2]

The student prepared two solutions.

• Solution A was made by mixing together 25 cm3 0.010 mol dm–3

aqueous sodium hydroxide with 50 cm3 0.010 mol dm–3 ethanoic acid, CH3COOH. Solution A is a buffer solution.

• Solution B was made by mixing together 25 cm3 0.020 mol dm–3

aqueous sodium hydroxide with 50 cm3 0.010 mol dm–3 ethanoic acid, CH3COOH. Solution B is not a buffer solution.

(i) What is meant by a buffer solution?

....................................................................................................................

....................................................................................................................

[1]

(ii) Explain why Solution A is a buffer solution whereas Solution B is not.

....................................................................................................................

....................................................................................................................

....................................................................................................................

....................................................................................................................

....................................................................................................................

....................................................................................................................

....................................................................................................................

[4]

[Turn over]

73 | P a g e

The student measured the pH of water as 7.0 at 25 °C. The student then warmed the water to 40 °C and measured the pH as 6.7.

What do these results tell you about the tendency of water to ionise as it gets warmer? Explain your reasoning in terms of equilibrium.

.............................................................................................................................

.............................................................................................................................

.............................................................................................................................

.............................................................................................................................

[2]

[Total 16 marks]

74 | P a g e


Most candidates obtained the first three marks. A few were thrown by being asked both questions together and so only discussed strong acids and did not go on to define a Brønsted-Lowry acid.

For the pH of nitric acid many candidates got full marks here, although some tried to manipulate the value for pH obtained in part (i) to work out the value for part (ii).

The majority of candidates were able to define Kw correctly but some then had difficulty

calculating the concentration of the sodium hydroxide solution. One of the more common errors here was to omit the negative sign when working out the proton concentration.

Apart from a few candidates who wanted the buffer to be capable of sustaining a constant pH, most candidates were able to give a correct definition.

On why solution A is a buffer and B is not, there is no doubt that this was an extremely challenging question, probably the hardest on the paper, but it was pleasing to see that a small group of candidates were able to give correct answers. Of the less able, many candidates fixed on the different concentrations of the solutions used to concoct A and B and reasoned that this lay behind the ability of only A to function as a buffer.

Others incorrectly identified the components of the buffer solution as ethanoic acid (as the weak acid) and sodium hydroxide (as the conjugate base). Such candidates then often tried to describe how this “buffer solution” would work in terms of the stock buffer theory. Candidates succeeded when identifying that a reaction had taken place between the ethanoic acid and sodium hydroxide to product a mixture of ethanoic acid and sodium ethanoate in A but only sodium ethanoate in B.

The last question on the pH of water at different temperatures was generally well answered. The majority of candidates noted that the degree of ionisation would increase with rising temperature but some failed to show that this conclusion had been based on noting that the proton concentration was increasing. Many candidates went further than required in their answer and discussed the energetics of the reaction. Here again candidates should be encouraged to give answers which stand alone. Many wrote, ‘it would ionise more’, which gains the mark but it would be much clearer were candidates to write, ‘water ionises more with increasing temperature’.

75 | P a g e

22. Neutralisation – titration curves



Interpret and sketch acid-base titration pH curves for strong and weak acids and bases.

Explain the choice of suitable indicators for acid-base titrations.


You will need to interpret titration curves and their relationship to acid / base strength and how this affects the choice of indicator used.

Notice that there are three sections to a titration curve:

1. A slight rise in pH

2. A sharp rise in pH

3. A slight rise in pH

In exams, many candidates make the mistake of showing section 1 and section 3 as horizontal lines.

76 | P a g e


A student titrated the 1.50 × 10–2 mol dm–3 methanoic acid with aqueous sodium hydroxide.

A 25.00 cm3 sample of the HCOOH(aq) was placed in a conical flask and the NaOH(aq) was added from a burette until the pH no longer changed.

(i) Write a balanced equation for the reaction between HCOOH(aq) and NaOH(aq).

....................................................................................................................

[1]

(ii) Part of the pH curve for this titration is shown below.

Calculate the concentration, in mol dm–3, of the aqueous sodium hydroxide.

concentration = ......................mol dm–3

[3]

[Turn over]

001234567pH89

1011121314

5 10 15 20 25

volume NaOH(aq) added cm3

30 35 40 45 50

77 | P a g e

(iii) Calculate the pH of the aqueous sodium hydroxide.

Kw = 1.00 × 10–14 mol dm–3

pH = .......................................

[2]

(iv) The pH ranges in which colour changes for three acid-base indicators are shown below.

indicator pH range

metacresol purple 2,4,6-trinitrotoluene

ethyl orange

7.4 – 9.0 11.5 – 13.0 3.4 – 4.8

Explain which of the three indicators is suitable for this titration.

....................................................................................................................

....................................................................................................................

....................................................................................................................

[2]

[Total 8 marks]

78 | P a g e


This neutralisation problem, based around a titration curve, created few problems for the average and better candidates.

(i) It was encouraging that most candidates were able to construct this straightforward equation, although it was relatively common to see the wrong H atom reacting to form NaCOOH.

(ii) Although most candidates were able to score some of the three available marks, the simple titration calculation did cause some unexpected problems, probably because

information had to be taken from the graph. 31 cm3 was a common error, producing a

calculated concentration of 0.0120 mol dm–3 instead of the correct concentration of 0.125

mol dm–3.

(iii) Most candidates could also calculate the pH of an alkali. Most used Kw, although a

sizeable minority using pOH. Either approach is suitable to produce the correct answer of 12.10. Some candidates spoilt matters by over-rounding this to a pH of 12.

(iv) The choice of indicator and justification was old territory and well answered by nearly everyone. Where a mistake did arise, it was usually with the choice of ethyl orange, presumably because this matches the near horizontal section of the graph. There were no comments about the use of TNT as an indicator.

79 | P a g e

23. Neutralisation – enthalpy changes



Define and use the term enthalpy change of neutralisation.

Calculate enthalpy changes from appropriate experimental results.


You may be asked about neutralisation.

One of the few definitions you need to know at A2 is the enthalpy of neutralisation.

You will need to be able to calculate enthalpy changes from experimental data using the Q = mcΔT equation (covered in AS Unit F322: Module 2 - Alcohols, halogenoalkanes and analysis).

80 | P a g e

Sample student answers

A2 Unit F325: Equilibria, energetics and elements

Module 4: Rates, equilibrium and pH

Question 1

Total marks: 15

(a) The reaction between 2-bromo–2-methylpropane (CH3)3C–Br and aqueous hydroxide ions OH

– takes place by the following outline mechanism:

Step 1 (CH3)3C–Br (CH3)3C

+ + Br

– (slow)

Step 2 (CH3)3C

+ + OH

– (CH3)3C–OH (fast)

(i) What is the order of reaction with respect to (CH3)3C–Br? (ii) What would be the effect of doubling the concentration of OH

– ions upon the rate of

the reaction? Explain your answer. (iii) What is the total order of reaction? (iv) Write the rate equation for this reaction. (v) Sketch a rate-concentration graph for the hydrolysis of (CH3)3C–Br.

Marks available: (i) 1 (ii) 2 (iii) 1 (iv) 1 (v) 1

Student answer:

(a) (i) One

(ii) It would have no effect because the order is zero with respect to OH– ions.

(iii) One (iv) Rate = k[(CH3)3C–Br]

1

(v)

81 | P a g e

Examiner comments:

(a) (i) This is because one molecule of (CH3)3C–Br is present in the slow rate determining step. (ii) This is because there are no OH

– ions present in the slow rate determining step.

(iii) This is calculated by adding together the orders of each of the reactants in the rate equation. (iv) This is because anything to the power zero = 1. (Allow rate = k[(CH3)3C–Br]

1 × [OH

–]0)

(v) As this is a sketch graph you don’t need any scales, however, you must label the axes.

82 | P a g e

(b) A series of experiments were carried out on the reaction between bromide ions and an acidified solution of bromated(V) ions, represented by the equation below: 5Br

–(aq) + BrO3

–(aq) + 6H

+(aq) 3Br2(aq) + 3H2O(l)

Trial number

Volume of

1.00 mol dm–3

BrO3

–(aq)/cm

3

Volume of

1.00 mol dm–3

Br

–(aq)/cm

3

Volume of

1.00 mol dm–3

H

+(aq)/cm

3

Volume of water/cm

3

Initial rate of formation of Br

2(aq)

/mol dm–3 s

–1

1 10 50 60 80 8.5 × 10–4

2 10 50 120 20 34.0 × 10–4

3 20 50 120 10 68.0 × 10–4

4 20 25 120 35 34.0 × 10–4

(ii) A student looked at the balanced equation for the reaction and said the rate equation was: rate = k[Br

–(aq)]

1 × [BrO3

–(aq)]

5 × [H

+(aq)]

6.

Explain why his reasoning was incorrect. (ii) Why can the volumes of reagents be used as relative concentrations of reagents? (iii) Use the data in the table above to determine the rate equation. (iv) Use the information from Trial 2 to calculate the value for the rate constant k. Give its units. (v) What would be the effect of increasing the temperature upon the rate constant?

Marks available: (i) 1 (ii) 1 (iii) 3 (iv) 3 (v) 1

Student answer:

(b) (i) Rate equations must be derived from experimental data. The balanced equation represents the overall change and is totally independent of the rate equation. (ii) Because the concentrations of reactants are identical and so is the final volume of solution used for each trial. (iii) rate = k[Br

–(aq)]

1 × [BrO3

–(aq)]

1 × [H

+(aq)]

2

(iv) In Trial 2:

83 | P a g e

[BrO3–] = 10 × 1.00 = 0.0500 mol dm

–3

200

[Br-] = 50 × 1.00 = 0.250 mol dm

–3

200

[H+] = 120 × 1.00 = 0.600 mol dm

–3

200

k (rate constant) = 34 × 10–4 mol dm

–3 s–1

0.0500 mol dm–3

× 0.250 mol dm–3

× [0.600 mol dm–3

]2

= 0.756 dm9 mol

–3 s–1

(v) As temperature increased, so too would k.

Examiner comments:

(b) (i) A common error is to assume that the rate equation can be derived from the balanced equation. (ii) Both points are needed for the full mark here. (iii) The results from Trials 1 and 2 tell you that by doubling [H

+] the rate quadruples – i.e.

second order. Comparing Trials 2 and 3 for BrO3– and Trials 3 and 4 for Br

- will also

show that rate [BrO3–]1 and rate [Br

–]1 respectively.

(iv) All concentrations must be converted to mol dm

–3 within the reaction solution. The total

volume of this solution was only 200 cm3.

Although the units look strange, they are correct! (v) The rate constant k is temperature-dependent. This is the only factor that could alter k.

84 | P a g e

Module 4: Rates, equilibrium and pH

Question 2

Total marks: 15

(a) The equation for the reaction between hydrogen and iodine at 400 °C in a sealed vessel is shown below:

H2(g) + I2(g) 2HI(g) (i) Write the equilibrium constant for this reaction.

(ii) The value of the equilibrium constant is 54.0 at 400 °C. Calculate the concentration

of hydrogen iodide when the concentration of H2 is 1.20 × 10–3 mol dm

–3 and the

concentration I2 is 2.00 × 10–3 mol dm

–3.

(iii) At room temperature, the value of the equilibrium constant for this reaction is approximately 800. Justify why this information tells you the reaction is exothermic in the forward direction.

Marks available: (i) 1 (ii) 2 (iii) 2

Student answer:

(a) (i) Kc = [HI(g)]2

[H2] × [I2]

(ii) [HI(g)]2 = 1.20 × 10

–3 mol dm

–3 × 2.00 × 10

–3 mol dm

–3 × 54.0

[HI(g)] = √(1.296 × 10–4 mol

2 dm

–6)

[HI(g)] = 0.0114 mol dm–3

(3sf) (iii) If Kc has increased then the concentration of HI relative to H2 and I2 must have increased. This means cooling the mixture has shifted the equilibrium in the exothermic (forward) direction.

Examiner comments:

(a) (i) Kc (unlike the rate equation) can be determined from the balanced equation. (ii) Always show as much working as possible and give your answer to the appropriate number of significant figures. (iii) Kc is temperature-dependent. When the temperature increases, Kc will increase if the forward reaction is endothermic or decrease if the forward reaction is exothermic.

85 | P a g e

(b) What is meant by the term pH?

Marks available: 1

Student answer:

(b) pH = – log [H+]

Examiner comments:

(b) Make sure you know how to put this data into your calculator.

(c) Calculate the pH of the following:

(i) 0.00100 mol dm–3

HCl(aq)

(ii) 0.0500 mol dm–3

NaOH

(iii) 0.100 mol dm–3

CH2ClCOOH(aq) where Ka of CH2ClCOOH(aq)

= 1.40 × 10–3 mol dm

–3.

Marks available: (i) 1 (ii) 1 (iii) 2

Student answer:

(c) (i) pH = – (log 0.00100) = 3.0

(ii) [H+]= Kw = 1.00 × 10

–14 = 2.00 × 10

–13 mol dm

–3

[OH–] 0.0500

pH = –(log 2.00 × 10

–13) = 12.7

(iii) [H

+]2 = 1.40 × 10

–3 × 0.100

[H+] = √1.4 × 10

–4 = 0.0118

pH = 1.93

Examiner comments:

(c) (i) Remember not to leave the value as –3.0; it should be a positive number. This is because the logs of numbers less than one are negative, so the minus sign in –log[H

+]

makes most pH values positive. (ii) There are other ways of calculating pH from OH

– ion concentration, but this is one of the

simplest. (iii) This type of calculation is commonly asked in examinations – you must learn the method needed to do the calculation. Ka = [H

+]2

[acid]

86 | P a g e

(d) A mixture of aqueous ethanoic acid CH3COOH(aq) and sodium ethanoate CH3COONa(aq) will act as a buffer solution. (i) What is meant by the term buffer? (ii) Describe how this mixture can act as a buffer.

Marks available: (i) 1 (ii) 4

Student answer:

(d) (i) A solution which will undergo minimal changes in pH when a small amount of acid or alkali is added. (ii) Two processes take place:

Process 1 CH3COOH(aq) ⇌ CH3COO–(aq) + H

+(aq)

Process 2 CH3COONa(aq) CH3COO–(aq) + Na

+(aq)

If H

+ ions (acid) are added, they will react with CH3COO

–(aq) ions. Process 2 provides a

plentiful supply, so any significant increase in [H+(aq)] is prevented.

If OH

– ions (alkali) are added, they will react with the H

+ ions produced from Process 1.

The equilibrium in Process 1 is pushed to the right-hand side, meaning that any new OH

– ions will have H

+ ions to react with, thus preventing a significant reduction in

[H+(aq)].

Examiner comments:

(d) (i) Avoid the common misconception of stating the pH is constant. Also it must be realised that buffers will only work if relatively small amounts of acid or alkali are added. (ii) The trick to explaining this is to realise that the reaction occurring in Process 1 is an equilibrium.

87 | P a g e

The Periodic Table of the Elements

Th

e P

eri

od

ic T

ab

le o

f th

e E

lem

en

ts

1

2

3

4

5

6

7

0

Ke

y

1.0

H

h

yd

roge

n

1

4.0

H

e

heliu

m

2

6.9

L

i lit

hiu

m

3

9.0

B

e

bery

lliu

m

4

rela

tive a

tom

ic m

ass

ato

mic

sym

bo

l n

am

e

ato

mic

(pro

ton)

num

ber

10.8

B

b

oro

n

5

12.0

C

ca

rbo

n

6

14.0

N

nitro

gen

7

16.0

O

oxygen

8

19.0

F

fluo

rin

e

9

20.2

N

e

neon

10

23.0

N

a

sodiu

m

11

24.3

M

g

ma

gnesiu

m

12

27

.0

Al

alu

min

ium

13

28.1

S

isili

co

n

14

31.0

P

p

hosp

horu

s

15

32.1

S

sulfur

16

35.5

C

l ch

lorine

17

39.9

A

rarg

on

18

39.1

K

pota

ssiu

m

19

40.1

C

a

calc

ium

20

45.0

S

c

sca

nd

ium

21

47.9

T

i

tita

niu

m

22

50.9

V

vana

diu

m

23

52.0

C

rchro

miu

m

24

54

.9

Mn

ma

ngane

se

25

55

.8

Fe

iro

n

26

58.9

C

ocoba

lt

27

58.7

N

inic

kel

28

63

.5

Cu

co

ppe

r

29

65.4

Z

n

zin

c

30

69

.7

Ga

ga

llium

31

72.6

G

ege

rma

niu

m

32

74.9

A

sars

enic

33

79.0

S

esele

niu

m

34

79.9

B

rb

rom

ine

35

83.8

K

rkry

pto

n

36

85.5

R

b

rubid

ium

37

87.6

S

r str

ontium

38

88.9

Y

ytt

rium

39

91.2

Z

r

zirco

niu

m

40

92.9

N

b

nio

biu

m

41

95.9

M

om

oly

bdenum

42

[98]

Tc

tech

ne

tiu

m

43

101

.1

Ru

ru

the

niu

m

44

102

.9

Rh

rhod

ium

45

10

6.4

P

d

palla

diu

m

46

10

7.9

A

g

silv

er

47

112.4

C

d

cad

miu

m

48

114

.8

Inin

diu

m

49

118

.7

Sn

tin

50

121.8

S

ban

tim

on

y

51

127

.6

Te

tellu

rium

52

126

.9

I io

din

e

53

131.3

X

exen

on

54

13

2.9

C

s

cae

siu

m

55

137

.3

Ba

b

arium

56

138

.9

La*

lan

thanum

57

178.5

H

f

ha

fniu

m

72

18

0.9

Ta

tan

talu

m

73

18

3.8

W

tu

ngste

n

74

186.2

R

e

rheniu

m

75

190

.2

Os

osm

ium

76

192

.2

Ir

irid

ium

77

19

5.1

P

t

pla

tinum

78

19

7.0

A

u

gold

79

20

0.6

H

g

merc

ury

80

20

4.4

T

l th

alli

um

81

20

7.2

P

ble

ad

82

209.0

B

ib

ism

uth

83

[209

] P

op

olo

niu

m

84

[210

] A

ta

sta

tine

85

[222

] R

nra

do

n

86

[22

3]

Fr

franciu

m

87

[226

] R

a

rad

ium

88

[227

] A

c*

actiniu

m

89

[261]

Rf

ruth

erf

ord

ium

104

[262]

Db

d

ubniu

m

105

[26

6]

Sg

seab

org

ium

10

6

[264]

Bh

b

oh

rium

107

[27

7]

Hs

ha

ssiu

m

10

8

[26

8]

Mt

me

itne

rium

10

9

[271]

Ds

darm

sta

dtium

11

0

[272

] R

g

roen

tge

niu

m

111

Ele

men

ts w

ith

ato

mic

nu

mbe

rs 1

12–

116 h

ave b

een

rep

ort

ed b

ut n

ot fu

lly

au

then

ticate

d

140.1

C

e

ce

rium

58

14

0.9

P

r pra

seo

dym

ium

59

14

4.2

N

dne

odym

ium

60

144.9

P

mpro

meth

ium

61

150

.4

Sm

sam

arium

62

152

.0

Eu

euro

piu

m

63

15

7.2

G

dga

do

liniu

m

64

15

8.9

T

b

terb

ium

65

16

2.5

D

ydysp

rosiu

m

66

16

4.9

H

oh

olm

ium

67

16

7.3

E

re

rbiu

m

68

168.9

T

mth

uliu

m

69

173

.0

Yb

ytt

erb

ium

70

175

.0

Lu

lute

tium

71

232.0

T

h

tho

rium

90

[231]

Pa

pro

tactiniu

m

91

23

8.1

U

ura

niu

m

92

[237]

Np

nep

tun

ium

93

[24

2]

Pu

plu

toniu

m

94

[24

3]

Am

am

ericiu

m

95

[247]

Cm

curi

um

96

[245

] B

k

be

rkeliu

m

97

[251]

Cf

ca

liforn

ium

98

[25

4]

Es

ein

ste

iniu

m

99

[253]

Fm

ferm

ium

100

[256

] M

dm

ende

levium

101

[254

] N

on

ob

eliu

m

102

[257

] L

rla

wre

nciu

m

103

• • • • • •

Documents

Unit 2 Equilibria, energetics and elements (F325) Rates ...UNIT+2+Module+1+Rat… · 1 | P a g e Unit 2 Equilibria, energetics and elements (F325) Rates, equilibrium and pH Topics