CIE AS Chemistry

Topic Questions

Syllabus Edition

First teaching 2020

Last exams 2024

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1.5 Chemical Energetics

11 mark

Which quantity gives the best indication of the relative strengths of the hydrogen bonds between water molecules in the liquid state?

  • Enthalpy changes of vaporisation

  • Bond dissociation energies

  • Enthalpy of formation

  • Activation energy

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21 mark

Enthalpy changes that are difficult to measure directly can often be determined using Hess’ Law to construct an enthalpy cycle.

Which enthalpy change is indicated by X in the enthalpy cycle shown?

1-5-e-2

  • +1 x Enthalpy of formation of water

  • -1 x Enthalpy of formation of water

  • +3 x Enthalpy of formation of water

  • -3 x Enthalpy of formation of water

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31 mark

Which equation below can represent both an enthalpy change of formation and combustion?

  • CH4 (g) + 2O2 (g) → CO2 (g) + 2H2O (l)

  • 4Na (s) + O2 (g) → 2Na2O (s)

  • HCl (aq) + NaOH (aq) → NaCl (aq) + H2O (l)

  • CO (g) + C (s) → CO2 (g)

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41 mark

The reaction pathway for a reversible reaction is shown below:

1-5-diagram-q4e

Which statement is correct?

  • The activation energy of the reverse reaction is +90 kJ mol–1 

  • The activation energy of the forward reaction is +20 kJ mol–1

  • The activation of the reverse reaction is +20 kJ mol–1

  • The enthalpy change of forwards reaction is - 70 kJ mol–1

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51 mark

The reaction pathway for a reversible reaction is shown below.

1-5-diagram-q5e

Which statements are correct?

1

The forward reaction is endothermic

2

The enthalpy change for the backward reaction is –30 kJ mol–1

3

The activation energy for the forward reaction is +90 kJ mol–1

  • 1 only

  • 1 and 2

  • 2 and 3

  • 1, 2 and 3

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61 mark

Which statement best describes why a reaction is said to be thermodynamically possible yet kinetically controlled?

  • The enthalpy of the reactants is higher than the products, and the reaction is very slow.

  • The enthalpy of the reactants is lower than the products, and the reaction is fast.

  • The enthalpy of the reactants is lower than the products, and the reaction is very slow.

  • The reaction is exothermic and very fast.

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71 mark

Use of the Periodic table is relevant to this question.

A student carried out an experiment to determine the enthalpy change for the combustion of ethanol.

The following results were obtained by the student. The specific heat capacity of water is 4.18 J g−1 ˚C−1.

1-5-diagram-q7e

Start temperature of the water

21 °C

Final temperature of the water

54 °C

Mass of alcohol burner before burning

259.75 g

Mass of alcohol burner after burning

259.18 g

Mass of glass beaker plus water

150.00 g

Mass of glass beaker

50.0 g

How much of the heat energy produced by the burning of ethanol went into the water?

  • 6897 J

  • 20691 J

  • 22572 J

  • 13794 J

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81 mark

PCl5 (g) dissociates as follows

PCl5 (g) → PCl3 (g) + Cl2 (g)

The yield of the products increases as the temperature is raised from 80oC to 110oC.

Which row correctly describes the formation of PCl3 (g)?

 

Shape of PCl3 molecule

The reaction is

A

Pyramidal

Endothermic

B

Pyramidal

Exothermic

C

Trigonal

Endothermic

D

Trigonal

Exothermic

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91 mark

Hydrogen atoms bond covalently to Iodine atoms to form hydrogen iodide as shown in the equation below: 

H2 (g) + I2 (g) → 2HI (g)

Which statement best describes what is meant by the average HI bond enthalpy?

  • The energy stored in a covalent bond.

  • The energy required to break one covalent bond in the gas phase.

  • The energy required to break one mole of the HI bonds in the gas phase.

  • The energy released when two atoms form a covalent bond.

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101 mark

The enthalpy of atomisation of a compound can be calculated using a range of different enthalpy changes. 

Which statement below correctly describes the enthalpy change of atomisation?

  • The energy stored in a covalent bond.

  • The energy required to break one covalent bond in the gas phase.

  • The energy required to break all bonds in an element.

  • The energy released when two atoms form a covalent bond.

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11 mark

In the gas phase, phosphorus pentachloride can be thermally decomposed into gaseous phosphorus trichloride and chlorine. 

PCl5 →PCl3 + Cl2

The table below gives the relevant bond energies found in these compounds.

Bond

Bond energy / kJ mol–1

P – Cl (in both chlorides)

Cl – Cl 

328

241

What is the enthalpy change in the decomposition of the reaction?

  • – 415 kJ mol–1

  • + 415 kJ mol–1

  • + 95 kJ mol–1

  • – 95 kJ mol–1

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21 mark

The equations below show the formation of sulfur oxides from sulfur and oxygen.

S (s) + O2 (g) → SO2 (g)     Hf = –297 kJ mol–1

S (s) + 1½ O2 (g) → SO3 (g)    Hf = –395 kJ mol–1

What is the enthalpy change of reaction, Hr, of 2SO2 (g) + O2 (g) → 2SO3 (g)?

  • + 98 kJ mol–1

  • - 98 kJ mol–1

  • + 196 kJ mol–1

  • - 196 kJ mol–1

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31 mark

In a calorimetric experiment 2.50 g of a fuel is burnt in oxygen. 30 % of the energy released during the combustion is absorbed by 500 g of water, the temperature of which rises from 25 °C to 68 °C. 

The specific heat capacity of water is 4.2 J g–1 K–1.

What is the total energy released per gram of fuel burnt?

  • 25 284 J

  • 63 210 J

  • 120 400 J

  • 301 000 J

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41 mark

Which equation correctly shows how the bond energy for the covalent bond Y-Z can be calculated by dividing ΔH by n?

  • n YZ (g) → n Y (g) + n over 2 Z2 (g)

  • Z (g) + Y Zn-1 (g) → YZn (g)

  • 2 YZn (g) → 2 YZn-1 (g) + Y2 (g)

  • YZn (g) → Y (g) + nZ (g)

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51 mark

The diagram shows the skeletal formula of cyclobutane.

1-5-h-5

The enthalpy change of formation of cyclobutane is +75.1 kJ mol–1, and the enthalpy change of atomisation of graphite is +712 kJ mol–1.

The bond enthalpy of C–H is 390 kJ mol–1 and of H–H is 429 kJ mol–1.

What is the average bond enthalpy of the C–C bond in cyclobutane, rounded to the nearest whole number?

  • 236 kJ mol–1

  • 315 kJ mol–1

  • 342 kJ mol–1

  • 700 kJ mol–1

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61 mark

Some bond energy values are listed below.

Bond

Bond energy / kJ mol-1

  Br – Br  

Cl – Cl

C – H

C – Cl

194

247

412

338

These bond energy values relate to the following four reactions.

W

Br2 → 2Br

X

2Cl → Cl2

Y

CH3 + Cl → CH3Cl

Z

CH4 → CH3 + H

What is the correct order of enthalpy changes of the above reactions from most negative to most positive?

  • Y → Z → W → X

  • Z → W → X → Y

  • Y → X → W → Z

  • X → Y → Z → W

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71 mark

A student calculated the standard enthalpy change of formation of propane, C3H8, using a method based on standard enthalpy changes of combustion.

He used correct values for the standard enthalpy change of combustion of propane (–2220 kJ mol–1) and hydrogen (–286 kJ mol–1) but he used an incorrect value for the standard enthalpy change of combustion of carbon. He then performed his calculation correctly. His final answer was –158 kJ mol–1.

What did he use for the standard enthalpy change of combustion of carbon?

  • -1234 kJ mol-1 

  • -411.3 kJ mol-1 

  • -2084 kJ mol-1 

  • -694.7 kJ mol-1 

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81 mark

Given the following enthalpy changes,

                            I2 (s) → I2 (g)                          ∆HƟ = +38 kJ mol–1

                            I2 (g) + 3Cl2 (g) → 2ICl3 (s)    ∆HƟ = –214 kJ mol–1

What is the correct value for ∆HfƟ of iodine trichloride, ICl3 (s)?

  • –214 kJ mol–1

  • –176 kJ mol–1

  • –88 kJ mol–1

  • +176 kJ mol–1

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91 mark

Using the following information:

                       CO (g) + begin mathsize 16px style 1 half end style O2 (g)   → CO2 (g)            HƟ  = –283 kJ mol–1

                       H2 (g) + 1 half O2 (g)   → H2O (I)              HƟ  = –286 kJ mol–1

                       H2O (g)   → H2O (I)                              HƟ = –44 kJ mol–1

What is the enthalpy change, ∆HƟ, for the following reaction? 

CO2 (g) + H2 (g) → CO (g) + H2O (g)

  • -41 kJ mol-1

  • +41  kJ mol-1

  • -525  kJ mol-1

  • +525  kJ mol-1

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101 mark

The conversion of graphite into diamond is an endothermic reaction. 

(ΔH = +3 kJ mol–1).

C (graphite)    C (diamond)

1

The enthalpy change of atomisation of diamond is smaller than that of graphite.

2

The bond energy of the C–C bonds in graphite is greater than that in diamond.

3

The enthalpy change of combustion of diamond is greater than that of graphite.

  • 1, 2 and 3

  • 1 and 2

  • 2 and 3

  • 1 only

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11 mark

A student mixed 30.0 cm3 of 0.0250 mol dm-3 potassium hydroxide solution with 30.0 cm3 of 0.0250 mol dm-3 nitric acid. The temperature rose by 0.50 oC. Assume no heat was lost to the surroundings. 

The final mixture had a specific heat capacity of 4.20 J cm-3 K-1.

What is the molar enthalpy change for the reaction?

  • -0.126 kJ mol-1

  • -84 kJ mol-1

  • 168 kJ mol-1

  • -168 kJ mol-1

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21 mark

Propanol has the molecular formula C3H6O. Use the following information to calculate the enthalpy change for the formation of propanol?

The enthalpy change of combustion of hydrogen is –286 kJ mol-1

The enthalpy change of combustion of carbon is –394 kJ mol-1

The enthalpy change of combustion of propanol is – 1816 kJ mol-1

  • -3856 kJ mol-1

  • -224 kJ mol-1

  • +1154 kJ mol-1

  • -2478 kJ mol-1

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31 mark

An experiment was carried out to determine the approximate value for the molar enthalpy change of neutralisation. 

75 cm3 of 3.00 mol dm–3 hydrochloric acid was placed in a polystyrene beaker of negligible heat capacity. Its temperature was recorded, and then 75 cm3 of 3.00 mol dm−3 potassium hydroxide at the same temperature was quickly added, and the solution stirred. 

The temperature rose by 14 °C. The resulting solution may be considered to have a specific heat capacity of 4.2 J g-1 K-1.

Which calculation below is correct?

  • negative space fraction numerator left parenthesis 75 space cross times space 4.2 space cross times space 14 right parenthesis over denominator left parenthesis 0.150 space cross times space 6.0 right parenthesis end fraction space J space m o l to the power of negative 1 end exponent

  • negative space fraction numerator left parenthesis 150 space cross times space 4.2 space cross times space 14 right parenthesis over denominator left parenthesis 0.075 space cross times space 3.0 right parenthesis end fraction space J space m o l to the power of negative 1 end exponent

  • negative space fraction numerator left parenthesis 150 space cross times space 4.2 space cross times space 14 right parenthesis over denominator left parenthesis 75.0 space cross times space 3.0 right parenthesis end fraction space J space m o l to the power of negative 1 end exponent

  • negative space fraction numerator left parenthesis 75 space cross times space 4.2 space cross times space 17 right parenthesis over denominator left parenthesis 0.075 space cross times space 6.0 right parenthesis end fraction space J space m o l to the power of negative 1 end exponent

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41 mark

The first stage in the industrial production of nitric acid from ammonia can be represented by the following equation.

4NH3 (g) + 5O2 (g) 4NO (g) + 6H2O (g)

Using the following standard enthalpy change of formation data, what is the value of the standard enthalpy change Hf for this reaction?

Compound

Hf /kJ mol–1

NH3 (g)

-51.3

NO (g)

+92.2

H2O(g)

-239.6

  • + 863.6 kJ mol–1

  • - 1601.2 kJ mol–1

  • - 863.6 kJ mol–1

  • - 1274.0 kJ mol–1

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51 mark

The standard enthalpy change H for the below reaction is -1776 kJ.

2N (g) + 6F (g) → 2NF3 (g)

What is the bond energy of the N–F bond?

  • - 592

  • + 592

  • + 296

  • - 296

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61 mark

The complete combustion of ethyne, C2H2, is shown in the equation below.

H−C≡C−H + 2½ O═O → O−H−H + 2O═C═O

Using the average bond enthalpies given in the table, what is the enthalpy change of combustion of ethyne?


Bond

Average bond enthalpy / kJ mol-1

C−H

390

C≡C 

870

O═O

489

C═O

790

O−H 

510

O−C 

290

  • + 1307.5 kJ mol–1

  • - 1307.5 kJ mol–1

  • + 1390 kJ mol–1

  • - 1390 kJ mol–1

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71 mark

Compound R into compound T, it was found that the reaction proceeded by way of compound S, which could be isolated. The following steps were involved.

R → S;   ΔH is positive

S → T;   ΔH is negative

Which reaction profile fits these data?

1-5-m-7

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81 mark

Titanium occurs naturally as the mineral rutile, TiO2. One possible method of extraction of titanium is to reduce the rutile by heating with carbon.

TiO2 (s) + 2C (s) → Ti (s) + 2CO (g)

The standard enthalpy changes of formation of TiO2 (s) and CO (g) are –890 kJ mol-1 and –105 kJ mol-1 respectively.

What is the standard enthalpy change of the extraction of titanium?

  • + 450 kJ mol–1

  • + 680 kJ mol–1

  • - 450 kJ mol–1

  • - 660 kJ mol–1

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91 mark

The combustion of ethanol (C2H5OH) is increasingly being used to fuel cars. 

The standard enthalpy change of formation of carbon dioxide is –382 kJ mol–1

The standard enthalpy change of formation of water is –275 kJ mol–1.

The standard enthalpy change of formation of ethanol is −266kJ mol–1

What is the standard enthalpy change of combustion of ethanol?

  • - 1367 kJ mol–1

  • - 1323 kJ mol–1

  • - 391 kJ mol–1

  • - 948 kJ mol–1

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101 mark

The table below discusses three types of enthalpy change:

‘+’ means that this type of standard enthalpy change can only have positive values.

‘–’ means that this type of standard enthalpy change can only have negative values.

‘+ / –’ means that either positive or negative values are possible.

Which row is correct?

 

Formation

Combustion

Neutralisation

A

+

+

+ / –

B

+ / –

+

+ / –

C

+ / –

D

+

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