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R1.2 Energy cycles in reactions

Practice exam-style IB Chemistry questions for Energy cycles in reactions, aligned with the syllabus and grouped by topic.

Verified by Dennis M.
Verified by Dennis M.
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Question 1
SL • Paper 1A
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SL • Paper 1A
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A covalent reaction involves breaking bonds in the reactants and forming bonds in the products. What is the energy change associated with each process?

A.

Bond breaking releases energy; bond forming absorbs energy

B.

Both bond breaking and bond forming absorb energy

C.

Both bond breaking and bond forming release energy

D.

Bond breaking absorbs energy; bond forming releases energy

Question 2
SL • Paper 1A
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SL • Paper 1A
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A reaction enthalpy calculated using data-booklet average bond enthalpies differs from the experimental value. What is the best explanation?

A.

Average bond enthalpies apply only to ionic compounds in the solid state

B.

Average bond enthalpies are mean gas-phase values from bonds in different molecular environments

C.

Average bond enthalpies include the energy needed to change every substance into its standard state

D.

Average bond enthalpies are exact values for each bond of the same written type

Question 3
SL • Paper 1A
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SL • Paper 1A
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The carbon-halogen bond enthalpies in halogenoalkanes generally decrease from C−ClC-Cl to C−IC-I. What is the most likely consequence for nucleophilic substitution, other factors being similar?

A.

Iodoalkanes react faster because the C−IC-I bond is weaker and easier to break

B.

Iodoalkanes react slower because the C−IC-I bond releases more energy when broken

C.

Chloroalkanes react faster because the C−ClC-Cl bond has the largest bond enthalpy

D.

All halogenoalkanes react at the same rate because bond enthalpy is an average value

Question 4
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SL • Paper 1A
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Hess's law is valid because enthalpy is a function. What does this mean for a chemical reaction?

A.

The enthalpy change depends only on the initial and final states

B.

The enthalpy change changes sign when a catalyst is added

C.

The enthalpy change is zero for any complete reaction

D.

The enthalpy change depends only on the rate of the reaction

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Question 5
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What is the equation that represents the standard enthalpy change of formation of methanol, CH3OH(l)CH_3OH(l)?

A.

C(s, graphite)+2H2(g)+12O2(g)→CH3OH(l)C(s,\text{ graphite})+2H_2(g)+\frac{1}{2}O_2(g) \to CH_3OH(l)

B.

CH3OH(l)+32O2(g)→CO2(g)+2H2O(l)CH_3OH(l)+\frac{3}{2}O_2(g) \to CO_2(g)+2H_2O(l)

C.

C(s, diamond)+2H2(g)+12O2(g)→CH3OH(l)C(s,\text{ diamond})+2H_2(g)+\frac{1}{2}O_2(g) \to CH_3OH(l)

D.

CH4(g)+12O2(g)→CH3OH(l)CH_4(g)+\frac{1}{2}O_2(g) \to CH_3OH(l)

Question 6
HL • Paper 1A
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HL • Paper 1A
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Graphite and diamond are allotropes of carbon. What are the standard enthalpies of formation of graphite and diamond?

A.

Diamond is 0 kJ mol−10\ \text{kJ mol}^{-1}; graphite is not necessarily 0 kJ mol−10\ \text{kJ mol}^{-1}

B.

Both graphite and diamond are 0 kJ mol−10\ \text{kJ mol}^{-1} because both are elements

C.

Graphite is 0 kJ mol−10\ \text{kJ mol}^{-1}; diamond is not necessarily 0 kJ mol−10\ \text{kJ mol}^{-1}

D.

Neither graphite nor diamond can have a standard enthalpy of formation

Question 7
HL • Paper 1A
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HL • Paper 1A
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In a Born-Haber cycle for magnesium oxide, what is the step represented by Mg+(g)→Mg2+(g)+e−Mg^+(g) \to Mg^{2+}(g)+e^-?

A.

First electron affinity of magnesium

B.

Enthalpy of atomization of magnesium

C.

First ionization energy of magnesium

D.

Second ionization energy of magnesium

Question 8
SL • Paper 2
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SL • Paper 2
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Bond enthalpy data can be used to interpret energy changes during reactions.

A

Define bond enthalpy.

[1]
B

why bond breaking is endothermic whereas bond forming is exothermic.

[2]

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Question 9
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Methanol has the formula CH3OHCH_3OH.

A

Write the equation, including symbols, for the standard enthalpy change of formation of liquid methanol.

[1]
B

why ΔHf⊖\Delta H_f^\ominus for graphite is zero but ΔHf⊖\Delta H_f^\ominus for diamond is not zero.

[1]
Question 10
SL • Paper 1A
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SL • Paper 1A
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The reaction between methane and chlorine is shown.

CH4(g)+Cl2(g)→CH3Cl(g)+HCl(g)CH_4(g)+Cl_2(g) \to CH_3Cl(g)+HCl(g)

Average bond enthalpies are: C−H=413 kJ mol−1C-H=413\ \text{kJ mol}^{-1}, Cl−Cl=242 kJ mol−1Cl-Cl=242\ \text{kJ mol}^{-1}, C−Cl=338 kJ mol−1C-Cl=338\ \text{kJ mol}^{-1}, H−Cl=431 kJ mol−1H-Cl=431\ \text{kJ mol}^{-1}.

What is the enthalpy change for the reaction?

A.

−114 kJ mol−1-114\ \text{kJ mol}^{-1}

B.

+356 kJ mol−1+356\ \text{kJ mol}^{-1}

C.

+114 kJ mol−1+114\ \text{kJ mol}^{-1}

D.

−356 kJ mol−1-356\ \text{kJ mol}^{-1}

Question 11
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Use the thermochemical equations.

C(s)+O2(g)→CO2(g)C(s)+O_2(g) \to CO_2(g) ΔH=−394 kJ mol−1\Delta H=-394\ \text{kJ mol}^{-1}

CO(g)+12O2(g)→CO2(g)CO(g)+\frac{1}{2}O_2(g) \to CO_2(g) ΔH=−283 kJ mol−1\Delta H=-283\ \text{kJ mol}^{-1}

What is ΔH\Delta H for the reaction below?

C(s)+12O2(g)→CO(g)C(s)+\frac{1}{2}O_2(g) \to CO(g)

A.

+111 kJ mol−1+111\ \text{kJ mol}^{-1}

B.

−111 kJ mol−1-111\ \text{kJ mol}^{-1}

C.

−677 kJ mol−1-677\ \text{kJ mol}^{-1}

D.

+677 kJ mol−1+677\ \text{kJ mol}^{-1}

Question 12
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HL • Paper 1A
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Use the standard enthalpies of combustion.

ΔHc⊖[CH3CH2OH(l)]=−1367 kJ mol−1\Delta H_c^\ominus[CH_3CH_2OH(l)]=-1367\ \text{kJ mol}^{-1}

ΔHc⊖[CH3COOH(l)]=−875 kJ mol−1\Delta H_c^\ominus[CH_3COOH(l)]=-875\ \text{kJ mol}^{-1}

What is ΔH⊖\Delta H^\ominus for the reaction below?

CH3CH2OH(l)+O2(g)→CH3COOH(l)+H2O(l)CH_3CH_2OH(l)+O_2(g) \to CH_3COOH(l)+H_2O(l)

A.

+492 kJ mol−1+492\ \text{kJ mol}^{-1}

B.

+2242 kJ mol−1+2242\ \text{kJ mol}^{-1}

C.

−492 kJ mol−1-492\ \text{kJ mol}^{-1}

D.

−2242 kJ mol−1-2242\ \text{kJ mol}^{-1}

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Question 13
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Use the standard enthalpies of formation.

ΔHf⊖[CaCO3(s)]=−1207 kJ mol−1\Delta H_f^\ominus[CaCO_3(s)]=-1207\ \text{kJ mol}^{-1}

ΔHf⊖[CaO(s)]=−635 kJ mol−1\Delta H_f^\ominus[CaO(s)]=-635\ \text{kJ mol}^{-1}

ΔHf⊖[CO2(g)]=−394 kJ mol−1\Delta H_f^\ominus[CO_2(g)]=-394\ \text{kJ mol}^{-1}

What is ΔH⊖\Delta H^\ominus for the reaction below?

CaO(s)+CO2(g)→CaCO3(s)CaO(s)+CO_2(g) \to CaCO_3(s)

A.

+178 kJ mol−1+178\ \text{kJ mol}^{-1}

B.

+2236 kJ mol−1+2236\ \text{kJ mol}^{-1}

C.

−178 kJ mol−1-178\ \text{kJ mol}^{-1}

D.

−2236 kJ mol−1-2236\ \text{kJ mol}^{-1}

Question 14
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SL • Paper 2
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Methane reacts with chlorine in a substitution reaction.

CH4(g)+Cl2(g)→CH3Cl(g)+HCl(g)CH_4(g)+Cl_2(g) \to CH_3Cl(g)+HCl(g)

The table gives average bond enthalpy data for the bonds involved.

Bond

Average bond enthalpy / kJ mol−1^{-1}

C-H

413

Cl-Cl

243

C-Cl

338

H-Cl

432

A

the bonds broken and the bonds formed in this reaction.

[1]
B

the enthalpy change, in kJ mol−1\text{kJ mol}^{-1}, using the average bond enthalpy data.

[2]
Question 15
SL • Paper 2
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SL • Paper 2
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A student uses average bond enthalpy data to ΔH\Delta H for the combustion of liquid ethanol. The experimental value is more exothermic than the estimated value.

A

two reasons why the estimated value differs from the experimental value.

[2]
Question 16
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SL • Paper 2
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A reaction can occur directly from AA to BB, or by a two-step route through an intermediate CC.

A→CΔH1=−84 kJ mol−1A \to C \qquad \Delta H_1=-84\ \text{kJ mol}^{-1}

C→BΔH2=+27 kJ mol−1C \to B \qquad \Delta H_2=+27\ \text{kJ mol}^{-1}

A

Hess's law.

[1]
B

the enthalpy change for the direct route A→BA \to B.

[1]

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Question 17
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The carbon-halogen bond affects the rate of nucleophilic substitution in halogenoalkanes. The table gives bond length and bond enthalpy data for three carbon-halogen bonds.

Bond

Bond length / pm

Bond enthalpy / kJ mol−1^{-1}

C-Cl

177

338

C-Br

194

276

C-I

214

238

A

why iodoalkanes generally react faster than chloroalkanes in nucleophilic substitution reactions.

[2]
Question 18
HL • Paper 2
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HL • Paper 2
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Ethene can be hydrogenated to ethane.

C2H4(g)+H2(g)→C2H6(g)C_2H_4(g)+H_2(g) \to C_2H_6(g)

The table gives standard enthalpy changes of combustion.

Substance

ΔH∘^\circc / kJ mol⁻¹

C2H4C_2H_4(g)

-1411

H2H_2(g)

-286

C2H6C_2H_6(g)

-1560

A

ΔH⊖\Delta H^\ominus for the hydrogenation of ethene using the combustion data.

[2]
Question 19
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HL • Paper 2
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Nitrogen monoxide reacts with oxygen to form nitrogen dioxide.

2NO(g)+O2(g)→2NO2(g)2NO(g)+O_2(g) \to 2NO_2(g)

The table gives standard enthalpy changes of formation.

Species

ΔH∘^\circf / kJ mol−1\text{mol}^{-1}

NO(g)NO(g)

90.3

O2(g)O_2(g)

0

NO2(g)NO_2(g)

33.2

A

the standard enthalpy change for the reaction.

[2]
Question 20
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The lattice enthalpy of MgOMgO is much larger than that of NaClNaCl when both are defined as the energy required to separate the solid into gaseous ions.

A

this difference in terms of ionic charge and ionic radius.

[2]

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Question 21
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A series of halogenoalkanes, CH3XCH_3X, undergoes nucleophilic substitution with hydroxide ions. Data for the carbon-halogen bond and the relative initial rate are shown.

Halogenoalkane

C-X bond length / pm

C-X bond enthalpy / kJ mol−1\text{kJ mol}^{-1}

Relative initial rate / arbitrary units

CH3FCH_3F

139

485

0.01

CH3ClCH_3Cl

177

338

1.0

CH3BrCH_3Br

194

276

20

CH3ICH_3I

214

238

100

A

the relationship between carbon-halogen bond length and carbon-halogen bond enthalpy.

[1]
B

the trend in relative initial rate from CH3ClCH_3Cl to CH3ICH_3I.

[2]
C

why bond enthalpy alone may not fully the rate of substitution for all CH3XCH_3X compounds.

[1]
Question 22
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SL • Paper 1B
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Chloromethane can be produced by the reaction of methane with chlorine:

CH4(g)+Cl2(g)→CH3Cl(g)+HCl(g)CH_4(g) + Cl_2(g) \to CH_3Cl(g) + HCl(g)

Average bond enthalpy data are provided.

Bond

Average bond enthalpy / kJ mol−1^{-1}

C-H

414

Cl-Cl

243

C-Cl

338

H-Cl

431

A

Identify the bonds broken in one mole of reaction as written.

[1]
B

the enthalpy change, in kJ mol−1\text{kJ mol}^{-1}, using the bond enthalpy data.

[2]
C

what the sign of the enthalpy change shows about the reaction.

[1]
Question 23
HL • Paper 1B
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HL • Paper 1B
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Standard enthalpies of formation data are given in the table.

Species

ΔH∘^\circf / kJ mol−1\text{mol}^{-1}

H2H_2(g)

0

O2O_2(g)

0

H2OH_2O(l)

-286

H2O2H_2O_2(l)

-188

A

ΔH\Delta H for H2O2(l)→H2O(l)+12O2(g)H_2O_2(l) \to H_2O(l) + \frac{1}{2}O_2(g).

[2]
B

why the enthalpy cycle can be used to the decomposition enthalpy.

[1]
Question 24
HL • Paper 1A
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HL • Paper 1A
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A Born-Haber cycle for NaCl(s)NaCl(s) uses the following data.

ΔHf⊖[NaCl(s)]=−411 kJ mol−1\Delta H_f^\ominus[NaCl(s)]=-411\ \text{kJ mol}^{-1}

Na(s)→Na(g)Na(s)\to Na(g) +108 kJ mol−1+108\ \text{kJ mol}^{-1}

12Cl2(g)→Cl(g)\frac{1}{2}Cl_2(g)\to Cl(g) +122 kJ mol−1+122\ \text{kJ mol}^{-1}

Na(g)→Na+(g)+e−Na(g)\to Na^+(g)+e^- +496 kJ mol−1+496\ \text{kJ mol}^{-1}

Cl(g)+e−→Cl−(g)Cl(g)+e^-\to Cl^-(g) −349 kJ mol−1-349\ \text{kJ mol}^{-1}

Using the IB definition of lattice enthalpy as separation into gaseous ions, what is ΔHlattice⊖\Delta H_{\text{lattice}}^\ominus for NaCl(s)NaCl(s)?

A.

+788 kJ mol−1+788\ \text{kJ mol}^{-1}

B.

−34 kJ mol−1-34\ \text{kJ mol}^{-1}

C.

+34 kJ mol−1+34\ \text{kJ mol}^{-1}

D.

−788 kJ mol−1-788\ \text{kJ mol}^{-1}

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Question 25
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Carbon monoxide can be formed from carbon and oxygen.

C(s, graphite)+O2(g)→CO2(g)ΔH=−394 kJ mol−1C(s,\text{ graphite})+O_2(g) \to CO_2(g) \qquad \Delta H=-394\ \text{kJ mol}^{-1}

CO(g)+12O2(g)→CO2(g)ΔH=−283 kJ mol−1CO(g)+\frac{1}{2}O_2(g) \to CO_2(g) \qquad \Delta H=-283\ \text{kJ mol}^{-1}

A

Write the target equation for the formation of one mole of carbon monoxide from its elements.

[1]
B

Use Hess's law to ΔH\Delta H for the target equation.

[2]
Question 26
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HL • Paper 2
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The data refer to a Born-Haber cycle for the formation of NaCl(s)NaCl(s) from Na(s)Na(s) and 12Cl2(g)\frac{1}{2}Cl_2(g). Lattice enthalpy is defined as the enthalpy change when one mole of ionic solid is separated into gaseous ions.

A

why the lattice enthalpy, as defined in the stem, has a positive value.

[1]
B

the lattice enthalpy of NaCl(s)NaCl(s) using the Born-Haber data shown.

[2]
Question 27
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Ethene reacts with hydrogen chloride to form chloroethane. Average bond enthalpy data and the displayed formulae for the reactants and product are provided.

Bond

Average bond enthalpy / kJ mol−1^{-1}

C=C

612

H-Cl

431

C-C

346

C-H

414

C-Cl

338

A

the bonds that are broken and the new bonds that are formed in the reaction.

[2]
B

the enthalpy change, in kJ mol−1\text{kJ mol}^{-1}, for the reaction using the data.

[2]
C

one reason why the value calculated from average bond enthalpies may differ from an experimental enthalpy change.

[1]
Question 28
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SL • Paper 1B
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A reaction from A to B can occur directly or through an intermediate C. An enthalpy cycle and two enthalpy changes are shown.

A

the enthalpy change for the step from C to B.

[2]
B

Hess's law.

[1]
C

why the direct route and the route through C can be compared even if the mechanism is different.

[1]

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Question 29
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Standard enthalpy of formation data are provided for ethanol and its complete combustion products at 298 K and 100 kPa.

Species

ΔH∘^\circf / kJ mol−1\text{mol}^{-1}

CO2(g)CO_2\text{(g)}

-394

H2O(l)H_2O\text{(l)}

-286

C2H5OH(l)C_2H_5OH\text{(l)}

-278

A

Write the equation for the standard enthalpy of formation of ethanol, C2H5OH(l)C_2H_5OH(l).

[1]
B

ΔHc⊖\Delta H_c^\ominus for ethanol using the formation data.

[2]
C

why using H2O(g)H_2O(g) instead of H2O(l)H_2O(l) would change the calculated value.

[1]
Question 30
HL • Paper 1B
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HL • Paper 1B
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Propene can be hydrogenated to propane:

C3H6(g)+H2(g)→C3H8(g)C_3H_6(g) + H_2(g) \to C_3H_8(g)

Standard enthalpies of combustion are shown.

Substance

Standard enthalpy of combustion / kJ mol−1^{-1}

C3H6(g)C_3H_6(g)

-2058

H2(g)H_2(g)

-286

C3H8(g)C_3H_8(g)

-2220

A

ΔH⊖\Delta H^\ominus for the hydrogenation reaction using the combustion data.

[2]
B

why reactant combustion enthalpies are subtracted from product combustion enthalpies in the reverse order to formation data.

[1]
C

why this value is usually more reliable than one calculated from average bond enthalpies.

[1]
Question 31
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HL • Paper 1B
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Limestone contains calcium carbonate, which decomposes on strong heating:

CaCO3(s)→CaO(s)+CO2(g)CaCO_3(s) \to CaO(s) + CO_2(g)

Standard enthalpy of formation data are shown.

Substance

ΔHf∘^\circ / kJ mol−1\text{mol}^{-1}

CaCO3(s)CaCO_3(s)

-1207

CaO(s)CaO(s)

-635

CO2(g)CO_2(g)

-394

A

ΔH⊖\Delta H^\ominus for the decomposition reaction.

[2]
B

whether the decomposition is endothermic or exothermic.

[1]
C

why the symbols in the equation and data table are important.

[1]
Question 32
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Graphite and diamond are allotropes of carbon. Their standard enthalpies of combustion to CO2(g)CO_2(g) are shown.

Carbon allotrope

ΔH∘^\circc / kJ mol−1\text{mol}^{-1}

graphite

-394

diamond

-396

A

ΔH⊖\Delta H^\ominus for C(s, graphite)→C(s, diamond)C(s,\text{ graphite}) \to C(s,\text{ diamond}).

[2]
B

why ΔHf⊖\Delta H_f^\ominus is zero for graphite but not for diamond.

[1]

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Question 33
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The Born-Haber cycle for MgO(s)MgO(s) includes atomization of magnesium, atomization of oxygen, the first and second ionization energies of magnesium, the first and second electron affinities of oxygen, and lattice enthalpy of formation.

Lattice enthalpy is defined as:

Mg2+(g)+O2−(g)→MgO(s)Mg^{2+}(g)+O^{2-}(g) \to MgO(s)

A

why both the first and second ionization energies of magnesium are included in the cycle.

[1]
B

the lattice enthalpy of formation of MgO(s)MgO(s) using the following data: ΔHf⊖[MgO(s)]=−602 kJ mol−1\Delta H_f^\ominus[MgO(s)]=-602\ \text{kJ mol}^{-1}, ΔHat⊖[Mg]=+150 kJ mol−1\Delta H_{at}^\ominus[Mg]=+150\ \text{kJ mol}^{-1}, ΔHat⊖[O]=+249 kJ mol−1\Delta H_{at}^\ominus[O]=+249\ \text{kJ mol}^{-1}, IE1=+738 kJ mol−1IE_1=+738\ \text{kJ mol}^{-1}, IE2=+1451 kJ mol−1IE_2=+1451\ \text{kJ mol}^{-1}, EA1=−141 kJ mol−1EA_1=-141\ \text{kJ mol}^{-1} and EA2=+744 kJ mol−1EA_2=+744\ \text{kJ mol}^{-1}.

[3]
Question 34
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The thermal decomposition of magnesium carbonate is difficult to measure directly. Two reactions with hydrochloric acid were measured in a calorimeter and used in a Hess cycle.

Reaction with HCl

ΔH\Delta H / kJ mol−1^{-1}

MgCO3(s)+2HCl(aq)→MgCl2(aq)+CO2(g)+H2O(l)\text{MgCO}_3\text{(s)} + 2\text{HCl(aq)} \to \text{MgCl}_2\text{(aq)} + \text{CO}_2\text{(g)} + \text{H}_2\text{O(l)}

-25

MgO(s)+2HCl(aq)→MgCl2(aq)+H2O(l)\text{MgO(s)} + 2\text{HCl(aq)} \to \text{MgCl}_2\text{(aq)} + \text{H}_2\text{O(l)}

-151

A

the enthalpy change for the decomposition of MgCO3(s)MgCO_3(s).

[2]
B

why the same final solution must appear in both acid-reaction pathways in the Hess cycle.

[1]
C

one experimental factor that could make the calculated decomposition enthalpy unreliable.

[1]
Question 35
HL • Paper 1B
Hard
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HL • Paper 1B
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A Born-Haber cycle for sodium chloride is shown. In this course, lattice enthalpy is defined as the enthalpy change when one mole of ionic solid is separated into gaseous ions.

Step

ΔH\Delta H / kJ mol−1^{-1}

Na(s)→Na(g)\text{Na}(s) \to \text{Na}(g)

+108

12Cl2(g)→Cl(g)\frac{1}{2}\text{Cl}_2(g) \to \text{Cl}(g)

+121

Na(g)→Na+(g)+e−\text{Na}(g) \to \text{Na}^+(g) + e^-

+496

Cl(g)+e−→Cl−(g)\text{Cl}(g) + e^- \to \text{Cl}^-(g)

−349

Na(s)+12Cl2(g)→NaCl(s)\text{Na}(s) + \frac{1}{2}\text{Cl}_2(g) \to \text{NaCl}(s)

−411

NaCl(s)→Na+(g)+Cl−(g)\text{NaCl}(s) \to \text{Na}^+(g) + \text{Cl}^-(g)

?

A

Identify one endothermic step in the cycle, other than lattice separation.

[1]
B

the lattice enthalpy of sodium chloride using the values in the cycle.

[2]
C

the positive sign of the lattice enthalpy value in this definition.

[1]
Question 36
SL • Paper 2
Hard
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SL • Paper 2
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Hydrogen chloride can be formed from hydrogen and chlorine.

H2(g)+Cl2(g)→2HCl(g)H_2(g) + Cl_2(g) \to 2HCl(g)

Average bond enthalpies are: H−H=436 kJ mol−1H-H = 436\ \text{kJ mol}^{-1}, Cl−Cl=242 kJ mol−1Cl-Cl = 242\ \text{kJ mol}^{-1} and H−Cl=431 kJ mol−1H-Cl = 431\ \text{kJ mol}^{-1}.

A

The reaction involves breaking bonds and forming bonds.

I.

why bond breaking is endothermic and bond forming is exothermic.

[2]
II.

the enthalpy change, in kJ mol−1\text{kJ mol}^{-1}, for the reaction using the average bond enthalpies.

[2]
B

The experimental value for this reaction is close to −185 kJ mol−1-185\ \text{kJ mol}^{-1}. For many other reactions, values calculated using average bond enthalpies differ more from experimental values. two reasons for these differences.

[3]

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Question 37
SL • Paper 2
Hard
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SL • Paper 2
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Ethene reacts with hydrogen bromide to form bromoethane.

CH2=CH2(g)+HBr(g)→CH3CH2Br(g)CH_2{=}CH_2(g) + HBr(g) \to CH_3CH_2Br(g)

Average bond enthalpies are: C=C=614 kJ mol−1C{=}C = 614\ \text{kJ mol}^{-1}, H−Br=366 kJ mol−1H-Br = 366\ \text{kJ mol}^{-1}, C−C=346 kJ mol−1C-C = 346\ \text{kJ mol}^{-1}, C−H=414 kJ mol−1C-H = 414\ \text{kJ mol}^{-1} and C−Br=276 kJ mol−1C-Br = 276\ \text{kJ mol}^{-1}.

A

Use the structures to account for the bonds broken and formed.

I.

the bonds broken in this reaction.

[2]
II.

the new bonds formed in this reaction.

[2]
B

the enthalpy change, in kJ mol−1\text{kJ mol}^{-1}, for the reaction using the average bond enthalpies.

[3]
C

why drawing displayed formulae before using bond enthalpies reduces errors in this calculation.

[1]
Question 38
SL • Paper 2
Hard
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SL • Paper 2
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The enthalpy change for converting carbon monoxide to carbon dioxide can be found indirectly.

CO(g)+12O2(g)→CO2(g)CO(g) + \frac{1}{2}O_2(g) \to CO_2(g)

Given:

C(s, graphite)+12O2(g)→CO(g)C(s,\text{ graphite}) + \frac{1}{2}O_2(g) \to CO(g), ΔH=−111 kJ mol−1\Delta H = -111\ \text{kJ mol}^{-1}

C(s, graphite)+O2(g)→CO2(g)C(s,\text{ graphite}) + O_2(g) \to CO_2(g), ΔH=−394 kJ mol−1\Delta H = -394\ \text{kJ mol}^{-1}

A

Use the given equations to obtain the target reaction.

I.

the change made to the equation forming CO(g)CO(g) before adding equations.

[1]
II.

ΔH\Delta H, in kJ mol−1\text{kJ mol}^{-1}, for the target reaction.

[2]
B

why the symbol of carbon is written as graphite in these equations.

[1]
C

why an indirect Hess cycle is useful for this reaction.

[2]
Question 39
HL • Paper 1B
Hard
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HL • Paper 1B
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A Born-Haber cycle for magnesium oxide is shown. The lattice enthalpy is defined as the enthalpy change for MgO(s)→Mg2+(g)+O2−(g)MgO(s) \to Mg^{2+}(g) + O^{2-}(g).

Born-Haber step

ΔH\Delta H / kJ mol−1^{-1}

Mg(s)→Mg(g)\text{Mg}(s) \to \text{Mg}(g) (sublimation)

+148

Mg(g)→Mg+(g)+e−\text{Mg}(g) \to \text{Mg}^+(g) + e^- (1st ionization energy)

+738

Mg+(g)→Mg2+(g)+e−\text{Mg}^+(g) \to \text{Mg}^{2+}(g) + e^- (2nd ionization energy)

+1451

12O2(g)→O(g)\frac{1}{2}\text{O}_2(g) \to \text{O}(g) (atomisation)

+249

O(g)+e−→O−(g)\text{O}(g) + e^- \to \text{O}^-(g) (1st electron affinity)

-141

O−(g)+e−→O2−(g)\text{O}^-(g) + e^- \to \text{O}^{2-}(g) (2nd electron affinity)

+844

Mg(s)+12O2(g)→MgO(s)\text{Mg}(s) + \frac{1}{2}\text{O}_2(g) \to \text{MgO}(s) (standard enthalpy of formation)

-602

A

the lattice enthalpy of magnesium oxide using the cycle.

[3]
B

why the second electron affinity of oxygen is endothermic.

[1]
C

why the lattice enthalpy of magnesium oxide is much larger than that of sodium chloride.

[1]
Question 40
SL • Paper 2
Hard
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SL • Paper 2
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The rate of nucleophilic substitution of halogenoalkanes is affected by the strength of the carbon-halogen bond. The average bond enthalpies are shown.

C−Cl=338 kJ mol−1C-Cl = 338\ \text{kJ mol}^{-1}, C−Br=276 kJ mol−1C-Br = 276\ \text{kJ mol}^{-1}, C−I=238 kJ mol−1C-I = 238\ \text{kJ mol}^{-1}.

A

The carbon-halogen bond is broken during substitution.

I.

Identify the strongest carbon-halogen bond from the data.

[1]
II.

the relationship between carbon-halogen bond enthalpy and the energy required to break the bond.

[2]
B

and the expected rates of substitution of chloroethane, bromoethane and iodoethane, using the bond enthalpy data.

[3]
C

one other factor, besides bond enthalpy, that can affect a bond enthalpy value.

[1]

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Question 41
SL • Paper 2
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SL • Paper 2
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A student investigates the enthalpy change of decomposition of hydrated copper(II) sulfate using two measured hydration reactions.

Reaction 1: CuSO4(s)+aq→CuSO4(aq)CuSO_4(s) + aq \to CuSO_4(aq), ΔH1=−66.1 kJ mol−1\Delta H_1 = -66.1\ \text{kJ mol}^{-1}

Reaction 2: CuSO4⋅5H2O(s)+aq→CuSO4(aq)+5H2O(l)CuSO_4\cdot 5H_2O(s) + aq \to CuSO_4(aq) + 5H_2O(l), ΔH2=+11.7 kJ mol−1\Delta H_2 = +11.7\ \text{kJ mol}^{-1}

Target reaction: CuSO4⋅5H2O(s)→CuSO4(s)+5H2O(l)CuSO_4\cdot 5H_2O(s) \to CuSO_4(s) + 5H_2O(l)

A

Hess's law can be applied to this system.

I.

Hess's law.

[1]
II.

why Hess's law is an application of conservation of energy.

[2]
B

the enthalpy change, in kJ mol−1\text{kJ mol}^{-1}, for the target reaction.

[2]
C

two experimental limitations that could affect the reliability of this Hess cycle determination.

[3]
Question 42
SL • Paper 2
Hard
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SL • Paper 2
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Methanol combusts according to the equation:

CH3OH(g)+32O2(g)→CO2(g)+2H2O(g)CH_3OH(g) + \frac{3}{2}O_2(g) \to CO_2(g) + 2H_2O(g)

Average bond enthalpies are: C−H=414 kJ mol−1C-H = 414\ \text{kJ mol}^{-1}, C−O=358 kJ mol−1C-O = 358\ \text{kJ mol}^{-1}, O−H=463 kJ mol−1O-H = 463\ \text{kJ mol}^{-1}, O=O=498 kJ mol−1O{=}O = 498\ \text{kJ mol}^{-1} and C=OC{=}O in CO2=804 kJ mol−1CO_2 = 804\ \text{kJ mol}^{-1}.

A

The calculation can be performed by counting bonds in the balanced equation.

I.

the total bond enthalpy for the bonds broken.

[2]
II.

the total bond enthalpy for the bonds formed.

[2]
B

the enthalpy change, in kJ mol−1\text{kJ mol}^{-1}, for the combustion of methanol as written.

[1]
C

The data booklet gives a standard enthalpy of combustion for CH3OH(l)CH_3OH(l) rather than CH3OH(g)CH_3OH(g). why this standard value would differ from the value calculated using average bond enthalpies.

[3]
Question 43
HL • Paper 2
Hard
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HL • Paper 2
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Propanone, CH3COCH3(l)CH_3COCH_3(l), is a volatile solvent. The equation for its complete combustion under standard conditions is:

CH3COCH3(l)+4O2(g)→3CO2(g)+3H2O(l)CH_3COCH_3(l) + 4O_2(g) \to 3CO_2(g) + 3H_2O(l)

Standard enthalpies of formation are: ΔHf⊖[CH3COCH3(l)]=−249 kJ mol−1\Delta H_f^\ominus[CH_3COCH_3(l)] = -249\ \text{kJ mol}^{-1}, ΔHf⊖[CO2(g)]=−394 kJ mol−1\Delta H_f^\ominus[CO_2(g)] = -394\ \text{kJ mol}^{-1} and ΔHf⊖[H2O(l)]=−286 kJ mol−1\Delta H_f^\ominus[H_2O(l)] = -286\ \text{kJ mol}^{-1}.

A

Use the formation data to the standard enthalpy change of combustion of propanone.

I.

why ΔHf⊖[O2(g)]\Delta H_f^\ominus[O_2(g)] is not included in the calculation.

[1]
II.

ΔHc⊖\Delta H_c^\ominus, in kJ mol−1\text{kJ mol}^{-1}, for propanone.

[3]
B

the standard formation equation for liquid propanone.

[2]
C

one advantage of using standard enthalpy of formation data rather than average bond enthalpies for this calculation.

[2]
Question 44
HL • Paper 2
Hard
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HL • Paper 2
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Ethene can be hydrogenated to ethane.

C2H4(g)+H2(g)→C2H6(g)C_2H_4(g) + H_2(g) \to C_2H_6(g)

Standard enthalpies of combustion are: ΔHc⊖[C2H4(g)]=−1411 kJ mol−1\Delta H_c^\ominus[C_2H_4(g)] = -1411\ \text{kJ mol}^{-1}, ΔHc⊖[H2(g)]=−286 kJ mol−1\Delta H_c^\ominus[H_2(g)] = -286\ \text{kJ mol}^{-1} and ΔHc⊖[C2H6(g)]=−1560 kJ mol−1\Delta H_c^\ominus[C_2H_6(g)] = -1560\ \text{kJ mol}^{-1}.

A

The combustion data can be used because reactants and products combust to the same final substances.

I.

Write the equation for the complete combustion of ethene.

[1]
II.

the expression, using combustion data, for calculating the standard enthalpy change of the hydrogenation reaction.

[2]
B

ΔH⊖\Delta H^\ominus, in kJ mol−1\text{kJ mol}^{-1}, for the hydrogenation reaction.

[2]
C

why reversing a thermochemical equation changes the sign of its enthalpy change.

[2]

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Question 45
HL • Paper 2
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HL • Paper 2
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Carbon exists as the allotropes graphite and diamond. Graphite is the standard of carbon at 298 K298\ \text{K} and 100 kPa100\ \text{kPa}.

The standard enthalpy change for C(s, graphite)→C(s, diamond)C(s,\text{ graphite}) \to C(s,\text{ diamond}) is +1.9 kJ mol−1+1.9\ \text{kJ mol}^{-1}.

A

Allotropes have different structures and bonding.

I.

Define standard enthalpy change of formation.

[2]
II.

the standard enthalpy of formation of graphite.

[1]
B

the standard enthalpy of formation of diamond and your answer.

[2]
C

why symbols and allotropes must be specified when using standard enthalpy data.

[2]
Question 46
HL • Paper 2
Hard
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HL • Paper 2
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The enthalpy change for the reaction between methane and steam can be calculated using either formation data or combustion data.

CH4(g)+H2O(g)→CO(g)+3H2(g)CH_4(g) + H_2O(g) \to CO(g) + 3H_2(g)

Standard enthalpies of formation are: CH4(g)=−75 kJ mol−1CH_4(g) = -75\ \text{kJ mol}^{-1}, H2O(g)=−242 kJ mol−1H_2O(g) = -242\ \text{kJ mol}^{-1} and CO(g)=−111 kJ mol−1CO(g) = -111\ \text{kJ mol}^{-1}.

A

Use the formation data to the enthalpy change for the reaction.

I.

why H2(g)H_2(g) does not contribute to the formation-data sum.

[1]
II.

ΔH⊖\Delta H^\ominus, in kJ mol−1\text{kJ mol}^{-1}, for the reaction.

[3]
B

the use of formation data with the use of combustion data in Hess law calculations.

[2]
C

whether the reaction is endothermic or exothermic and relate this to the feasibility of carrying it out at high temperature.

[2]
Question 47
HL • Paper 2
Hard
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HL • Paper 2
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A Born-Haber cycle is used to the lattice enthalpy of magnesium oxide, defined in this question as:

MgO(s)→Mg2+(g)+O2−(g)MgO(s) \to Mg^{2+}(g) + O^{2-}(g)

Data are given in kJ mol−1\text{kJ mol}^{-1}:

ΔHf⊖[MgO(s)]=−602\Delta H_f^\ominus[MgO(s)] = -602; atomization of Mg(s)=+148Mg(s) = +148; atomization of oxygen, 12O2(g)→O(g)=+249\frac{1}{2}O_2(g) \to O(g) = +249; IE1(Mg)=+738IE_1(Mg) = +738; IE2(Mg)=+1451IE_2(Mg) = +1451; EA1(O)=−141EA_1(O) = -141; EA2(O)=+744EA_2(O) = +744.

A

Interpret the steps in the Born-Haber cycle.

I.

why two ionization energies of magnesium are included.

[1]
II.

why the second electron affinity of oxygen is endothermic.

[2]
B

the lattice enthalpy of magnesium oxide as defined in this question.

[3]
C

why the lattice enthalpy of magnesium oxide is much larger than that of sodium chloride.

[2]
Question 48
HL • Paper 2
Hard
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HL • Paper 2
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The lattice enthalpy of potassium bromide can be determined from a Born-Haber cycle. Lattice enthalpy is defined here as:

KBr(s)→K+(g)+Br−(g)KBr(s) \to K^+(g) + Br^-(g)

Data are given in kJ mol−1\text{kJ mol}^{-1}:

ΔHf⊖[KBr(s)]=−394\Delta H_f^\ominus[KBr(s)] = -394; atomization of K(s)=+89K(s) = +89; atomization of bromine, 12Br2(l)→Br(g)=+112\frac{1}{2}Br_2(l) \to Br(g) = +112; IE1(K)=+419IE_1(K) = +419; EA1(Br)=−325EA_1(Br) = -325.

A

The cycle includes atomization, ionization, electron affinity and lattice enthalpy steps.

I.

Write the equation for the first electron affinity of bromine.

[1]
II.

why the first ionization energy of potassium is endothermic.

[2]
B

the lattice enthalpy of potassium bromide as defined in this question.

[3]
C

why different data sources may quote lattice enthalpies with opposite signs.

[2]

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R1.1 Measuring enthalpy changes

R1.3 Energy from fuels