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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.
Paper
Difficulty
Status
Level
Question 1
SL ‱ Paper 1A
Easy
Calculator Permitted

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
Easy
Calculator Permitted

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
Easy
Calculator Permitted

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
SL ‱ Paper 1A
Easy
Calculator Permitted

Hess's law is valid because enthalpy is a state 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

Question 5
HL ‱ Paper 1A
Easy
Calculator Permitted

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
Easy
Calculator Permitted

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
Easy
Calculator Permitted

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
Easy
Calculator Permitted

Bond enthalpy data can be used to interpret energy changes during reactions.

A

Define bond enthalpy.

[1]
Write your answer here...
B

Explain why bond breaking is endothermic whereas bond forming is exothermic.

[2]
Write your answer here...

0

Question 9
HL ‱ Paper 2
Easy
Calculator Permitted

Methanol has the formula CH3OHCH_3OH.

A

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

[1]
Write your answer here...
B

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

[1]
Write your answer here...

0

Question 10
SL ‱ Paper 1A
Medium
Calculator Permitted

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
SL ‱ Paper 1A
Medium
Calculator Permitted

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
HL ‱ Paper 1A
Medium
Calculator Permitted

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}

Question 13
HL ‱ Paper 1A
Medium
Calculator Permitted

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
SL ‱ Paper 2
Medium
Calculator Permitted

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.

BondAverage bond enthalpy / kJ mol⁻Âč
C-H413
Cl-Cl243
C-Cl338
H-Cl432
A

State the bonds broken and the bonds formed in this reaction.

[1]
Write your answer here...
B

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

[2]
Write your answer here...

0

Question 15
SL ‱ Paper 2
Medium
Calculator Permitted

A student uses average bond enthalpy data to estimate ΔH\Delta H for the combustion of liquid ethanol. The experimental value is more exothermic than the estimated value.

A

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

[2]
Write your answer here...

0

Question 16
SL ‱ Paper 2
Medium
Calculator Permitted

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 simple Hess cycle showing A at the left, B at the right and intermediate C below. A top arrow goes from A to B and is labelled as the unknown enthalpy change. Two arrows show the alternative route from A to C and from C to B, labelled with the two given enthalpy changes but without additional explanatory text.
A

State Hess's law.

[1]
Write your answer here...
B

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

[1]
Write your answer here...

0

Question 17
SL ‱ Paper 2
Medium
Calculator Permitted

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.

BondBond length / pmBond enthalpy / kJ mol^-1
C-Cl177338
C-Br194276
C-I214238
A

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

[2]
Write your answer here...

0

Question 18
HL ‱ Paper 2
Medium
Calculator Permitted

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.

SubstanceΔH°c / kJ mol⁻Âč
C₂H₄(g)-1411
H₂(g)-286
C₂H₆(g)-1560
A

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

[2]
Write your answer here...

0

Question 19
HL ‱ Paper 2
Medium
Calculator Permitted

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.

SpeciesΔH°f / kJ mol^-1
NO(g)90.3
O2(g)0
NO2(g)33.2
A

Calculate the standard enthalpy change for the reaction.

[2]
Write your answer here...

0

Question 20
HL ‱ Paper 2
Medium
Calculator Permitted

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

Explain this difference in terms of ionic charge and ionic radius.

[2]
Write your answer here...

0

Question 21
SL ‱ Paper 1B
Medium
Calculator Permitted

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.

HalogenoalkaneC-X bond length / pmC-X bond enthalpy / kJ mol^-1Relative initial rate / arbitrary units
CH3F1394850.01
CH3Cl1773381.0
CH3Br19427620
CH3I214238100
A

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

[1]
Write your answer here...
B

Explain the trend in relative initial rate from CH3ClCH_3Cl to CH3ICH_3I.

[2]
Write your answer here...
C

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

[1]
Write your answer here...

0

Question 22
SL ‱ Paper 1B
Medium
Calculator Permitted

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.

BondAverage bond enthalpy / kJ mol^-1
C-H414
Cl-Cl243
C-Cl338
H-Cl431
A

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

[1]
Write your answer here...
B

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

[2]
Write your answer here...
C

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

[1]
Write your answer here...

0

Question 23
HL ‱ Paper 1B
Medium
Calculator Permitted

Standard enthalpies of formation data are given in the table.

SpeciesΔH°f / kJ mol^-1
H2(g)0
O2(g)0
H2O(l)-286
H2O2(l)-188
A

Calculate 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]
Write your answer here...
B

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

[1]
Write your answer here...

0

Question 24
HL ‱ Paper 1A
Medium
Calculator Permitted

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}

Question 25
SL ‱ Paper 2
Medium
Calculator Permitted

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]
Write your answer here...
B

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

[2]
Write your answer here...

0

Question 26
HL ‱ Paper 2
Medium
Calculator Permitted

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 Born-Haber cycle for sodium chloride. It shows Na(s) plus half Cl2(g) leading to NaCl(s) with the standard enthalpy of formation, and an alternative route through Na(g), Cl(g), Na+(g), Cl-(g), and gaseous ions. The lattice enthalpy arrow is from NaCl(s) to Na+(g) plus Cl-(g) and is labelled as unknown. The other arrows are labelled as atomization of sodium, atomization of chlorine, first ionization energy of sodium and first electron affinity of chlorine.
A

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

[1]
Write your answer here...
B

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

[2]
Write your answer here...

0

Question 27
SL ‱ Paper 1B
Medium
Calculator Permitted

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

BondAverage bond enthalpy / kJ mol^-1
C=C612
H-Cl431
C-C346
C-H414
C-Cl338
A

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

[2]
Write your answer here...
B

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

[2]
Write your answer here...
C

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

[1]
Write your answer here...

0

Question 28
SL ‱ Paper 1B
Medium
Calculator Permitted

A reaction from A to B can occur directly or through an intermediate C. An enthalpy cycle and two enthalpy changes are shown.

Bar chart of relative enthalpy levels for states A, C and B.
A

Determine the enthalpy change for the step from C to B.

[2]
Write your answer here...
B

State Hess's law.

[1]
Write your answer here...
C

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

[1]
Write your answer here...

0

Question 29
HL ‱ Paper 1B
Medium
Calculator Permitted

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

SpeciesΔH°f / kJ mol^-1
CO2(g)-394
H2O(l)-286
C2H5OH(l)-278
A

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

[1]
Write your answer here...
B

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

[2]
Write your answer here...
C

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

[1]
Write your answer here...

0

Question 30
HL ‱ Paper 1B
Medium
Calculator Permitted

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.

SubstanceStandard enthalpy of combustion / kJ mol^-1
C3H6(g)-2058
H2(g)-286
C3H8(g)-2220
A

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

[2]
Write your answer here...
B

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

[1]
Write your answer here...
C

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

[1]
Write your answer here...

0

Question 31
HL ‱ Paper 1B
Medium
Calculator Permitted

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.

SubstanceΔHf° / kJ mol^-1
CaCO3(s)-1207
CaO(s)-635
CO2(g)-394
A

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

[2]
Write your answer here...
B

Deduce whether the decomposition is endothermic or exothermic.

[1]
Write your answer here...
C

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

[1]
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0

Question 32
HL ‱ Paper 1B
Medium
Calculator Permitted

Graphite and diamond are allotropes of carbon. Their standard enthalpies of combustion to CO2(g)CO_2(g) are shown.

Carbon allotropeΔH°c / kJ mol^-1
graphite-394
diamond-396
A

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

[2]
Write your answer here...
B

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

[1]
Write your answer here...

0

Question 33
HL ‱ Paper 2
Medium
Calculator Permitted

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 Born-Haber cycle for magnesium oxide. It shows $Mg(s)$ plus half $O_2(g)$ forming $MgO(s)$ and an alternative route through $Mg(g)$, $O(g)$, $Mg^+(g)$, $Mg^{2+}(g)$, $O^-(g)$, $O^{2-}(g)$, and gaseous $Mg^{2+}$ plus $O^{2-}$ ions. Arrows are labelled for atomization, first and second ionization energies, first and second electron affinities, enthalpy of formation and an unknown lattice enthalpy, $L$, for formation of $MgO(s)$ from gaseous ions.
A

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

[1]
Write your answer here...
B

Determine 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]
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0

Question 34
SL ‱ Paper 1B
Hard
Calculator Permitted

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 HClΔH / kJ mol^-1
MgCO3(s) + 2HCl(aq) → MgCl2(aq) + CO2(g) + H2O(l)-25
MgO(s) + 2HCl(aq) → MgCl2(aq) + H2O(l)-151
A

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

[2]
Write your answer here...
B

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

[1]
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C

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

[1]
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0

Question 35
HL ‱ Paper 1B
Hard
Calculator Permitted

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.

StepΔH / kJ mol^-1
Na(s) → Na(g)+108
1/2 Cl₂(g) → Cl(g)+121
Na(g) → Naâș(g) + e⁻+496
Cl(g) + e⁻ → Cl⁻(g)−349
Na(s) + 1/2 Cl₂(g) → NaCl(s)−411
NaCl(s) → Naâș(g) + Cl⁻(g)?
A

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

[1]
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B

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

[2]
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C

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

[1]
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0

Question 36
SL ‱ Paper 2
Hard
Calculator Permitted

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.

Explain why bond breaking is endothermic and bond forming is exothermic.

[2]
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II.

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

[2]
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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. Discuss two reasons for these differences.

[3]
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Question 37
SL ‱ Paper 2
Hard
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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}.

Displayed structural formulae for ethene, hydrogen bromide and bromoethane arranged left-to-right with a reaction arrow. All atoms and bonds are shown clearly so that the change from a carbon-carbon double bond to a carbon-carbon single bond and addition of H and Br can be identified. The diagram must not include bond enthalpy values or calculation steps.
A

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

I.

State the bonds broken in this reaction.

[2]
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II.

State the new bonds formed in this reaction.

[2]
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B

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

[3]
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C

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

[1]
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0

Question 38
SL ‱ Paper 2
Hard
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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.

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

[1]
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II.

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

[2]
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B

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

[1]
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C

Discuss why an indirect Hess cycle is useful for this reaction.

[2]
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0

Question 39
HL ‱ Paper 1B
Hard
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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 stepΔH / kJ mol^-1
Mg(s) → Mg(g) (sublimation)+148
Mg(g) → Mg+(g) + e− (1st ionization energy)+738
Mg+(g) → Mg2+(g) + e− (2nd ionization energy)+1451
1/2 O2(g) → O(g) (atomisation)+249
O(g) + e− → O−(g) (1st electron affinity)-141
O−(g) + e− → O2−(g) (2nd electron affinity)+844
Mg(s) + 1/2 O2(g) → MgO(s) (standard enthalpy of formation)-602
A

Determine the lattice enthalpy of magnesium oxide using the cycle.

[3]
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B

Explain why the second electron affinity of oxygen is endothermic.

[1]
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C

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

[1]
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0

Question 40
SL ‱ Paper 2
Hard
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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]
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II.

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

[2]
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B

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

[3]
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C

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

[1]
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0

Question 41
SL ‱ Paper 2
Hard
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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 cycle with hydrated copper(II) sulfate at the top left, anhydrous copper(II) sulfate plus water at the top right, and aqueous copper(II) sulfate plus water as the common lower state. Arrows show the two dissolution routes downward to the common lower state and the unknown top reaction. Enthalpy labels are present as symbols only, not numerical substitutions.
A

Hess's law can be applied to this system.

I.

State Hess's law.

[1]
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II.

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

[2]
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B

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

[2]
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C

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

[3]
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0

Question 42
SL ‱ Paper 2
Hard
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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.

Calculate the total bond enthalpy for the bonds broken.

[2]
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II.

Calculate the total bond enthalpy for the bonds formed.

[2]
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B

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

[1]
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C

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

[3]
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0

Question 43
HL ‱ Paper 2
Hard
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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 calculate the standard enthalpy change of combustion of propanone.

I.

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

[1]
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II.

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

[3]
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B

Deduce the standard formation equation for liquid propanone.

[2]
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C

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

[2]
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0

Question 44
HL ‱ Paper 2
Hard
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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 Hess cycle for hydrogenation of ethene to ethane. Ethene plus hydrogen appear at the top left, ethane at the top right, and common combustion products carbon dioxide and liquid water at the bottom. Arrows from both top positions point down to the common products and the top arrow is the hydrogenation reaction. Enthalpy labels are symbols only.
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]
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II.

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

[2]
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B

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

[2]
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C

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

[2]
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0

Question 45
HL ‱ Paper 2
Hard
Calculator Permitted

Carbon exists as the allotropes graphite and diamond. Graphite is the standard state 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]
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II.

State the standard enthalpy of formation of graphite.

[1]
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B

Deduce the standard enthalpy of formation of diamond and explain your answer.

[2]
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C

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

[2]
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0

Question 46
HL ‱ Paper 2
Hard
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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 calculate the enthalpy change for the reaction.

I.

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

[1]
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II.

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

[3]
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B

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

[2]
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C

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

[2]
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0

Question 47
HL ‱ Paper 2
Hard
Calculator Permitted

A Born-Haber cycle is used to determine 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 Born-Haber cycle for magnesium oxide showing formation of MgO(s) from Mg(s) and half a mole of oxygen, atomization to Mg(g) and O(g), ionization to Mg2+(g), electron affinity steps to O2-(g), and lattice separation from MgO(s) to gaseous ions. Arrows and labels are symbolic; numerical values are not inserted into the diagram.
A

Interpret the steps in the Born-Haber cycle.

I.

State why two ionization energies of magnesium are included.

[1]
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II.

Explain why the second electron affinity of oxygen is endothermic.

[2]
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B

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

[3]
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C

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

[2]
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0

Question 48
HL ‱ Paper 2
Hard
Calculator Permitted

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 Born-Haber cycle for potassium bromide showing K(s) plus half Br2(l) forming KBr(s), atomization to gaseous atoms, ionization of K, electron affinity of Br, and lattice separation to gaseous ions. Arrows are clearly labelled by process names but do not contain numerical answers.
A

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

I.

Write the equation for the first electron affinity of bromine.

[1]
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II.

Explain why the first ionization energy of potassium is endothermic.

[2]
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B

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

[3]
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C

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

[2]
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0


R1.1 Measuring enthalpy changes

R1.3 Energy from fuels