R1.2 Energy cycles in reactions - IB Chemistry Question Bank

Question 1

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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

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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

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The carbon-halogen bond enthalpies in halogenoalkanes generally decrease from $C-Cl$ to $C-I$ . What is the most likely consequence for nucleophilic substitution, other factors being similar?

A.

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

B.

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

C.

Chloroalkanes react faster because the $C-Cl$ 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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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

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

A.

$C(s,\text{ graphite})+2H_2(g)+\frac{1}{2}O_2(g) \to CH_3OH(l)$

B.

$CH_3OH(l)+\frac{3}{2}O_2(g) \to CO_2(g)+2H_2O(l)$

C.

$C(s,\text{ diamond})+2H_2(g)+\frac{1}{2}O_2(g) \to CH_3OH(l)$

D.

$CH_4(g)+\frac{1}{2}O_2(g) \to CH_3OH(l)$

Question 6

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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 $0\ \text{kJ mol}^{-1}$ ; graphite is not necessarily $0\ \text{kJ mol}^{-1}$

B.

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

C.

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

D.

Neither graphite nor diamond can have a standard enthalpy of formation

Question 7

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In a Born-Haber cycle for magnesium oxide, what is the step represented by $Mg^+(g) \to Mg^{2+}(g)+e^-$ ?

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

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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

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

[2]

Question 9

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Methanol has the formula $CH_3OH$ .

A

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

[1]

B

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

[1]

Question 10

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The reaction between methane and chlorine is shown.

$CH_4(g)+Cl_2(g) \to CH_3Cl(g)+HCl(g)$

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

What is the enthalpy change for the reaction?

A.

$-114\ \text{kJ mol}^{-1}$

B.

$+356\ \text{kJ mol}^{-1}$

C.

$+114\ \text{kJ mol}^{-1}$

D.

$-356\ \text{kJ mol}^{-1}$

Question 11

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Use the thermochemical equations.

$C(s)+O_2(g) \to CO_2(g)$ $\Delta H=-394\ \text{kJ mol}^{-1}$

$CO(g)+\frac{1}{2}O_2(g) \to CO_2(g)$ $\Delta H=-283\ \text{kJ mol}^{-1}$

What is $\Delta H$ for the reaction below?

$C(s)+\frac{1}{2}O_2(g) \to CO(g)$

A.

$+111\ \text{kJ mol}^{-1}$

B.

$-111\ \text{kJ mol}^{-1}$

C.

$-677\ \text{kJ mol}^{-1}$

D.

$+677\ \text{kJ mol}^{-1}$

Question 12

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Use the standard enthalpies of combustion.

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

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

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

$CH_3CH_2OH(l)+O_2(g) \to CH_3COOH(l)+H_2O(l)$

A.

$+492\ \text{kJ mol}^{-1}$

B.

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

C.

$-492\ \text{kJ mol}^{-1}$

D.

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

Question 13

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Use the standard enthalpies of formation.

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

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

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

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

$CaO(s)+CO_2(g) \to CaCO_3(s)$

A.

$+178\ \text{kJ mol}^{-1}$

B.

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

C.

$-178\ \text{kJ mol}^{-1}$

D.

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

Question 14

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Methane reacts with chlorine in a substitution reaction.

$CH_4(g)+Cl_2(g) \to CH_3Cl(g)+HCl(g)$

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

Bond	Average bond enthalpy / kJ mol $^{-1}$
C-H	413
Cl-Cl	243
C-Cl	338
H-Cl	432

A

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

[1]

B

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

[2]

Question 15

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A student uses average bond enthalpy data to estimate $\Delta H$ 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]

Question 16

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A reaction can occur directly from $A$ to $B$ , or by a two-step route through an intermediate $C$ .

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

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

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]

B

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

[1]

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 $^{-1}$
C-Cl	177	338
C-Br	194	276
C-I	214	238

A

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

[2]

Question 18

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Ethene can be hydrogenated to ethane.

$C_2H_4(g)+H_2(g) \to C_2H_6(g)$

The table gives standard enthalpy changes of combustion.

Substance	ΔH $^\circ$ c / kJ mol⁻¹
$C_2H_4$ (g)	-1411
$H_2$ (g)	-286
$C_2H_6$ (g)	-1560

A

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

[2]

Question 19

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Nitrogen monoxide reacts with oxygen to form nitrogen dioxide.

$2NO(g)+O_2(g) \to 2NO_2(g)$

The table gives standard enthalpy changes of formation.

Species	ΔH $^\circ$ f / kJ $\text{mol}^{-1}$
$NO(g)$	90.3
$O_2(g)$	0
$NO_2(g)$	33.2

A

Calculate the standard enthalpy change for the reaction.

[2]

Question 20

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The lattice enthalpy of $MgO$ is much larger than that of $NaCl$ 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]

Question 21

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A series of halogenoalkanes, $CH_3X$ , 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 / $\text{kJ mol}^{-1}$	Relative initial rate / arbitrary units
$CH_3F$	139	485	0.01
$CH_3Cl$	177	338	1.0
$CH_3Br$	194	276	20
$CH_3I$	214	238	100

A

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

[1]

B

Explain the trend in relative initial rate from $CH_3Cl$ to $CH_3I$ .

[2]

C

Suggest why bond enthalpy alone may not fully predict the rate of substitution for all $CH_3X$ compounds.

[1]

Question 22

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Chloromethane can be produced by the reaction of methane with chlorine:

$CH_4(g) + Cl_2(g) \to CH_3Cl(g) + HCl(g)$

Average bond enthalpy data are provided.

Bond	Average bond enthalpy / kJ mol $^{-1}$
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

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

[2]

C

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

[1]

Question 23

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Standard enthalpies of formation data are given in the table.

Species	ΔH $^\circ$ f / kJ $\text{mol}^{-1}$
$H_2$ (g)	0
$O_2$ (g)	0
$H_2O$ (l)	-286
$H_2O_2$ (l)	-188

A

Calculate $\Delta H$ for $H_2O_2(l) \to H_2O(l) + \frac{1}{2}O_2(g)$ .

[2]

B

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

[1]

Question 24

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A Born-Haber cycle for $NaCl(s)$ uses the following data.

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

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

$\frac{1}{2}Cl_2(g)\to Cl(g)$ $+122\ \text{kJ mol}^{-1}$

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

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

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

A.

$+788\ \text{kJ mol}^{-1}$

B.

$-34\ \text{kJ mol}^{-1}$

C.

$+34\ \text{kJ mol}^{-1}$

D.

$-788\ \text{kJ mol}^{-1}$

Question 25

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Carbon monoxide can be formed from carbon and oxygen.

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

$CO(g)+\frac{1}{2}O_2(g) \to CO_2(g) \qquad \Delta H=-283\ \text{kJ mol}^{-1}$

A

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

[1]

B

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

[2]

Question 26

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The data refer to a Born-Haber cycle for the formation of $NaCl(s)$ from $Na(s)$ and $\frac{1}{2}Cl_2(g)$ . 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]

B

Determine the lattice enthalpy of $NaCl(s)$ 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 $^{-1}$
C=C	612
H-Cl	431
C-C	346
C-H	414
C-Cl	338

A

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

[2]

B

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

[2]

C

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

[1]

Question 28

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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.

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]

B

State Hess's law.

[1]

C

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

[1]

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 $^\circ$ f / kJ $\text{mol}^{-1}$
$CO_2\text{(g)}$	-394
$H_2O\text{(l)}$	-286
$C_2H_5OH\text{(l)}$	-278

A

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

[1]

B

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

[2]

C

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

[1]

Question 30

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Propene can be hydrogenated to propane:

$C_3H_6(g) + H_2(g) \to C_3H_8(g)$

Standard enthalpies of combustion are shown.

Substance	Standard enthalpy of combustion / kJ mol $^{-1}$
$C_3H_6(g)$	-2058
$H_2(g)$	-286
$C_3H_8(g)$	-2220

A

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

[2]

B

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

[1]

C

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

[1]

Question 31

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Limestone contains calcium carbonate, which decomposes on strong heating:

$CaCO_3(s) \to CaO(s) + CO_2(g)$

Standard enthalpy of formation data are shown.

Substance	ΔHf $^\circ$ / kJ $\text{mol}^{-1}$
$CaCO_3(s)$	-1207
$CaO(s)$	-635
$CO_2(g)$	-394

A

Calculate $\Delta H^\ominus$ for the decomposition reaction.

[2]

B

Deduce whether the decomposition is endothermic or exothermic.

[1]

C

Suggest why the state 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 $CO_2(g)$ are shown.

Carbon allotrope	ΔH $^\circ$ c / kJ $\text{mol}^{-1}$
graphite	-394
diamond	-396

A

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

[2]

B

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

[1]

Question 33

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The Born-Haber cycle for $MgO(s)$ 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:

$Mg^{2+}(g)+O^{2-}(g) \to MgO(s)$

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]

B

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

[3]

Question 34

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SL • Paper 1B

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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	$\Delta H$ / kJ mol $^{-1}$
$\text{MgCO}_3\text{(s)} + 2\text{HCl(aq)} \to \text{MgCl}_2\text{(aq)} + \text{CO}_2\text{(g)} + \text{H}_2\text{O(l)}$	-25
$\text{MgO(s)} + 2\text{HCl(aq)} \to \text{MgCl}_2\text{(aq)} + \text{H}_2\text{O(l)}$	-151

A

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

[2]

B

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

[1]

C

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

[1]

Question 35

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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	$\Delta H$ / kJ mol $^{-1}$
$\text{Na}(s) \to \text{Na}(g)$	+108
$\frac{1}{2}\text{Cl}_2(g) \to \text{Cl}(g)$	+121
$\text{Na}(g) \to \text{Na}^+(g) + e^-$	+496
$\text{Cl}(g) + e^- \to \text{Cl}^-(g)$	−349
$\text{Na}(s) + \frac{1}{2}\text{Cl}_2(g) \to \text{NaCl}(s)$	−411
$\text{NaCl}(s) \to \text{Na}^+(g) + \text{Cl}^-(g)$	?

A

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

[1]

B

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

[2]

C

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

[1]

Question 36

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Hydrogen chloride can be formed from hydrogen and chlorine.

$H_2(g) + Cl_2(g) \to 2HCl(g)$

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

A

The reaction involves breaking bonds and forming bonds.

I.

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

[2]

II.

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

[2]

B

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

[3]

Question 37

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Ethene reacts with hydrogen bromide to form bromoethane.

$CH_2{=}CH_2(g) + HBr(g) \to CH_3CH_2Br(g)$

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

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]

II.

State the new bonds formed in this reaction.

[2]

B

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

[3]

C

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

[1]

Question 38

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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.

$CO(g) + \frac{1}{2}O_2(g) \to CO_2(g)$

Given:

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

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

A

Use the given equations to obtain the target reaction.

I.

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

[1]

II.

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

[2]

B

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

[1]

C

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

[2]

Question 39

HL • Paper 1B

Hard

Calculator Permitted

HL • Paper 1B

Hard

Calculator Permitted

A Born-Haber cycle for magnesium oxide is shown. The lattice enthalpy is defined as the enthalpy change for $MgO(s) \to Mg^{2+}(g) + O^{2-}(g)$ .

Born-Haber step	$\Delta H$ / kJ mol $^{-1}$
$\text{Mg}(s) \to \text{Mg}(g)$ (sublimation)	+148
$\text{Mg}(g) \to \text{Mg}^+(g) + e^-$ (1st ionization energy)	+738
$\text{Mg}^+(g) \to \text{Mg}^{2+}(g) + e^-$ (2nd ionization energy)	+1451
$\frac{1}{2}\text{O}_2(g) \to \text{O}(g)$ (atomisation)	+249
$\text{O}(g) + e^- \to \text{O}^-(g)$ (1st electron affinity)	-141
$\text{O}^-(g) + e^- \to \text{O}^{2-}(g)$ (2nd electron affinity)	+844
$\text{Mg}(s) + \frac{1}{2}\text{O}_2(g) \to \text{MgO}(s)$ (standard enthalpy of formation)	-602

A

Determine the lattice enthalpy of magnesium oxide using the cycle.

[3]

B

Explain why the second electron affinity of oxygen is endothermic.

[1]

C

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

[1]

Question 40

SL • Paper 2

Hard

Calculator Permitted

SL • Paper 2

Hard

Calculator Permitted

The rate of nucleophilic substitution of halogenoalkanes is affected by the strength of the carbon-halogen bond. The average bond enthalpies are shown.

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

A

The carbon-halogen bond is broken during substitution.

I.

Identify the strongest carbon-halogen bond from the data.

[1]

II.

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

[2]

B

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

[3]

C

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

[1]

Question 41

SL • Paper 2

Hard

Calculator Permitted

SL • Paper 2

Hard

Calculator Permitted

A student investigates the enthalpy change of decomposition of hydrated copper(II) sulfate using two measured hydration reactions.

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

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

Target reaction: $CuSO_4\cdot 5H_2O(s) \to CuSO_4(s) + 5H_2O(l)$

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]

II.

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

[2]

B

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

[2]

C

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

[3]

Question 42

SL • Paper 2

Hard

Calculator Permitted

SL • Paper 2

Hard

Calculator Permitted

Methanol combusts according to the equation:

$CH_3OH(g) + \frac{3}{2}O_2(g) \to CO_2(g) + 2H_2O(g)$

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

A

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

I.

Calculate the total bond enthalpy for the bonds broken.

[2]

II.

Calculate the total bond enthalpy for the bonds formed.

[2]

B

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

[1]

C

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

[3]

Question 43

HL • Paper 2

Hard

Calculator Permitted

HL • Paper 2

Hard

Calculator Permitted

Propanone, $CH_3COCH_3(l)$ , is a volatile solvent. The equation for its complete combustion under standard conditions is:

$CH_3COCH_3(l) + 4O_2(g) \to 3CO_2(g) + 3H_2O(l)$

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

A

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

I.

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

[1]

II.

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

[3]

B

Deduce the standard formation equation for liquid propanone.

[2]

C

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

[2]

Question 44

HL • Paper 2

Hard

Calculator Permitted

HL • Paper 2

Hard

Calculator Permitted

Ethene can be hydrogenated to ethane.

$C_2H_4(g) + H_2(g) \to C_2H_6(g)$

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

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]

II.

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

[2]

B

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

[2]

C

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

[2]

Question 45

HL • Paper 2

Hard

Calculator Permitted

HL • Paper 2

Hard

Calculator Permitted

Carbon exists as the allotropes graphite and diamond. Graphite is the standard state of carbon at $298\ \text{K}$ and $100\ \text{kPa}$ .

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

A

Allotropes have different structures and bonding.

I.

Define standard enthalpy change of formation.

[2]

II.

State the standard enthalpy of formation of graphite.

[1]

B

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

[2]

C

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

[2]

Question 46

HL • Paper 2

Hard

Calculator Permitted

HL • Paper 2

Hard

Calculator Permitted

The enthalpy change for the reaction between methane and steam can be calculated using either formation data or combustion data.

$CH_4(g) + H_2O(g) \to CO(g) + 3H_2(g)$

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

A

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

I.

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

[1]

II.

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

[3]

B

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

[2]

C

Evaluate 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

Calculator Permitted

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:

$MgO(s) \to Mg^{2+}(g) + O^{2-}(g)$

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

$\Delta H_f^\ominus[MgO(s)] = -602$ ; atomization of $Mg(s) = +148$ ; atomization of oxygen, $\frac{1}{2}O_2(g) \to O(g) = +249$ ; $IE_1(Mg) = +738$ ; $IE_2(Mg) = +1451$ ; $EA_1(O) = -141$ ; $EA_2(O) = +744$ .

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]

II.

Explain why the second electron affinity of oxygen is endothermic.

[2]

B

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

[3]

C

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

[2]

Question 48

HL • Paper 2

Hard

Calculator Permitted

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:

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

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

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

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]

II.

Explain why the first ionization energy of potassium is endothermic.

[2]

B

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

[3]

C

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