DP Chemistry HL R1.2 Energy Cycles in Reactions

DP Chemistry HL R1.2 Energy Cycles in Reactions - Full notes

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DP Chemistry HL | R1.2 Energy Cycles in Reactions

DP CHEMISTRY · REACTIVITY

R1.2

Energy Cycles in Reactions

Full notes for HL · IB-style

Contents

1	Bond breaking and bond making
2	Average bond enthalpies
3	Calculating ΔH from bond enthalpies
4	Hess's law
5	Using Hess's law with given equations
6	Enthalpies of formation and combustion (HL only)
7	Born-Haber cycles and lattice enthalpy (HL only)
8	Bond enthalpy: definition and averages
9	Calculating ΔH from bond enthalpies
10	Why bond enthalpy values differ from experiment
11	Hess's law: the principle
12	Hess's law: manipulating equations
13	Standard enthalpy of formation
14	Using enthalpies of formation
15	Using enthalpies of combustion
16	Drawing energy cycles
17	Enthalpy of hydration of a salt (experimental Hess)
18	Enthalpy of reaction from experiment with Hess
19	Born–Haber cycles (HL)
20	Mixed review: worked problems
21	Additional exam-style practice
22	Estimating ΔH for reactions with liquids
23	Energy cycles in industry and nature
24	Graded practice: energy cycle calculations
25	Common pitfalls in Hess's law questions
26	Common misconceptions and links
27	Exam-style question with markscheme
28	Check your understanding
29	Summary and topic map

These are full teaching notes. Read each section in order, try the Think about it question before you read on, and do every Quick check before moving to the next section.

Before you start

Rate yourself on each statement now, and again after you've worked through the notes. Shade R (not yet), A (getting there) or G (confident).

I can…	R	A	G
explain that bond breaking is endothermic and bond making exothermic
define average bond enthalpy and explain its limitations
calculate ΔH from bond enthalpies
state Hess's law and explain why it follows from conservation of energy
use Hess's law to combine equations and calculate ΔH
calculate ΔH from standard enthalpies of formation and combustion
construct Born-Haber cycles and calculate lattice enthalpy

Sections marked HL only (Sections 6 and 7) are Additional Higher Level content. Every other section is common to SL and HL.

Guiding questions

How can we calculate enthalpy changes that cannot be measured directly? How is the energy of a reaction related to the bonds broken and formed?

How this topic is examined

R1.2 is examined through bond enthalpy calculations and Hess's law cycles. HL questions include formation and combustion data and Born-Haber cycles.

In bond enthalpy calculations, always use: ΔH = Σ(bonds broken) − Σ(bonds formed). Show the bonds counted.

Command term	What the examiner wants
State / Identify	a brief answer, no explanation
Outline / Describe	a brief or detailed account, without reasons
Explain	a detailed account with causes, reasons or mechanisms
Calculate / Determine	a numerical answer with working, correct significant figures and units
Deduce / Predict	reach a conclusion from given information
Draw / Sketch	a clear labelled diagram, structure or graph
Evaluate / Discuss	weigh strengths and limitations; reach a judgement

How to use these notes

Each section teaches one part of the syllabus in full. You'll see these boxes throughout; each has one job:

Box	What it's for
Think about it	a question to get you thinking before a section; the answer comes in the section
Key point	the one idea you must not forget
Common mistake	an error examiners see every year, and how to avoid it
In the real world	where the chemistry shows up outside the classroom
Try this	something quick you can do yourself to test the idea
Worked example	full solutions with IB-style marks (M1 method, A1 answer, R1 reasoning)
Quick check	short questions to test yourself at the end of each section

1 Bond breaking and bond making

Breaking needs energy; making releases it.

Think about it

Burning methane releases a lot of energy, yet the first step must be breaking strong C–H and O=O bonds. How can a reaction release energy overall?

Every chemical reaction involves breaking bonds in the reactants and forming new bonds in the products. Breaking bonds is endothermic: energy must be supplied to overcome the attraction between atoms. Forming bonds is exothermic: energy is released when atoms are attracted together.

The overall enthalpy change depends on the balance: if more energy is released in forming bonds than is absorbed in breaking bonds, the reaction is exothermic; if less, it is endothermic.

Common mistake

Saying 'bond breaking releases energy'. Breaking always requires energy; it is bond formation that releases energy.

2 Average bond enthalpies

The energy to break a bond, averaged over many compounds.

The bond enthalpy is the energy needed to break one mole of a covalent bond in gaseous molecules, forming gaseous atoms: for example, Cl2(g) → 2Cl(g), ΔH = +242 kJ mol−1. Values are always positive.

For bonds that occur in many different compounds (such as C–H), the strength varies slightly with the environment, so data booklets give average bond enthalpies. Because these are averages, and because they apply only to the gas phase, calculated ΔH values are only approximate and may differ from experimental values.

Bond	Average bond enthalpy / kJ mol−1	Bond	Average bond enthalpy / kJ mol−1
C–H	414	O=O	498
C–C	346	C=O	804
C=C	614	O–H	463
H–H	436	N≡N	945
Cl–Cl	242	N–H	391
H–Cl	431	C–Cl	324

Quick check

1. Define bond enthalpy.

2. Why are calculated ΔH values from bond enthalpies approximate?

Answers: 1. energy to break one mole of a bond in gaseous molecules 2. values are averages; apply only to gases

3 Calculating ΔH from bond enthalpies

Sum of bonds broken minus sum of bonds formed.

ΔH = Σ (bond enthalpies of bonds broken) − Σ (bond enthalpies of bonds formed)

WORKED EXAMPLE 1 Combustion of methane

Use bond enthalpies to estimate ΔH for CH4(g) + 2O2(g) → CO2(g) + 2H2O(g).

Solution

Bonds broken: 4 C–H + 2 O=O = 4(414) + 2(498) = 2652 kJM1

Bonds formed: 2 C=O + 4 O–H = 2(804) + 4(463) = 3460 kJM1

ΔH = 2652 − 3460 = −808 kJ mol−1A1

Experimental value (−890 kJ mol−1, water as liquid) differs: average values; H2O(l) not (g)R1

WORKED EXAMPLE 2 Hydrogenation of ethene

Estimate ΔH for C2H4(g) + H2(g) → C2H6(g).

Solution

Broken: C=C + H–H = 614 + 436 = 1050 kJ (the four C–H bonds are unchanged)M1

Formed: C–C + 2 C–H = 346 + 828 = 1174 kJM1

ΔH = 1050 − 1174 = −124 kJ mol−1A1

Quick check

1. What is ΔH for H2 + Cl2 → 2HCl?

2. Why are only some bonds counted in the second example?

Answers: 1. (436 + 242) − 2(431) = −184 kJ mol−1 2. unchanged bonds cancel out

4 Hess's law

The route doesn't matter.

Hess's law states that the enthalpy change for a reaction is independent of the route taken, provided the initial and final conditions are the same. It follows from the law of conservation of energy: if two routes from the same reactants to the same products gave different enthalpy changes, energy could be created by going one way and returning the other.

Hess's law allows enthalpy changes that are difficult or impossible to measure directly (because the reaction is too slow, produces side products, or cannot be carried out) to be calculated from other known enthalpy changes using an energy cycle.

Figure 1. Hess's law: the total enthalpy change is the same by either route.

5 Using Hess's law with given equations

Rearrange and add equations like algebra.

To find an unknown ΔH, manipulate the given equations so they add up to the target equation: reverse an equation (change the sign of ΔH); multiply an equation (multiply ΔH by the same factor); then add the equations, cancelling species that appear on both sides.

WORKED EXAMPLE 3 Combining equations

Calculate ΔH for C(s) + ½O2(g) → CO(g), given: (1) C(s) + O2(g) → CO2(g), ΔH = −394 kJ mol−1; (2) CO(g) + ½O2(g) → CO2(g), ΔH = −283 kJ mol−1.

Solution

Keep (1): C + O2 → CO2, −394M1

Reverse (2): CO2 → CO + ½O2, +283M1

Add: C + ½O2 → CO; ΔH = −394 + 283 = −111 kJ mol−1A1

This cannot be measured directly because some CO2 always formsR1

WORKED EXAMPLE 4 Hydration of copper(II) sulfate

ΔH for dissolving anhydrous CuSO4 is −66 kJ mol−1 and for dissolving CuSO4·5H2O is +11 kJ mol−1. Calculate ΔH for CuSO4(s) + 5H2O(l) → CuSO4·5H2O(s).

Solution

Route: anhydrous → solution (−66); hydrated → solution (+11), so solution → hydrated is −11M1

ΔH = −66 − 11 = −77 kJ mol−1A1

Quick check

1. State Hess's law.

2. What happens to ΔH when an equation is reversed?

Answers: 1. ΔH is independent of the route taken 2. its sign changes

6 Enthalpies of formation and combustion (HL only)

Two standard routes for Hess's law calculations.

HL only

This section is Additional Higher Level content. SL students do not need it.

The standard enthalpy of formation (ΔHf⊖) is the enthalpy change when one mole of a compound is formed from its elements in their standard states under standard conditions. For example: C(s) + 2H2(g) → CH4(g). By definition, ΔHf⊖ of an element in its standard state is zero.

Using formation data: ΔHr = ΣΔHf(products) − ΣΔHf(reactants). Using combustion data: ΔHr = ΣΔHc(reactants) − ΣΔHc(products) (note the reversed order, because combustion goes from the substances to common products CO2 and H2O).

WORKED EXAMPLE 5 Using enthalpies of formation

Calculate ΔH for C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(l). ΔHf: C3H8 −104, CO2 −394, H2O(l) −286 kJ mol−1.

Solution

Σ products = 3(−394) + 4(−286) = −2326M1

Σ reactants = −104 + 0 = −104M1

ΔH = −2326 − (−104) = −2222 kJ mol−1A1

WORKED EXAMPLE 6 Using enthalpies of combustion

Calculate ΔHf of ethanol from ΔHc: C(s) −394, H2(g) −286, C2H5OH(l) −1367 kJ mol−1. Target: 2C + 3H2 + ½O2 → C2H5OH.

Solution

ΣΔHc(reactants) = 2(−394) + 3(−286) = −1646M1

ΔH = −1646 − (−1367) = −279 kJ mol−1A1

7 Born-Haber cycles and lattice enthalpy (HL only)

An energy cycle for ionic compounds.

HL only

This section is Additional Higher Level content. SL students do not need it.

The lattice enthalpy of an ionic compound (S2.1) cannot be measured directly, but can be calculated using a Born-Haber cycle, an application of Hess's law. The cycle links the formation of the ionic solid from its elements (ΔHf) to a series of steps that form gaseous ions.

For sodium chloride, the steps are: atomization of sodium, Na(s) → Na(g) (+107); first ionization energy of sodium, Na(g) → Na+(g) + e− (+496); atomization of chlorine, ½Cl2(g) → Cl(g) (+121); first electron affinity of chlorine, Cl(g) + e− → Cl−(g) (−349); and the reverse of lattice enthalpy, Na+(g) + Cl−(g) → NaCl(s). By Hess's law: ΔHf = sum of all the steps.

WORKED EXAMPLE 7 Lattice enthalpy of NaCl

ΔHf(NaCl) = −411 kJ mol−1. Using the values above, calculate the lattice enthalpy.

Solution

ΔHf = ΔHat(Na) + IE(Na) + ΔHat(Cl) + EA(Cl) − ΔHlatM1

−411 = 107 + 496 + 121 − 349 − ΔHlatM1

ΔHlat = 107 + 496 + 121 − 349 + 411 = +786 kJ mol−1A1

Figure 2. The steps of the Born-Haber cycle for NaCl.

Quick check

1. Define standard enthalpy of formation.

2. What is ΔHf of O2(g)?

3. Why is a Born-Haber cycle needed?

Answers: 1. ΔH when one mole of compound forms from elements in standard states 2. zero 3. lattice enthalpy cannot be measured directly

8 Bond enthalpy: definition and averages

Breaking one mole of bonds.

Bond enthalpy is the energy needed to break one mole of a particular covalent bond in gaseous molecules. Values in data booklets are averages over many compounds (except for diatomic molecules), so calculated ΔH values are approximate. Bond breaking is endothermic (+); bond making is exothermic (−).

Bond	kJ mol−1	Bond	kJ mol−1
H–H	436	C=O	804
C–H	414	O–H	463
O=O	498	C–C	346
C=C	614	N≡N	945
N–H	391	H–Cl	432
Cl–Cl	242	C–Cl	324

WORKED EXAMPLE 8 Definition (basic)

Write an equation representing the bond enthalpy of Cl–Cl.

Solution

Cl2(g) → 2Cl(g)A1

ΔH = +242 kJ mol−1A1

Common mistake

Using bond enthalpies for reactions involving liquids or solids without adjustment. Bond enthalpies apply to gaseous species only.

Quick check

1. Why is the C–H bond enthalpy an average value?

2. Is bond breaking exothermic or endothermic?

Answers: 1. the C–H bond strength varies slightly between molecules 2. endothermic

9 Calculating ΔH from bond enthalpies

Bonds broken minus bonds formed.

WORKED EXAMPLE 9 Hydrogen chloride (standard)

Calculate ΔH for H2(g) + Cl2(g) → 2HCl(g).

Solution

Broken: 436 + 242 = 678A1

Formed: 2 × 432 = 864A1

ΔH = 678 − 864 = −186 kJ mol−1A1

WORKED EXAMPLE 10 Methane combustion (exam level)

Calculate ΔH for CH4(g) + 2O2(g) → CO2(g) + 2H2O(g).

Solution

Broken: 4(414) + 2(498) = 2652A1

Formed: 2(804) + 4(463) = 3460A1

ΔH = −808 kJ mol−1A1

Quick check

1. Calculate ΔH for N2 + 3H2 → 2NH3.

2. Calculate ΔH for C2H4 + H2 → C2H6.

Answers: 1. 945 + 1308 − 2346 = −93 kJ mol−1 2. 614 + 436 − (346 + 2 × 414) = −124 kJ mol−1

10 Why bond enthalpy values differ from experiment

Averages and states.

Calculated values differ from experimental ones because bond enthalpies are averages, and because some species in the experiment may be liquids (energy is needed to vaporize them, which bond enthalpy calculations ignore).

WORKED EXAMPLE 11 Comparing (standard)

The calculated ΔHc of methane is −808 kJ mol−1, but the data value is −891 kJ mol−1. Explain.

Solution

Bond enthalpies are averagesR1

Data value forms H2O(l); condensation releases extra energyR1

11 Hess's law: the principle

Route independence.

Hess's law: the total enthalpy change for a reaction is independent of the route taken, provided the initial and final conditions are the same. It follows from the conservation of energy. It lets us calculate enthalpy changes that cannot be measured directly.

WORKED EXAMPLE 12 Simple cycle (standard)

C(s) + O2 → CO2, ΔH = −394; CO + ½O2 → CO2, ΔH = −283. Find ΔH for C(s) + ½O2 → CO.

Solution

Route: C → CO2 directly equals C → CO → CO2M1

ΔH = −394 − (−283) = −111 kJ mol−1A1

Quick check

1. Why can ΔH for C(s) → CO(g) not be measured directly?

Answers: 1. some CO2 always forms; CO cannot be made pure from carbon

12 Hess's law: manipulating equations

Reverse, multiply, add.

Arrange the given equations so that, when added, they give the target equation. Reversing an equation changes the sign of ΔH; multiplying it multiplies ΔH. Species appearing on both sides cancel.

WORKED EXAMPLE 13 Combining equations (exam level)

Find ΔH for 2C(s) + 3H2(g) → C2H6(g) using ΔHc: C −394, H2 −286, C2H6 −1560 kJ mol−1.

Solution

2(−394) + 3(−286) − (−1560)M1 A1

= −86 kJ mol−1A1

Quick check

1. Given S + O2 → SO2 (−297) and 2SO2 + O2 → 2SO3 (−198), find ΔH for S + 1.5O2 → SO3.

Answers: 1. −297 − 99 = −396 kJ mol−1

13 Standard enthalpy of formation

Forming one mole from elements.

ΔHf⦵ is the enthalpy change when one mole of a compound is formed from its elements in their standard states under standard conditions. By definition, ΔHf of an element in its standard state is zero.

WORKED EXAMPLE 14 Formation equations (standard)

Write the equation for ΔHf of ethanol.

Solution

2C(s) + 3H2(g) + ½O2(g) → C2H5OH(l)A2

Quick check

1. Write the equation for ΔHf of calcium carbonate.

2. What is ΔHf of O2(g)?

Answers: 1. Ca(s) + C(s) + 1.5O2(g) → CaCO3(s) 2. 0

14 Using enthalpies of formation

Products minus reactants.

WORKED EXAMPLE 15 Formation data (exam level)

Calculate ΔH for Fe2O3(s) + 3CO(g) → 2Fe(s) + 3CO2(g). ΔHf: Fe2O3 −824, CO −111, CO2 −394 kJ mol−1.

Solution

Products: 3(−394) = −1182A1

Reactants: −824 + 3(−111) = −1157A1

ΔH = −25 kJ mol−1A1

Quick check

1. Calculate ΔHc of methane: ΔHf CH4 −75, CO2 −394, H2O(l) −286.

Answers: 1. −394 + 2(−286) − (−75) = −891 kJ mol−1

15 Using enthalpies of combustion

Reactants minus products.

WORKED EXAMPLE 16 Combustion data (exam level)

Find ΔHf of propane from ΔHc: C −394, H2 −286, C3H8 −2219 kJ mol−1.

Solution

3(−394) + 4(−286) − (−2219)M1 A1

= −107 kJ mol−1A1

Common mistake

Mixing up the two formulas. With formation data it is products minus reactants; with combustion data it is reactants minus products.

Quick check

1. Find ΔH for C2H4 + H2 → C2H6 using ΔHc: C2H4 −1411, H2 −286, C2H6 −1560.

Answers: 1. −1411 − 286 + 1560 = −137 kJ mol−1

16 Drawing energy cycles

Visual Hess's law.

Draw the target reaction across the top, and a common set of species (elements for formation data, combustion products for combustion data) below. Arrows show the two routes; follow them in the right directions, changing signs when going against an arrow.

WORKED EXAMPLE 17 Cycle (standard)

Describe the cycle used to find ΔHf of ethanol from combustion data.

Solution

Top: elements → ethanol; bottom: CO2 and H2OR1

Elements → combustion products (ΔHc of C and H2); ethanol → combustion products (ΔHc of ethanol)R1

17 Enthalpy of hydration of a salt (experimental Hess)

Indirect measurement.

The enthalpy change of converting anhydrous CuSO4 into CuSO4·5H2O cannot be measured directly, but both dissolve in water. Measure ΔHsol of each and combine: ΔH = ΔHsol(anhydrous) − ΔHsol(hydrated).

WORKED EXAMPLE 18 Hydration (exam level)

ΔHsol(CuSO4) = −66.5 kJ mol−1; ΔHsol(CuSO4·5H2O) = +11.7 kJ mol−1. Find ΔH for CuSO4(s) + 5H2O(l) → CuSO4·5H2O(s).

Solution

−66.5 − 11.7M1

= −78.2 kJ mol−1A1

18 Enthalpy of reaction from experiment with Hess

Decomposition of CaCO3.

The decomposition CaCO3 → CaO + CO2 requires a high temperature and is hard to measure. Instead, measure ΔH of CaCO3 + 2HCl and CaO + 2HCl, then subtract.

WORKED EXAMPLE 19 Decomposition (exam level)

ΔH1 (CaCO3 + 2HCl) = −17 kJ mol−1; ΔH2 (CaO + 2HCl) = −195 kJ mol−1. Find ΔH for the decomposition.

Solution

ΔH = ΔH1 − ΔH2M1

= +178 kJ mol−1A1

19 Born–Haber cycles (HL)

Lattice enthalpy from a cycle.

A Born–Haber cycle relates the lattice enthalpy of an ionic compound to atomization, ionization, electron affinity and formation enthalpies. For NaCl: ΔHf = ΔHat(Na) + IE1(Na) + ΔHat(Cl) + EA(Cl) − ΔHlattice.

WORKED EXAMPLE 20 Lattice enthalpy (exam level)

Find the lattice enthalpy of NaCl: ΔHf −411, ΔHat(Na) +107, IE(Na) +496, ΔHat(Cl) +121, EA(Cl) −349 kJ mol−1.

Solution

Lattice = 107 + 496 + 121 − 349 + 411M1 A1

= +786 kJ mol−1A1

20 Mixed review: worked problems

Bringing the ideas together.

WORKED EXAMPLE 21 Mixed problem (exam level)

Hydrazine burns: N2H4(g) + O2(g) → N2(g) + 2H2O(g). Using bond enthalpies (N–N 158, N–H 391, O=O 498, N≡N 945, O–H 463), calculate ΔH.

Solution

Broken: 158 + 4(391) + 498 = 2220A1

Formed: 945 + 4(463) = 2797A1

ΔH = −577 kJ mol−1A1

Quick check

1. Use ΔHf (NH3 −46, NO −(+90), H2O(g) −242) to find ΔH for 4NH3 + 5O2 → 4NO + 6H2O.

Answers: 1. 4(90) + 6(−242) − 4(−46) = −908 kJ

21 Additional exam-style practice

Paper 1 and Paper 2.

WORKED EXAMPLE 22 Paper 1 question (standard)

Which has ΔHf⦵ = 0? A. H2O(l) B. O3(g) C. Br2(l) D. Br2(g)

Solution

C: bromine's standard state is liquidA1

WORKED EXAMPLE 23 Paper 2 question (exam level)

(a) Define standard enthalpy of combustion. (b) Calculate ΔHf of ethanol from ΔHc: C −394, H2 −286, C2H5OH −1367.

Solution

(a) enthalpy change when one mole of a substance is completely burned in oxygen under standard conditionsA1

(b) 2(−394) + 3(−286) + 1367 = −279 kJ mol−1M1 A1

22 Estimating ΔH for reactions with liquids

Including vaporization.

To use bond enthalpies for liquids, add the enthalpy of vaporization to convert them to gases first. For example, for a reaction forming H2O(l), subtract 44 kJ mol−1 for each mole of water condensed.

WORKED EXAMPLE 24 Correction (standard)

Correct the bond-enthalpy value for methane combustion (−808 kJ to H2O(g)) to H2O(l).

Solution

2 × (−44) = −88M1

−896 kJ mol−1, close to the data value −891A1

23 Energy cycles in industry and nature

Why they matter.

Hess's law calculations predict the heat released by reactions in industrial plants, design of fuels and explosives, and energy balances in biochemistry (for example, the energy from oxidizing glucose).

WORKED EXAMPLE 25 Glucose (standard)

Find ΔHc of glucose from ΔHf: glucose −1273, CO2 −394, H2O(l) −286.

Solution

6(−394) + 6(−286) + 1273M1

= −2807 kJ mol−1A1

In the real world

The energy content printed on food labels is found by Hess's law-type calculations and bomb calorimetry: about 17 kJ g−1 for carbohydrates and proteins and 37 kJ g−1 for fats.

24 Graded practice: energy cycle calculations

Build fluency.

Work through these in order. Each uses one of the three methods: bond enthalpies, formation data or combustion data. Decide the method first, then set out the calculation clearly.

WORKED EXAMPLE 26 Bond enthalpies (standard)

Calculate ΔH for CH4(g) + Cl2(g) → CH3Cl(g) + HCl(g).

Solution

Broken: C–H 414 + Cl–Cl 242 = 656A1

Formed: C–Cl 324 + H–Cl 432 = 756A1

ΔH = −100 kJ mol−1A1

WORKED EXAMPLE 27 Formation data (standard)

Calculate ΔH for 2NO2(g) → N2O4(g). ΔHf: NO2 +33, N2O4 +9 kJ mol−1.

Solution

9 − 2(33)M1

= −57 kJ mol−1A1

Quick check

1. Calculate ΔH for C2H5OH(l) + 3O2(g) → 2CO2(g) + 3H2O(l) using ΔHf: ethanol −278, CO2 −394, H2O(l) −286.

2. Calculate ΔH for H2(g) + ½O2(g) → H2O(g) using bond enthalpies.

Answers: 1. 2(−394) + 3(−286) + 278 = −1368 kJ mol−1 2. 436 + 249 − 926 = −241 kJ mol−1

25 Common pitfalls in Hess's law questions

A checklist.

Check: correct state symbols; balanced equations; multiply ΔH values by coefficients; elements have ΔHf = 0; signs reversed when reversing equations; formation (products − reactants) versus combustion (reactants − products); and final units (kJ mol−1).

WORKED EXAMPLE 28 Error spotting (standard)

A student calculates ΔH for 2H2 + O2 → 2H2O(l) as −286 kJ. Identify the error.

Solution

Did not multiply by 2: ΔH = −572 kJR1 A1

26 Common misconceptions and links

Students often think…	Actually…
bond breaking releases energy	bond breaking is endothermic
bond enthalpy calculations are exact	they use averages for gases
ΔH changes depending on the route	Hess: independent of route
ΔHf of elements is never zero	zero for elements in standard states

How R1.2 connects to the rest of the course

Topic	Connection
R1.1 Measuring enthalpy	calorimetry data
S2.2 Covalent model	bond order and strength
S2.1 Ionic model	lattice enthalpy
R1.4 Entropy (HL)	Gibbs energy

27 Exam-style question with markscheme

Try this under exam conditions (about 15 minutes) before reading the markscheme.

Text A. Standard enthalpies of combustion: C(s) −394, H2(g) −286, C2H6(g) −1560 kJ mol−1.

(a) Use bond enthalpies to estimate ΔH for 2H2(g) + O2(g) → 2H2O(g). [3]

(b) State why this is an estimate. [1]

(d) Calculate the lattice enthalpy of KCl: ΔHf −437, ΔHat(K) +89, IE(K) +419, ΔHat(Cl) +121, EA(Cl) −349 kJ mol−1. [2]

Markscheme

Part	Answer	Marks
(a)	broken: 2(436) + 498 = 1370	M1
	formed: 4(463) = 1852	M1
	ΔH = −482 kJ mol−1	A1
(b)	average bond enthalpies used	A1
(c)	2(−394) + 3(−286) = −1646	M1
	minus (−1560)	M1
	= −86 kJ mol−1	A1
(d)	89 + 419 + 121 − 349 + 437	M1
	= +717 kJ mol−1	A1

Examiner's comments. In (c), many answers subtracted in the wrong order. Draw the cycle to check the direction of each arrow.

28 Check your understanding

Section A: multiple choice

1. Bond breaking is

A. endothermic B. exothermic C. neither D. always fast

2. ΔH = Σ bonds broken − Σ bonds formed gives

A. exact values B. ΔH for solids C. ΔH for gas-phase reactions approximately D. activation energy

3. Hess's law follows from

A. collision theory B. conservation of energy C. Le Chatelier's principle D. the ideal gas law

4. Reversing an equation

A. doubles ΔH B. has no effect C. makes ΔH zero D. changes the sign of ΔH

5. The bond enthalpy of Cl–Cl refers to

A. Cl2(l) → 2Cl(g) B. Cl2(g) → 2Cl(g) C. 2Cl → Cl2 D. Cl + e− → Cl−

6. If bonds formed release more energy than bonds broken absorb, the reaction is

A. exothermic B. endothermic C. slow D. impossible

7. ΔHf of an element in its standard state is

A. positive B. negative C. equal to its bond enthalpy D. zero

8. In a Born-Haber cycle the electron affinity of Cl is

A. endothermic B. zero C. exothermic D. the lattice enthalpy

Section B: structured

9. Explain why some reactions are exothermic and others endothermic in terms of bonds. [3]

10. State Hess's law and explain how it is used. [3]

11. Estimate ΔH for N2 + 3H2 → 2NH3 using bond enthalpies. [3]

12. Explain why MgO has a larger lattice enthalpy than NaCl, using a Born-Haber cycle idea. [3]

Answers

1. A 2. C 3. B 4. D 5. B 6. A 7. D 8. C

9. Breaking endothermic, forming exothermic; balance determines sign.

10. ΔH independent of route; construct cycle with known values; calculate unknown.

11. Broken 945 + 3(436) = 2253; formed 6(391) = 2346; ΔH = −93 kJ mol−1.

12. Higher charges, smaller ions; stronger attraction; larger lattice enthalpy calculated from cycle.

29 Summary and topic map

Idea	Remember
Bonds	breaking endothermic; forming exothermic
Bond enthalpy	gaseous; average values; approximate
ΔH from bonds	Σ broken − Σ formed
Hess's law	route independent; conservation of energy
Combining equations	reverse: change sign; multiply: scale
HL	ΔHf and ΔHc cycles; Born-Haber; lattice enthalpy

Before the exam, make sure you…

•list bonds broken and formed
•check signs when reversing
•draw energy cycles
•explain why bond enthalpy values are approximate

Energy Cycles on one page

Cover this page with a sheet of paper and try to rebuild it from memory. Whatever you can't recall is what to revise next.

Figure 3. Topic map for Energy Cycles.

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