Chemsheets A2 033 Thermodynamicsmchem.weebly.com/uploads/9/2/0/5/9205056/thermodynamics.pdf · © 17-Jul-12 Chemsheets A2 033 2 AS THERMODYNAMICS REVISION What is enthalpy? It is

© www.CHEMSHEETS.co.uk 17-Jul-12 Chemsheets A2 033 1


AS THERMODYNAMICS REVISION

What is enthalpy? It is a measure of the heat content of a substance

Enthalpy change (∆H) = Change in heat content at constant pressure

Standard conditions (∆H) = 100 kPa and a stated temperature

Exothermic reactions

Endothermic reactions

Standard enthalpy change of formation (∆∆∆∆Hf) (“enthalpy of formation”)

Enthalpy change when 1 mole of a substance is formed from its constituent elements with all reactants and products in standard states under standard conditions.

e.g. CH4(g) ��. H2O(l) ��. NH3(g) ��. C2H5OH(l) ��. CH3Br(l) ��. Na2O(s) ��.

Note: re ∆Hf of an element in its standard state = 0 by definition

Standard enthalpy change of combustion (∆∆∆∆Hc) (“enthalpy of combustion”)

Enthalpy change when 1 mole of a substance is completely burned in oxygen with all reactants and products in standard states under standard conditions.

e.g. CH4(g) ��. H2(g) ��. C2H6(g) ��. C2H5OH(l) ��.

Na(s) ��.

C6H14(l) ��.


Hess’s Law Calculations

The enthalpy change for a reaction is independent of the route taken

e.g. the enthalpy change to go from A → B direct is

the same as going from A → C → B

1) Calculations involving enthalpies of formation (“Type 1 questions”)

• If the enthalpy of formation for the reactants and products in a reaction are known, the overall enthalpy change is easy to

calculate.

∆∆∆∆H = [SUM of ∆∆∆∆Hf products] – [SUM ∆∆∆∆Hf reactants]

• Remember that ∆Hf of all elements is zero.

• Watch for the very frequent mistake of doing reactants – products, rather than products – reactants.

• If the overall enthalpy change for a reaction is known along with the enthalpy of formation of all but one of the reactants/products, then this equation can be used to find the missing enthalpy of formation.

Example

Calculate the overall enthalpy change for this reaction: CH4(g) + 2 O2(g) → CO2(g) + 2 H2O(l)

∆Hf CH4(g) = -75, CO2(g) = -393, H2O(l) = -286 kJ/mol

2) Calculations involving enthalpies of combustion (“Type 2 questions”)

• Questions that involve enthalpies of combustion can usually be done using the cycle shown below.

• The reaction involved across the top is often an enthalpy of formation (from elements to a compound).

• The sum of the clockwise arrows equals the sum of the anticlockwise arrows.

• Be careful when drawing your cycle to ensure that arrows are going in the right direction and the number of moles is correct.

• If you use a cycle like this, there is no need to worry about getting the number of oxygen molecules in the downward arrows.

reactants products

oxides

∆H

∆Hc ∆Hc

A B

C


Example

Calculate the enthalpy of formation of ethanol (C2H5OH) given the following enthalpies of combustion.

∆Hc C(s) = -393, H2(g) = -286, C2H5OH(l) = -1371 kJ/mol

3) Calculations involving bond enthalpies (“Type 3 questions”)

• Bond enthalpy is the enthalpy change to break one mole of covalent bonds in the gas phase.

• For most bonds (e.g. C-H, C-C, C=O, O-H, etc.) the value for the bond enthalpy is an average taken from a range of molecules as the exact value varies from compound to compound. For some bond enthalpies (e.g. H-H, H-Cl, O=O, etc) the value is exact as only one molecule contains that bond.

• Enthalpies of reaction that have been calculated using mean bond enthalpies are not as accurate as they might be because the values used are averages and not the specific ones for that compound.

• This cycle works for any question that involves bond enthalpies, whether to find a bond enthalpy or ∆H for a reaction.

• Remember that substances must be in the gas state before bonds

are broken, and so ∆H to go to the gas state is needed for solids and

liquids. (Note - ∆H vaporisation is the enthalpy change to convert a liquid to a gas)

• As with other cycles, the sum of the clockwise arrows equals the sum of the anticlockwise arrows. Be careful to ensure that arrow directions and number of moles are correct.

reactants products

gas atoms

∆H

(∆H→gas) + bond enthalpies

(∆H→gas) +

bond enthalpies

Example

Ethanol has the formula C2H5OH and is used as a fuel (e.g. for cars in Brazil). It burns in the following reaction for which the enthalpy change is -1015 kJ/mol.

C2H5OH(l) + 3 O2(g) → 2 CO2(g) + 3 H2O(g)

Calculate the C-C bond enthalpy in ethanol given the following bond enthalpies and enthalpy of vaporisation of ethanol.

Bond enthalpies: C-H 412, O-H 463, C-O 360, C=O 743, O=O 498 kJ/mol

Enthalpy of vaporisation of ethanol, C2H5OH(l) = 44 kJ/mol

-1015 + 4(743) + 6(463) = 44 + (C-C) + 5 (412) + 463 + 360 + 3 (498) (C-C) = -1015 + 4(743) + 6(463) - 44 - 5(412) - 463 - 360 - 3(498) = +314 kJ mol

-1

C2H5OH(l) + 3 O2(g) 2 CO2(g) + 3 H2O(g)

2 C(g) + 6 H(g) + 7 O(g)

-1015

44C-C

5(412)463360

3(498)

4(743) 6(463)


TASK 1 – A MIXTURE OF SIMPLE AS THERMODYNAMICS QUESTIONS

1 Find ∆Hf of butane given that the following data.

∆Hc: C4H10(g) = –2877, C(s) = –394, H2(g) = –286 kJ mol-1

2 Find ∆H for the following reaction using the data below. C3H8(g) + 5 O2(g) → 3 CO2(g) + 4 H2O(l)

∆Hf: C3H8(g) = –104, CO2(g) = –394, H2O(l) = –286 kJ mol-1

3 Find ∆H for the following reaction using the bond enthalpy data below. C2H6(g) + 3½ O2(g) → 2 CO2(g) + 3 H2O(g)

C-C = 348, C-H = 412, O=O = 496, C=O = 743, O-H = 463 kJ mol-1

4 Find ∆Hc of propan-2-ol given that the following data.

∆Hf: CH3CH(OH)CH3(l) = –318, ∆Hc: C(s) = –394, H2(g) = –286 kJ mol-1

5 Calculate ∆Hf of CCl4(l) given the following data.

CCl4(l) → CCl4(g) = +31 kJ mol-1

C(s) → C(g) = +715 kJ mol-1

Bond enthalpy (Cl-Cl) = +242 kJ mol-1

Bond enthalpy (C-Cl) = +338 kJ mol-1

6 Find ∆H for the hydrogenation of propene using the data below. CH3CH=CH2(g) + H2(g) → CH3CH2CH3(g)

∆Hc: CH3CH=CH2(g) = –2059, H2(g) = –286, CH3CH2CH3(g) = –2220 kJ mol-1

7 0.55 g of propanone was burned in a calorimeter containing 80 g of water. The temperature rose by 47.3ºC.

Calculate ∆Hc for propanone given the specific heat capacity of water is 4.18 J mol-1

K-1

.

8 25 cm3 of 2.0 mol dm

-3 nitric acid was reacted with 25 cm

3 of 2.0 mol dm

-3 potassium hydroxide is an insulated cup.

The temperature rose from 20.2ºC to 33.9ºC. Calculate ∆H for the reaction given the specific heat capacity of water is 4.18 J mol

-1 K

-1.

9 The engines of the lunar module of Apollo 11 used methylhydrazine (CH3NHNH2) and dinitrogen tetraoxide. They react as follows:

4 CH3NHNH2(l) + 5 N2O4(l) → 4 CO2(g) + 12 H2O(l) + 9 N2(g)

Calculate the enthalpy change for the reaction using the following data:

∆Hf: CH3NHNH2(l) = +53, N2O4 (l) = -20, CO2(g) = -393, H2O(l) = -286 kJmol-1

10) a) State Hess’s law.

b) Calculate ∆Hf of the following using the ∆Hc values provided. i) ethyne (C2H2(g)) ii) ethanol (C2H5OH(l))

∆Hc (kJmol-1

): C(s) = -393; H2(g) = -286; C2H2(g) = -1300; C2H5OH(l) = -1367

11 What is the sign of the following enthalpy changes (+ or -)?

a) combustion of magnesium b) freezing water c) melting ice

12 Calculate the average C-C bond enthalpy in benzene (C6H6) given the following data.

∆Hf (C6H6(l)) = +49 kJmol-1

C(s) → C(g) ∆H = +715 kJmol-1

H2(g) → 2 H(g) ∆H = +436 kJmol-1

C6H6(l) → C6H6(g) ∆H = +31 kJmol-1

E (C-H) = +413 kJmol-1


TASK 2 – EXTRA BOND ENTHALPY CALCULATIONS

Use this data in the questions that follow.

Bond kJmol-1

Bond kJmol-1

Bond kJmol-1

Bond kJmol-1

H-H 436 C=C 612 O=O 496 N≡N 944

Cl-Cl 242 C-C 348 Br-Br 193 C-O 360

H-Cl 431 C-H 412 C-Br 276 S=O 743

∆Ha (kJ mol-1

): C(s) +715; S(s) +223

∆Hf (kJ mol-1

): NH3(g) –46; H2O(l) –286; C2H5OH(l) –278; SO2(g) –297

∆Hvap (kJ mol-1

): H2O(l) +42; C2H5OH(l) +44

1) Calculate ∆H for the following reaction using the bond enthalpies.

a) H2(g) + Cl2(g) → 2 HCl(g)

b) C2H4(g) + Br2(g) → C2H4Br2(g)

c) ∆Hf of C3H8(g)

2) Calculate the average N-H bond enthalpy in ammonia.

3) Calculate the mean O-H bond enthalpy in water.

4) Calculate the O-H bond strength in C2H5OH(l).

5) Calculate the mean S=O bond strength in SO2(g).


DEFINITIONS

Enthalpy change Definition Exo/endothermic Example

Enthalpy of

formation (∆Hf):

Enthalpy change when one mole of a substance is formed from its constituent elements with all substances in their standard states

Exothermic (-ve) for most substances

e.g. Na2O(s) 2 Na(s) + ½ O2(g) → Na2O(s)

Enthalpy of

combustion (∆Hc)

Enthalpy change when one mole of a substance undergoes complete combustion in oxygen with all substances in standard states

Exothermic (-ve) e.g. hydrogen H2(g) + ½ O2(g) → H2O(g)

Ionisation enthalpy

(∆Hi)

The first ionisation energy is the enthalpy change when one mole of gaseous atoms loses one electron per atom to produce gaseous 1+ ions.

Endothermic (+ve) e.g. magnesium Mg(g) → Mg+(g) + e

-

The second ionisation energy is when one mole of gaseous 2+ ions is produced from one mole of 1+ ions.

Endothermic (+ve) e.g. magnesium Mg+(g) → Mg

2+(g) + e

-

Electron affinity

(∆Hea)

The first electron affinity is the enthalpy change when one mole of gaseous atoms gains one electron per atom to produce gaseous 1- ions.

Exothermic (-ve) for many non-metals

e.g. oxygen O(g) + e- → O

-(g)

The second electron affinity is the enthalpy change when one mole of gaseous 1- ions gains one electron per ion to produce gaseous 2- ions.

Endothermic (+ve) as adding –ve electron to –ve ion

e.g. oxygen O-(g) + e

- → O

2-(g)

Enthalpy of

atomisation (∆Ha)

Enthalpy change when one mole of gaseous atoms is produced from an element in its normal state.

Endothermic (+ve) e.g. iodine ½ I2(s) → I(g)

Hydration enthalpy

(∆Hhyd)

Enthalpy change when one mole of gaseous ions become hydrated (dissolved in water).

Exothermic (-ve) e.g. magnesium ions Mg2+

(g) + aq → Mg2+

(aq)

Enthalpy of solution

(∆Hsol)

Enthalpy change when one mole of an ionic solid dissolves in an amount of water large enough so that the dissolved ions are well separated and do not interact with each other.

Varies e.g. magnesium chloride MgCl2(s) + aq → Mg2+

(aq) + 2 Cl-(aq)

Bond dissociation

enthalpy (∆Hdis)

Enthalpy change when one mole of covalent bonds is broken in the gaseous state.

Endothermic (+ve) e.g. I-I bond I2(g) → 2 I(g)

Lattice enthalpy of

formation (∆HL)

Enthalpy change when one mole of a solid ionic compound is formed from into its constituent ions in the gas phase

Exothermic (-ve) e.g. magnesium chloride Mg2+

(g) + 2 Cl-(g) → MgCl2(s)

Lattice enthalpy of

disociation (∆HL)

Enthalpy change when one mole of a solid ionic compound is broken up into its constituent ions in the gas phase

Endothermic (+ve) e.g. magnesium chloride MgCl2(s) → Mg2+

(g) + 2 Cl-(g)


TASK 3 – DEFINITIONS

1) ∆Hf of C6H6(l) ��...��.

2) ∆Hf of CH3COOH(l) ��..

3) ∆Hc of H2(g) ��

4) ∆Hc of CH3COOH(l) ��..

5) 1st ionisation energy of aluminium ��..

6) 2nd ionisation energy of aluminium ��.

7) 3rd ionisation energy of aluminium ��..

8) 1st electron affinity of chlorine ��..

9) lattice enthalpy of formation of sodium oxide ��.

10) lattice enthalpy of dissociation of aluminium oxide ��

11) ∆Hhyd of sodium ions ��.

12) ∆Hhyd of oxide ions ��.

13) ∆Hsol of sodium oxide ��

14) ∆Hsol of magnesium carbonate ��

15) Bond dissociation enthalpy of water ��.

16) Bond dissociation enthalpy of hydrogen ��..

17) ∆Ha° of bromine ��

18) Bond dissociation enthalpy of bromine ��..��..

19) Enthalpy of vaporisation of bromine ��.��..��..

20) 1st electron affinity of bromine ��..


TASK 4 – MAKING SURE OF DEFINITIONS

For each of the following reactions, name the enthalpy change. There may be more than one possible answer to some. Some answers may involve more than one enthalpy change!

e.g. CH4(g) + 2 O2(g) → CO2(g) + 2 H2O(l) enthalpy of combustion of CH4(g)

e.g. 2 CH4(g) + 4 O2(g) → 2 CO2(g) + 4 H2O(l) 2 x enthalpy of combustion of CH4(g)

1) Ca(g) → Ca+(g) + e

- ��..��..

2) S(g) + 2 e- → S

2-(g)

��..��..

3) Al2O3(s) → 2 Al3+

(g) + 3 O2-

(g)

��..��..

4) NaBr(s) → Na+(aq) + Br

-(aq)

��..��..

5) I2(s) → 2 I(g)

��..��..

6) C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(l) ��..��..

7) 4 Al(s) + 3 O2(g) → 2 Al2O3(s)

��..��..

8) Ca2+

(g) + O2-

(g) → CaO(s)

��..��..

9) Na(s) → Na(g)

��..��..

10) P4(s) → 4 P(g)

��..��..

11) HCl(g) → H(g) + Cl(g)

��..��..

12) H2O(l) → 2 H(g) + O(g)

��..��..

13) Al(g) → Al3+

(g) + 3 e-

��..��..

14) Ca2+

(g) → Ca2+

(aq)

��..��..

15) Mg(s) → Mg2+

(g) + 2 e-

��..��..

16) Mg(s) + ½ O2(g) → MgO(s)

��..��..


SOLUTION CALCULATIONS

• This cycle works for questions involving enthalpies of solution. It is a simple cycle from a solid to gas ions to dissolved ions.

• Beware of whether the lattice enthalpy is formation or dissociation.

Ionic solid Dissolved ions

Gas ions

∆Hsol

Hydration enthalpies

Lattice enthalpy of formation

Example

Calculate the enthalpy of solution of magnesium chloride given that the lattice enthalpy of formation of magnesium chloride is -2493 kJmol

-1 and the enthalpies of hydration of magnesium and chloride ions are -1920 and -364 kJmol

-1 respectively

MgCl2(s) Mg2+

(aq) + 2 Cl-(aq)

∆Hsol

-1920 2(-364)

-2493

Mg2+

(g) + 2 Cl-(g)

-2493 + ∆Hsol = -1920 + 2 (-364)

∆Hsol = -1920 + 2 (-364) + 2493

∆Hsol = -155 kJ mol-1

TASK 5 – SOLUTION CALCULATIONS

1) Calculate the enthalpy of solution of NaCl given that the lattice enthalpy of formation of NaCl is -771 kJmol-1

and the enthalpies of hydration of sodium and chloride ions are -406 and -364 kJmol

-1 respectively.

2) Calculate the enthalpy of hydration of bromide ions given that the hydration enthalpy of barium ions is -1360 kJmol

-1,

the lattice enthalpy of formation for BaBr2 is -1937 kJmol-1

and the enthalpy of solution of BaBr2 = -38 kJmol-1

. 3) Calculate the lattice enthalpy of formation of calcium iodide given that its enthalpy of solution is -120 kJmol

-1 and the

enthalpies of hydration of calcium and iodide ions are -1650 and -293 kJmol-1

respectively.

4) Calculate the enthalpy of solution of the ammonium chloride using this data: ∆Hhyd (kJ mol-1

): NH4+

-301; Cl- -364;

Lattice enthalpy of formation (kJ mol-1

): ammonium chloride –640.


BORN-HABER CYCLES

• The lattice enthalpy of a compound is an indication of the strength of the ionic bonding. The greater the magnitude of the

lattice enthalpy, the stronger the bonding.

• Generally speaking, compounds with smaller ions and ions with higher charges have stronger attractions and so greater lattice enthalpy.

e.g. NaCl has a higher lattice enthalpy than KCl as they Na+ ion is smaller than the K

+ ion.

e.g. MgCl2 has a higher lattice enthalpy than NaCl as the Mg2+

ion has a higher charge and is smaller than the Na+ ion.

• A Born-Haber cycle is a cycle that includes all the enthalpy changes in the formation of an ionic compound.

SOLID IONIC COMPOUND

ELEMENTS IN NORMAL STATES

GASEOUS ATOMS

GASEOUS METAL ION(S) AND

GASEOUS NON-METAL ATOMS

GASEOUS IONS

enthalpy of formation

atomisation of

metal and non-metal

ionisation energy/ies of

metal atom(s)

electron

affinity/ies of non-metal atom(s)

lattice enthalpy of formation

Enthalpy of formation = sum of all the other enthalpy values

• Please note that:

• steps should have values and labels written on

• it is best to show each separate step (e.g. if both elements are atomised, show this as two steps)

• sometimes each separate ionisation enthalpy and/or electron affinity are shown as separate steps

Comparison of values of lattice enthalpy from Born-Haber cycles to theoretical values

• Lattice enthalpies can be calculated either by:

• a Born-Haber cycle (which uses experimentally measured values for the other enthalpy changes)

• theoretical calculation based on the charge and size of positive and negative ions

• The closer the experimentally measured value, using a Born-Haber cycle, to the theoretical value, the purer the ionic bonding. If the difference is large, then the compound has some covalent character.

• The more polarising the positive ion, the more the negative ion is distorted in shape and the more covalent character. The smaller the positive ion, and the higher its charge, the more polarising it is. Larger negative ions are polarised more easily.

Substance LiCl LiI NaI MgO

Experimental value (from Born-Haber cycle) / kJ mol-1

846 744 684 3889

Theoretical value / kJ mol-1 833 728 684 3929

Difference / kJ mol-1 13 16 2 40


Example

Calculate the lattice enthalpy of formation of magnesium chloride given the following data:

Enthalpy of formation of magnesium chloride = –642 kJ mol-1

Enthalpy of atomisation of magnesium = +150 kJ mol-1

Enthalpy of atomisation of chlorine = +121 kJ mol-1

1st ionisation enthalpy of magnesium = +736 kJ mol-1

2nd ionisation enthalpy of magnesium = +1450 kJ mol-1

1st electron affinity of chlorine = –364 kJ mol-1

Enthalpy of formation = sum of the others

–642 = 150 + 2(121) + 736 + 1450 + 2(–364) + LE

–642 = 1850 + LE

LE = –642 – 1850 = –2492 kJ mol-1

MgCl2(s)

Mg(s) + Cl2(g)

Mg(g) + Cl2(g)

Mg2+(g) + 2 Cl

-(g)

Enthalpy of formation of MgCl2 –642

Enthalpy of atomisation of Mg 150

lattice enthalpy of formation of MgCl2

Mg(g) + 2 Cl(g)

2 x Enthalpy of atomisation of Cl 2(121)

Mg+(g) + 2 Cl(g) + e

-

1st ionisation enthalpy of Mg 736

Mg2+(g) + 2 Cl(g) + 2 e

-

2nd ionisation enthalpy of Mg 1450

2(-364) 2 x electron affinity of Cl

Example

Calculate the enthalpy of formation of sodium bromide given the following data:

Lattice enthalpy of formation of sodium bromide = –733 kJ mol-1

Enthalpy of atomisation of sodium = +109 kJ mol-1

Enthalpy of atomisation of bromine = +112 kJ mol-1

1st ionisation enthalpy of sodium = +494 kJ mol-1

1st electron affinity of bromine = –342 kJ mol-1


TASK 6 – BORN-HABER CYCLES


kJmol-1 K Ca Al Co Cu Br I O Cl

enthalpy of atomisation 90 193 314 427 112 107 248 121

1st ionisation energy 418 590 577 757 745

2nd ionisation energy 3070 1150 1820 1640 1960

3rd ionisation energy 4600 4940 2740 3230 3550

1st electron affinity -342 -142 -364

2nd electron affinity +844

1) Calculate the enthalpy of formation of potassium chloride given that the lattice enthalpy of formation of potassium

chloride is -710 kJmol-1

.

2) Calculate the enthalpy of formation of calcium bromide given that the lattice enthalpy of formation of calcium bromide

is -2125 kJmol-1

.

3) Calculate the lattice enthalpy of formation of aluminium oxide given that the enthalpy of formation of aluminium oxide

is -1669 kJmol-1

.

4) Calculate the lattice enthalpy of formation of calcium oxide given that the enthalpy of formation of calcium oxide is -

635 kJmol-1

.

5) Calculate the first electron affinity of iodine given that the lattice enthalpy of dissociation of calcium iodide is +2054

kJmol-1

and its enthalpy of formation is -535 kJmol-1

.

6) Calculate the enthalpy of atomisation of copper given that the enthalpy of formation of CuO is -155 kJmol

-1 and its

lattice enthalpy of formation is -4149 kJmol-1

.

7) The lattice enthalpy of formation of the three possible chlorides of cobalt are given:

CoCl -700; CoCl2 -2624; CoCl3 -5350 kJmol-1

. a) Using Born-Haber cycles, calculate the enthalpy of formation of each chloride.

b) Which of these chlorides is energetically stable with respect to their elements under standard conditions.

c) Which compound is likely to be formed when cobalt and chlorine react under normal conditions?


TASK 7 – A VARIETY OF ∆∆∆∆H QUESTIONS

1 a) The symbol ∆H is often used. Explain what the ∆H and each mean. (2) b) Define standard enthalpy of formation. (3) c) Write an equation representing the standard enthalpy of formation of:

i) water ii) propan-1-ol (2)

d) Define standard enthalpy of combustion. (2) e) Write an equation representing the standard enthalpy of combustion of:

i) hydrogen ii) propan-1-ol (2)

(Total 11)

2 a) State Hess’s law. (1) b) Calculate the enthalpy of formation of HCl using the data below.

H2(g) → 2 H(g) ∆H = +436 kJ mol-1

Cl2(g) → 2 Cl(g) ∆H = +242 kJ mol-1

HCl(g) → H(g) + Cl(g) ∆H = +431 kJ mol-1

(3) c) What will be the first step in the mechanism for the reaction for the formation of HCl? (1) d) Explain your answer to (c). (1)

(Total 5)

3 Calculate a value for the mean bond enthalpy of the O-H bond in water using the data below.

H2(g) → 2 H(g) ∆H = +436 kJ mol-1

O2(g) → 2 O(g) ∆H = +496 kJ mol-1

H2(g) + ½ O2(g) → H2O(g) ∆H = -242 kJ mol-1

(Total 3)

4 Hydrogen sulphide, H2S, is a liquid at room temperature. Calculate the average H-S bond enthalpy in hydrogen sulphide given that the enthalpy of formation of hydrogen sulphide is –20 kJ mol

-1 and the following data.

H2(g) → 2 H(g) ∆H = +436 kJ mol-1

S(s) → S(g) ∆H = +223 kJ mol-1

H2S(l) → H2S(g) ∆H = +19 kJ mol-1

(Total 3)

5 a) The standard enthalpy of formation of propane is -104 kJ mol-1

. Given that the standard enthalpies of formation of carbon dioxide and water are -394 and -286 kJ mol

-1 respectively, calculate the standard enthalpy change of

combustion of propane. (3)

b) Using bond enthalpies only, calculate the standard enthalpy of combustion of propane.

C-C 348, C-H 412, O=O 496, C-O 360, C=O 743, H-O 463 kJ mol-1

(3)

c) Compare your answers to (a) and (b) and comment on any difference. (2)

(Total 8)

6 What is the sign (+ or -) of the following standard enthalpy changes (justify your answer)?

a) condensing steam b) combustion of octane c) boiling ethanol (Total 3)


7 Write equations representing each of the following: a) 1st electron affinity of iodine e) bond dissociation energy of iodine b) lattice enthalpy of formation of magnesium iodide f) enthalpy of solution of magnesium iodide c) enthalpy of atomisation of iodine g) 2nd ionisation energy of magnesium d) enthalpy of solution of sodium carbonate h) enthalpy of hydration of calcium ions (Total 8)

8 The standard enthalpy change of combustion of propanoic acid is -1574 kJ mol-1

. Given that the standard enthalpies of formation of carbon dioxide and water are -393 and -286 kJ mol

-1 respectively, calculate the

standard enthalpy change of formation of propanoic acid. (Total 3)

9 Calculate the enthalpy of hydration of bromide ions given that the hydration enthalpy of barium ions is -1360 kJ mol

-1, the lattice enthalpy for BaBr2 is -1937 kJ mol

-1 and the enthalpy of solution of BaBr2 =

-38 kJmol-1

. (Total 3)

10 Use the data below to calculate the enthalpy of solution of

a) magnesium sulphate b) barium sulphate Lattice enthalpy of formation (kJ mol

-1): MgSO4 = -2833; BaSO4 = -2474

Hydration enthalpy (kJ mol-1

): Mg2+

= -1891; Ba2+

= -1360; SO42-

= -1004 (Total 5)

11 a) Define lattice enthalpy of formation. (1)

b) For each pair of compounds, predict which will have the lattice enthalpy of greatest magnitude, and explain why.

i) LiF or NaF ii) CaO or MgO iii) NaCl or MgCl2 (6)

(Total 7)

12 Show why calcium and chlorine react together to form CaCl2 rather than CaCl or CaCl3 (you will need three Born-Haber cycles to do this). The lattice enthalpies of formation of the chlorides are: CaCl = -719; CaCl2 = -2218; CaCl3 = -4650 kJ mol

-1. You will also need to use data in the table in question 5.

(Total 10)

13 In each of the following questions you should draw a clear Born-Haber cycle and use data from the table below.

kJ mol-1 Na Al Hg Ca Cl Br I O

enthalpy of atomisation 109 314 61 193 121 112 107 248

1st ionisation energy 494 577 1010 590

2nd ionisation energy 4560 1820 1150

3rd ionisation energy 6940 2740 3300 4940

1st electron affinity -364 -314 -142

2nd electron affinity +844

a) Calculate the lattice enthalpy of formation of aluminium oxide given that its enthalpy of formation is -1669 kJ/mol. (6)

b) Calculate the enthalpy of formation of sodium iodide given that its lattice enthalpy of formation is -684 kJ mol-1

. (6)

c) Calculate the first electron affinity of bromine given that the lattice enthalpy of formation of calcium bromide is -2148 kJ mol

-1 and the enthalpy of formation is calcium bromide is -675 kJ mol

-1. (6)

d) Calculate the second ionisation energy of mercury given that the lattice enthalpy of dissociation of mercury (II) chloride is +2625 kJ mol

-1 and the enthalpy of formation is mercury (II) chloride is -230 kJ mol

-1. (6)

(Total 24)


14 Calculate the enthalpy of formation of magnesium sulphide given the following data.

Lattice enthalpy of formation of magnesium sulphide ∆H = - 3255 kJ mol-1

Mg(s) → Mg(g) ∆H = + 150 kJ mol-1

Mg(g) → Mg+(g) + e

- ∆H = + 736 kJ mol

-1

Mg+(g) → Mg

2+(g) + e

- ∆H = + 1450 kJ mol

-1

S(s) → S(g) ∆H = + 223 kJ mol-1

S(g) + e- → S

-(g) ∆H = - 200 kJ mol

-1

S-(g) + e

- → S

2-(g) ∆H = + 532 kJ mol

-1 (Total 6)

15 Lattice enthalpy gives us useful information about the structure of an ionic compound. a) Explain why the magnitude of the lattice enthalpy of:

i) KCl (-701) is less than CaCl2 (-2237)

ii) NaF (-902) is greater than KF (-801) (3) b) Using the sizes and charges of ions and the laws of electrostatics, theoretical values can be calculated for lattice

enthalpies. The table below shows a comparison between values measured using experimental data with Born-Haber cycles, and theoretical values.

NaCl NaBr NaI AgCl AgBr AgI

Experimental value (kJ mol-1) -776 -742 -699 -890 -877 -867

Theoretical value (kJ mol-1) -766 -731 -686 -768 -759 -736

For many substances, the measured and theoretical values are very similar, such the as the sodium halides, but

for some substances there is a significant difference, such as the silver halides. Account for this difference. (2)

(Total 5)

16 Use the following data to calculate the average C-S bond energy in CS2(l). Enthalpy of atomisation of sulphur = + 223 kJ mol

-1

Enthalpy of atomisation of carbon = + 715 kJ mol-1

Enthalpy of formation of CS2(l) = + 88 kJ mol-1

Enthalpy of vaporisation of CS2(l) = +27 kJ mol-1

i.e. CS2(l) → CS2(g) (Total 4)

17 The enthalpy of hydration of magnesium and sodium ions are -1920 and -406 kJ mol-1

respectively. Explain why the enthalpy of hydration of Mg

2+ has a greater magnitude than that of Na

+. (Total 2)


ENTROPY

What is entropy?

• Entropy (S) is disorder. The more disordered (or random) something is, the greater the entropy.

• Gases have the most entropy (particles are move rapidly and randomly) whereas solids have least entropy (particles vibrating about fixed positions).

• There is a tendency for entropy to increase, i.e. for things to become more disordered.

How does the entropy of a substance vary with temperature?

• The higher the temperature, the faster the particles vibrate/move and so the greater the entropy (disorder).

• Gases have the most entropy (particles are move rapidly and randomly) whereas solids have least entropy (particles vibrating about fixed positions).

S (gas) > S (liquid) > S (solid)

• The variation with temperature and state of a substance is shown. Note how:

• entropy increases with temperature (at 0 K, S = 0);

• there are big increases in entropy on state changes (melting and boiling);

• the entropy increase from liquid to gas is greater than that for solid to liquid due to the large amount of disorder in gases compared to solids and liquids.

S (Jmol-1

K-1

)

Temperature

SOLID

LIQUID

GAS

melting

boiling

How does the entropy of different substances compare? substance S (Jmol-1K

-1)

• Note the entropy units – J mol-1

K-1

, not kJ.

• The more ordered the structure the lower the entropy.

• Structures like diamond and graphite are very highly structured and so have very low entropies.

C (diamond)

C (graphite)

SiO2 (s)

H2O (l)

NaCl (s)

MgCl2 (s)

H2O (g)

CO2 (g)

2

6

42

70

72

90

189

214


Working out the entropy change for a reaction:

• We can calculate the entropy change for a reaction:

∆∆∆∆S = [SUM Sproducts] – [SUM Sreactants]

Example

Calculate the entropy change for this reaction: C2H5OH(l) + 3 O2(g) → 2 CO2(g) + 3 H2O(l)

Entropy: C2H5OH(l) = 161, O2(g) = 205, CO2(g) = 214, H2O(l) = 70 J mol-1

K-1

∆∆∆∆S ∆S = [SUM of S products] – [SUM S reactants]

∆S = [2(214) + 3(70)] – [161 + 3(205)] = -138 J mol-1 K

-1

Predicting the sign of an entropy change

• In the same way as a decrease in enthalpy (∆H -ve) is a driving force in making reactions go, so is an increase in entropy (∆S +ve), i.e., an increase in disorder is favourable.

• For many reactions it is possible to tell whether entropy will increase or decrease by looking at the equation.

e.g. NH4NO3(s) → N2O(g) + 2 H2O(g)

one mole of solid gives three moles of gas ∴ far more disorder ∴ ∆S +ve

TASK 8 – ENTROPY CALCULATIONS


substance

S (J mol

-1 K

-1)

∆Hf (kJ mol

-1)

substance

S (J mol

-1 K

-1)

∆Hf (kJ mol

-1)

H2O(g) 189 -242 Fe(s) 27

H2O(l) 70 O2(g) 205

H2(g) 131 Fe2O3(s) 90 -822

Cl2(g) 223 C(graphite) 5.7

HCl(g) 187 CO2(g) 214 -394

NaHCO3(s) 102 -948 N2(g) 192

Na2CO3(s) 136 -1131 NH3(g) 193

C3H8(g) 270

1) a) Predict the sign of ∆S for the following reactions, where possible, giving your reasoning for each one.

i) H2O(g) → H2O(l)

ii) H2(g) + Cl2(g) → 2 HCl(g)

iii) C3H8(g) + 5 O2(g) → 3 CO2(g) + 4 H2O(l)

b) Calculate ∆S for the above reactions.

2) a) Sketch a graph to show how the entropy of water varies with temperature.

b) Show clearly the melting and boiling points on your graph.


GIBBS FREE ENERGY & FEASIBLE REACTIONS

What is Gibbs free energy?

• Gibbs free energy (G) combines the two thermodynamic factors of enthalpy (H) and entropy (S).

∆∆∆∆G = ∆∆∆∆H - T∆∆∆∆S

What is the connection between Gibbs free energy change and thermodynamic feasibility?

• When ∆G ≤ 0, a reaction is thermodynamically feasible.

• This means the reaction can happen (it might not happen as the activation is too high).

• The term spontaneous is taken to mean the same thing as feasible, but it still means that a reaction can happen rather than that it actually does!

How does temperature affect feasibility?

∆H ∆S How ∆G varies with temperature How feasibility varies with temperature

-ve +ve ∆G always –ve Reaction feasible at all temperatures

-ve -ve ∆G –ve at low T, +ve at high T Reaction feasible at low T, not feasible at high T

+ve +ve ∆G +ve at low T, -ve at high T Reaction not feasible at low T, feasible at high T

+ve -ve ∆G always +ve Reaction not feasible at all temperatures

How does this link to changes of state?

Melting

• A substance cannot melt below its melting point as ∆G for melting is +ve below the melting point.

• At the melting point, ∆G = 0 and so melting is feasible and the substance melts.

Boiling

• A substance cannot boil below its boiling point as ∆G for boiling is +ve below the boiling point.

• At the boiling point, ∆G = 0 and so boiling is feasible and the substance boils.


Example 1

Zinc carbonate decomposes on heating: ZnCO3(s) → ZnO(s) + CO2(g)

ZnCO3(s) ZnO(s) CO2(g)

∆Hf (kJ mol-1) -812 -348 -394

S (J mol-1

K-1) 83 44 214

a) Calculate ∆H, ∆S and ∆G for this reaction at 298 K.

b) Is the reaction spontaneous at 298 K?

c) Give the temperature range in which the decomposition of ZnCO3 is spontaneous

a) ∆∆∆∆H ∆H = [SUM of ∆Hf products] – [SUM ∆Hf reactants]

∆H = [-348 - 394] – [-812] = +70 kJ mol-1

∆∆∆∆S ∆S = [SUM of S products] – [SUM S reactants]

∆S = [44 + 124] – [83] = +85 J mol-1 K

-1

∆∆∆∆S ∆G = ∆H - T∆S

∆H = 70 - 298 (85) = +44.7 J mol-1 K

-1

1000

b) Reaction is not spontaneous at 298 K as ∆G is +ve

c) ∆G = ∆H - T∆S

when ∆G = 0, ∆H - T∆S = 0 ∴ T = ∆H = +70 = 824 K

∆S (+85

/1000)

The reaction is spontaneous when T ≥ 824 K

Example 2

The enthalpy of vaporisation of ethanol is +43.5 kJ mol

-1. The boiling point of ethanol is 78°C. Calculate the entropy change

for the vaporisation of ethanol.

∆G = ∆H - T∆S

At boiling point, ∆G = 0, ∆H - T∆S = 0 ∴ ∆S = ∆H = +43.5 = 0.124 kJ mol-1 K

-1 = 124 J mol

-1 K

-1

T 351


TASK 9 – GIBBS FREE ENERGY CALCULATIONS


substance

S (J mol

-1 K

-1)

∆Hf (kJ mol

-1)

substance

S (J mol

-1 K

-1)

∆Hf (kJ mol

-1)

H2O(g) 189 -242 Fe(s) 27

H2O(l) 70 O2(g) 205

H2(g) 131 Fe2O3(s) 90 -822

Cl2(g) 223 C(graphite) 5.7

HCl(g) 187 CO2(g) 214 -394

NaHCO3(s) 102 -948 N2(g) 192

Na2CO3(s) 136 -1131 NH3(g) 193

C3H8(g) 270

1) a) Calculate ∆G° for the following reaction at 298 K.

H2(g) + Cl2(g) → 2 HCl(g) ∆H = -185 kJ mol-1

∆S = 20 J K-1

mol-1

b) Is this reaction feasible at 298 K?

c) Why does this reaction not happen at room temperature?

2) For the decomposition: 2 NaHCO3(s) → Na2CO3(s) + CO2(g) + H2O(g)

a) Calculate ∆H and ∆S.

b) Calculate ∆G° at 298 K.

c) Calculate the temperature range over which the reaction is feasible.

3) Graphite burns in oxygen to form carbon dioxide. C(graphite) + O2(g) → CO2(g)

a) Calculate ∆S at 298 K.

b) Calculate ∆G at 298 K.

4) For the reaction: 2 Fe(s) +

3/2 O2(g) → Fe2O3(s)

a) Calculate ∆H and ∆S.

b) Calculate ∆G at 298 K.


5) a) Calculate ∆S for this reaction at 298 K: Mg(s) + ½ O2(g) → MgO(s)

∆H = -602 kJ mol-1

∆G = -570 kJ mol-1

b) Explain the sign and magnitude of ∆S°.

6) a) Calculate ∆H, ∆S and ∆G for the following reaction at 298 K:

Fe2O3(s) + 3 H2(g) → 2 Fe(s) + 3 H2O(g)

b) Will this reaction be feasible at 298K?



7) a) Calculate ∆S° for the following process which takes place 91 K:

CH4(s) → CH4(l) ∆H = 0.94 kJmol-1

b) Calculate the boiling point of methane given:

CH4(l) → CH4(g) ∆H = 8.2 kJmol-1

∆S = 73.2 Jmol-1

K-1

c) Compare the magnitude of the ∆S values for melting and boiling methane and comment on the difference.

d) Explain why both melting and boiling are feasible at the temperatures shown, despite being endothermic.

8) The following enthalpy changes are at 298 K:

∆Hhyd (kJmol-1

): K+ -322; Cl

- -364 Lattice enthalpy KCl = -701 kJmol

-1

a) Calculate the standard enthalpy of solution of potassium chloride at 298 K.

b) Potassium chloride dissolves readily in water at 298 K. Deduce the sign of the entropy change for this reaction, and

explain your reasoning.

c) Explain, in terms of the behaviour of particles, why the entropy change has the sign given in your answer to (b).

d) Use your answer to (a) to calculate the smallest possible entropy change there must be when potassium chloride dissolves in water at 298 K, given that it is a feasible change.

9) Use the data in the table to answer this question.

CaCO3(s) CaO(s) CO2(g) MgCO3(s) MgO(s)

∆Hf (kJ mol-1) -1207 -635 -394 -1113 -602

S (J mol-1

K-1) 90 40 214 66 27

a) A reaction that can be made to take place by changing the conditions is:

CaCO3(s) → CaO(s) + CO2(g)

i) Calculate ∆H for this reaction. (1)

ii) Calculate ∆S for this reaction. (1)

iii) Calculate ∆G for this reaction at 298 K. (1)

iv) Is the reaction spontaneous at 298 K? (1)

v) Will increasing or decreasing the temperature help to make ∆G more negative? (1)

vi) At what temperature will ∆G equal zero? (3)

vii) Give the temperature range in which the decomposition of CaCO3 is spontaneous. (1)

viii) How can this reaction be spontaneous even though it is endothermic? (1)

b) Repeat your calculations for the decomposition of MgCO3 to find the temperature at which is undergoes similar decomposition. (4)

c) i) Use your results to compare the thermal stability of MgCO3 and CaCO3. (1)

ii) Suggest a reason for the difference in thermal stability of MgCO3 and CaCO3. (1)

(Total 16)

10) State the sign of the entropy change in the following reactions (or whether it stays roughly constant), justifying

your answer in each case.

a) CO(g) + H2O(g) → CO2(g) + H2(g) c) NaCl(s) + aq → Na+(aq) + Cl

-(aq)

b) N2(g) + 3 H2(g) → 2 NH3(g) d) I2(s) → I2(g) (Total 4)


FULL WORKED SOLUTIONS are available to subscribers of www.chemsheets.co.uk.

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C(s) + 2 H2(g) → CH4(g)

H2(g) + ½ O2(g) → H2O(l)

½ N2(g) + 3/2 H2(g) → NH3(g)

2 C(s) + 3 H2(g) + ½ O2(g) → C2H5OH(l)

C(s) + 3/2 H2(g) + ½ Br2(l) → CH3Br(l)

2 Na(s) + ½ O2(g) → Na2O(s)

CH4(g) + 2 O2(g) → CO2(g) + 2 H2O(l)

H2(g) + ½ O2(g) → H2O(l)

C2H6(g) + 3½ O2(g) → 2 CO2(g) + 3 H2O(l)

Na(s) + ¼ O2(g) → ½ Na2O(s)

C6H14(g) + 9½ O2(g) → 6 CO2(g) + 7 H2O(l)

TASK 1 – A MIXTURE OF SIMPLE AS THERMODYNAMICS QUESTIONS

1 -129 kJ mol-1

2 -2222 kJ mol-1

3 -1194 kJ mol-1

4 -2008 kJ mol-1

5 -184 kJ mol-1

6 -125 kJ mol-1

7 -1669 kJ mol-1

8 -57.3 kJ mol-1

9 -5116 kJ mol-1

10 +288, -277 kJ mol-1

12 +507 kJ mol-1

13 a -1823 kJ mol-1

b -1311 kJ mol-1

c -1690 kJ mol-1

TASK 2 – EXTRA BOND ENTHALPY CALCULATIONS

1 a -184 kJ mol-1

b -95 kJ mol-1

c -103 kJ mol-1

2 391 kJ mol-1

3 464 kJ mol-1

4 452 kJ mol-1

5 508 kJ mol-1

TASK 5 – SOLUTION CALCULATIONS

1 +1 kJ mol

-1 2 -307.5 kJ mol

-1 3 -2316 kJ mol

-1

4 -25 kJ mol-1

TASK 6 – BORN_HABER CYCLES

1 -445 kJ mol

-1 2 -652 kJ mol

-1 3 -15421 kJ mol

-1

4 -3518 kJ mol-1

5 -314 kJ mol-1

6 +339 kJ mol-1

7 CoCl = +241 kJ mol-1

, CoCl2 = -286 kJ mol-1

, CoCl3 = -25 kJ mol-1


TASK 7 – A VARIETY OF ∆∆∆∆H QUESTIONS 2b -92 kJ mol

-1 3 +463 kJ mol

-1 4 +330 kJ mol

-1

5a -2222 kJ mol-1

, b -1690 kJ mol-1

8 -463 kJ mol-1

9 -307.5 kJ mol-1

10a -62 kJ mol-1

, b +110 kJ mol-1

12 CaCl = -179, CaCl2 = -771, CaCl3 = +1494 kJ mol-1

13a -15421 kJ mol-1

, b -288 kJ mol-1

, c -342 kJ mol-1

, d = +1810 kJ mol-1

14 -364 kJ mol-1

16 +523 kJ mol-1

TASK 8 – ENTROPY CALCULATIONS

1b i -119 J mol

-1 K

-1 ii +20 J mol

-1 K

-1 iii -373 J mol

-1 K

-1

TASK 9 – GIBBS FREE ENERGY CALCULATIONS 1a -191 kJ mol

-1

2a ∆H = +129 kJ mol-1

, ∆S = +335 J mol-1

K-1

, b +29.2 kJ mol-1

, c T ≥ 385 K

3 ∆H = -394 kJ mol-1

, ∆S = +3.3 J mol-1

K-1

, ∆G = -395 kJ mol-1

4 ∆H = -822 kJ mol-1

, ∆S = -272 J mol-1

K-1

, ∆G = -741 kJ mol-1

, T ≤ 3028 K

5 ∆S = -107 J mol-1

K-1

6 ∆H = +96 kJ mol-1

, ∆S = +138 J mol-1

K-1

, ∆G = +54.9 kJ mol-1

, T ≥ 696 K

7a ∆S = +10.3 J mol-1

K-1

, b 112 K

8a ∆H = +15 kJ mol-1

, d ∆S = +50.3 J mol-1

K-1

9a ∆H = +178 kJ mol-1

, ∆S = +164 J mol-1

K-1

, ∆G = +129 kJ mol-1

, T ≥ 1085 K

9b ∆H = +117 kJ mol-1

, ∆S = +175 J mol-1

K-1

, ∆G = +64.9 kJ mol-1

, T ≥ 669 K

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Chemsheets A2 033 Thermodynamicsmchem.weebly.com/uploads/9/2/0/5/9205056/thermodynamics.pdf · © 17-Jul-12 Chemsheets A2 033 2 AS THERMODYNAMICS REVISION What is enthalpy? It is