Physical Unit 1.8: Thermodynamics

Question 1

define enthalpy of formation

Answer 1

the enthalpy change when 1 mole of a substance is formed from its constituent elements with all reactants & products in their standard states under standard conditions

e.g. 2Na(s) + 1/2O2(g) –> Na2O(s)

exothermic (-ve)

Question 2

define standard enthalpy of combustion

Answer 2

enthalpy change when 1 mole of a substance is completely burned in oxygen with all reactants & products in standard states under standard conditions
e.g. H2 + 1/2O2(g) –> H2O(g)

exothermic (-ve)

Question 3

define standard enthalpy of neutralisation

Answer 3

enthalpy change when 1 mole of water is formed in a reaction b/w an acid & an alkali under standard conditions

e.g.1/2H2SO4(aq) + NaOH(aq) –> 1/2NaSO4(aq) + H2O(l)

exothermic (-ve)

Question 4

define enthalpy of ionisation

Answer 4

first ionisation energy: enthalpy change when each atom in 1 mole of gaseous atoms loses 1 e- to form 1 mole of gaseous 1+ ions
e.g. Mg(g) –> Mg+(g) + e-
endothermic (+ve)

2nd: enthalpy change when each ion in 1 mole of gaseous 1+ ions loses 1 e- to form 1 mole of gaseous 2+ ions
e.g. Mg+(g) –> Mg2+(g) + e-
endothermic (+ve)

Question 5

define electron affinity

Answer 5

1st electron affinity: enthalpy change when each atom in one mole of gaseous atoms gains one e- to form one mole of gaseous 1- ions
e.g. O(g) + e- –> O-(g)
exothermic (-ve)

2nd electron affinity: enthalpy change when each ion in one mole of gaseous 1- ions gains one e- to form one mole of gaseous 2- ions
e.g. O-(g) + e- –> O2-(g)
endothermic (+ve) bc adding -ve e- to -ve ion, which repel

Question 6

define enthalpy of atomisation

Answer 6

enthalpy change when one mole of gaseous atoms is produced from an element in its standard state
e.g. 1/2I2(g) –> I(g)

endothermic (+ve)

Question 7

define hydration enthalpy

Answer 7

enthalpy change when one mole of gaseous ions become hydrated/dissolved in water
e.g. Mg2+(g) + aq –> Mg2+(aq)

exothermic (-ve)

Question 8

define enthalpy of solution

Answer 8

enthalpy change when one mole of an ionic solid dissolves in an amount of water large enough that the dissolved ions are well separated & do not interact with each other
e.g. MgCl2(s) –> Mg2+(aq) + 2Cl-(aq)

varies endo or exo

Question 9

more -ve enthalpy of solution =
more +ve enthalpy of solution =

Answer 9

more likely to dissolve/more soluble
less likely to dissolve/less soluble

Question 10

define bond dissociation enthalpy

Answer 10

enthalpy change when one mole of covalent bonds is broken in the gaseous state
e.g. I2(g) –> 2I(g)
endothermic (+ve)
roughly in the hundreds

Question 11

define lattice enthalpy of formation

Answer 11

enthalpy change when one mole of a solid ionic compound is formed from its constituent ions in the gas phase
e.g. Mg2+(g) + 2Cl-(g) –> MgCl2(s)
exothermic (-ve)

Question 12

define lattice enthalpy of dissociation

Answer 12

enthalpy change when one mole of a solid ionic compound is broken up into its constituent ions in the gas phase
e.g. MgCl2(s) –> Mg2+(g) + 2Cl-(g)
endothermic (+ve)

Question 13

define enthalpy of vaporisation

Answer 13

enthalpy change when one mole of a liquid is turned into a gas at the bp of the liquid
e.g. H2O(l) –> H2O(g)
endothermic (+ve)

Question 14

define enthalpy of fusion

Answer 14

enthalpy change when one mole of a solid is turned into a liquid
e.g. Mg(s) –> Mg(l)
endothermic (+ve)

Question 15

what is the enthalpy of formation of an element & why?

Answer 15

0
by definition

Question 16

Hess’s Law

Answer 16

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

Question 17

calculations involving enthalpy of formation data

Answer 17

arrows from elements to compounds

Question 18

calculations involving enthalpy of combustion data

Answer 18

arrows from compounds to oxides

Question 19

define mean bond enthalpy

Answer 19

enthalpy change when one mole of covalent bonds is broken in the gas phase often averaged over a range of different compounds

Question 20

why are enthalpies of reaction that have been calculated using mean bond enthalpy data not as accurate?

Answer 20

because the values used are averaged over a range of different compounds not specific values for the given compound

Question 21

calculations involving bond enthalpy data

Answer 21

arrows from compounds to gas atoms
sometimes need to also use enthalpy of vaporisation to convert liquid to gas

Question 22

what is one way to reduce heat loss in calorimetry?

Answer 22

measure the heat capacity of the calorimeter as a whole

Question 23

calorimetry calculations

Question 24

calculations involving enthalpy of solution

Answer 24

see booklet for cycle
gas ions
hydration enthalpy
lattice enthalpy of formation
do hydration enthalpy of each compound in separate steps

Question 25

the greater the magnitude of lattice enthalpy,

Answer 25

the stronger the bonding i.e. the more +ve the lattice enthalpy of dissociation or the more -ve the lattice enthalpy of formation

Question 26

what factors cause lattice enthalpy to be greater?

Answer 26

smaller ions higher charge on ions --> so stronger ionic attractions

Question 27

Born-Haber cycles

Answer 27

see booklet every line must balance for atoms & charges (i.e. no overall charge) first line is elements in standard states draw a separate step for each enthalpy change 2nd & 3rd electron affinities are endothermic so go up (not down) write numerical values on each step

Question 28

compare experimental lattice enthalpy (LE) to theoretical LE of a perfectly ionic model

Answer 28

experimental LE is calculated using a Born-Haber cycle with each ΔH value found by accurate measurement in experiments theoretical LE is calculated by a theoretical mathematical calculation that considers the size, charge & arrangement of ions in the lattice it is assumed that the structure is perfectly ionic (& that ions are perfectly spherical)

Question 29

which is the real LE value?

Answer 29

the experimental LE value

Question 30

describe covalent character

Answer 30

distortion of ions in ionic compounds (polarised) so ions are not perfectly spherical = covalent character +ve ions that are small &/or highly charged are very good at distorting -ve ions that are large &/or highly charged are easier to distort

Question 31

how does covalent character affect the properties of ionic compounds?

Answer 31

covalent character = lower solubility in water lower mp bc covalent character disrupts the ionic lattice lower electrical conductivity

Question 32

the bigger the % difference b/w the experimental & theoretical values of LE, PRACTICE QS ON THIS!!

Answer 32

the greater the covalent character of an ionic compound greater the distortion of ions, the greater the covalent character

Question 33

solubility is not just about covalent character

Answer 33

if a compound has a more negative LE of formation (<-1000), it is likely to be insoluble bc enthalpy of solution will be v large & +ve

Question 34

diagram of covalent character

Answer 34

see booklet

Question 35

define entropy

Answer 35

disorder the more disordered something is, the greater the entropy

Question 36

what are the units of entropy

Answer 36

Jmol-1K-1

Question 37

describe & explain the relative entropy of each state

Answer 37

gases have the most entropy as particles move rapidly & randomly solids have the least entropy as particles vibrate about fixed positions

Question 38

what is the tendency for entropy?

Answer 38

there is a tendency for entropy to increase = for things to become more disordered 2nd law of thermodynamics is that over time, entropy will naturally increase

Question 39

how does the entropy of a substance vary with temp.?

Answer 39

3rd law of thermodynamics: entropy of a substance is 0 at absolute 0 (bc particles do not move so are in perfect order) & increases with temp. the higher the temp., the faster the particles vibrate/move so the greater the entropy there are big increases in entropy on state changes (melting & boiling) entropy increase from liquid to gas is greater than increase from solid to liquid bc of the large amount of disorder in gases compared to l & s

Question 40

graph of temp. (x) vs entropy (y)

Answer 40

see booklet

Question 41

the more ordered the structure,

Answer 41

the lower the entropy structures like diamond are v highly ordered so have v low entropy smaller = more ordered

Question 42

calculating entropy changes

Answer 42

see booklet cycle arrows from 0 Kelvin to compounds that are +ve

Question 43

in reaction with an increase in entropy,

Answer 43

ΔS is +ve with a decrease in entropy, ΔS is -ve reactions with an increase in entropy are favourable

Question 44

explanation for why entropy increases/decreases/does not change

Answer 44

state # moles of each state increases: e.g. 1 mole of solid --> 3 moles of gas so ΔS will be +ve & large bc large increase in disorder decreases: e.g. 3 moles of gas --> 1 mole of solid so ΔS will be -ve bc increase in order no change: e.g. ΔS will be close to 0 bc no significant change in disorder 3 moles of gas --> 3 moles of gas

Question 45

define Gibbs free energy change (ΔG)

Answer 45

combines enthalpy change (ΔH) & entropy change (ΔS)

Question 46

what is the formula for ΔG & units?

Answer 46

ΔG = ΔH - TΔS ΔH is in kJmol-1 T is in K ΔS is in Jmol-1K-1

Question 47

decrease in enthalpy (a -ve ΔH) is more favourable increase in entropy (a +ve ΔS) is more favourable

Question 48

when is a reaction feasible?

Answer 48

when ΔG ≤ 0

Question 49

define feasible & when a reaction is feasible

Answer 49

a reaction can take place depends on temp. - can be feasible at certain temps but not at others the point at which a reaction switches from being feasible to not feasible is when ΔG = 0

Question 50

why might a feasible reaction not take place?

Answer 50

bc it might have a v high activation energy - need to look at kinetics

Question 51

spontaneous = feasible

Question 52

watch units: must divide tΔS by 1000 to get kJmol-1

Question 53

ΔG calculations

Answer 53

see booklet

Question 54

describe changes of state in terms of ΔG

Answer 54

below the mp of a substance, melting is not feasible bc ΔG is +ve at the mp, ΔG = 0 so melting becomes feasible (also freezing is feasible at the same T) below the bp of a substance, boiling is not feasible bc ΔG is +ve at the bp, ΔG = 0 so boiling becomes feasible (also condensing is feasible at the same T)

Question 55

what is ΔG at the mp/bp?

Answer 55

0

Question 56

calculation involving ΔG at a change of state

Answer 56

set ΔG to = 0 see booklet

Question 57

what are the units of ΔG?

Answer 57

kJmol-1