Thermodynamics

Question 1

∆H

Answer 1

Heat energy exchange, enthalpy

Question 2

Exothermic

Answer 2

-ve ∆H. Heat is released into surroundings

Therefore +ve∆T

Question 3

Endothermic

Answer 3

+ve ∆H

Question 4

First law

Answer 4

Total energy remains constant

Question 5

Standard Conditions

Answer 5

100kPa pressure
298K temp
1 mol dm-3 concentration

Question 6

∆Hr

Answer 6

For reactions in molar quantities in a chemical equation

Question 7

∆Hc

Answer 7

Change when 1 mol reacts completely with O2

Question 8

∆Hf

Answer 8

Change when 1 mol of compound is formed from its constituents

Question 9

∆Hneut

Answer 9

Change when acid us neutralised by base to form 1 mol H2O

Question 10

∆Hat

Answer 10

To form 1mol gaseous atoms

Question 11

∆Hi.e

Answer 11

To remove 1e- per atom in 1mol gas

Question 12

∆He.a

Answer 12

To add 1e- per atom to 1mol gaseous atoms

Question 13

∆Hsolution

Answer 13

To dissolve 1mole of solute to form solution

Question 14

∆Hhyd

Answer 14

To dissolve 1mole gaseous ions in water to form 1mole hydrated ions in solutions

Question 15

Lattice enthalpy, ∆Hl.e

Answer 15

For formation of 1mol ionic compound.
Exothermic
ID indirectly by Born-Haber Cycle, and applying Hess’ Law

Question 16

∆H Equation

Answer 16

Q=mc∆T
m= mass of surroundings
c = specific heat capacity of surroundings
∆T = temp change (final-initial temp)

Question 17

Hess’ Law

Answer 17

If a reaction can go via +1 route, final and initial conditions are same; total enthalpy change is the same

Question 18

∆Hc equation

Answer 18

∆H=∑∆Hc (reactants) - ∑∆Hc (products)

Question 19

∑∆Hf

Answer 19

∆H = ∑∆Hf (products) - (reactants)

For elements, ∑∆Hf=0

Question 20

Bond enthalpy

Answer 20

Change to break 1mol of bonds

Question 21

Bond breaking vs bond making

Answer 21

Breaking = endo ∑(bond enthalpies in reactants)
Making = exo -∑(bond enthalpies in products)

Question 22

Bond enthalpy equation

Answer 22

∆H = ∑(BEs in reactants) - ∑(BEs in products)

Question 23

Small BEs

Answer 23

Break first, therefore have quick reactions

Question 24

Entropy, S

Answer 24

Measures no. of ways to arrange molecules

Increased with disorder (0K= perfect crystals and 0 entropy

Question 25

Lattice Enthalpy factors

Answer 25

Ionic size and charge increase Size: decreases charge density Decreases between ion attraction Lattice energy becomes less negative

Question 26

Ionic Charge and LE

Answer 26

Cation: ↑charge = more attraction ↓size =more attraction Anion: ↑charge = more attraction ↑size = less attraction

Question 27

Hydration enthalpy factors

Answer 27

Charge: ↑ionic charge ↑H2O attraction

Question 28

Hydration enthalpy factors for size

Answer 28

Cations: ↓size ↑H2O attraction Anions: ↑size ↓H2O attraction

Question 29

BE equation

Answer 29

∆H = ∑(BEs of reactants) - ∑(product BEs)

Question 31

Entropy equation

Answer 31

∆S = ∑S(products) - ∑S(reactants)

Question 33

Free energy ∆G equation

Answer 33

∆G = ∆H-T∆S

Question 34

Spontaneous (feasible) process criteria

Answer 34

-ve ∆G

Question 35

-ve∆H + +ve∆S

Answer 35

-ve∆G

Question 36

+ve∆H + -ve∆S

Answer 36

+∆G

Question 37

-ve∆H + -ve∆S

Answer 37

-ve ∆G at low temps (∆H > T∆S)

Question 38

+ve∆H + +ve∆S

Answer 38

-∆G at high temps | T∆S > ∆H

Question 41

-ve ∆G and equil constant

Answer 41

Equil constant has large value (products predominate)

Question 42

+ve ∆G and equil

Answer 42

Equil constant is small value

Question 42

Enthalpy and entropy comparison

Answer 42

Lower enthalpy system = more stable one More disorderly system = more stable one -more moles also = more entropy

Question 43

∆Ssystem

Answer 43

∆Ssystem = ∑S(products) - ∑S(reactants)

Question 43

Positive v negative entropy

Answer 43

Positive = favour for a process

Question 44

∆Ssurroundings

Answer 44

∆Ssurroundings = -(∆H/T)

Question 44

∆G≤0 implications

Answer 44

∆G≤0 = feasible/spontaneous reaction Therefore ∆G MUST be negative

Question 45

∆Stotal

Answer 45

∆Ssystem + ∆Ssurroundings ∆Stotal >0 = feasible process When ∆Stotal increases, so does magnitude of equil constant: ∆S = RInK

Question 45

Temperature factors

Answer 45

K, not C ∆G = 0, temp is at point where reaction can become spontaneous (T = ∆H/∆S)