Thermodynamics

Question 1

Standard Enthalpy of Formation

Answer 1

enthalpy change when one mole of a compound is formed from its elements under standard conditions, all reactants and products in their standard states.

e.g. Na(s) + ½Cl2(g) → NaCl(s)

Question 2

Standard Enthalpy of Atomisation

Answer 2

enthalpy change when one mole of gaseous atoms is formed from an element in its standard state.

e.g. 1/2Br2(l) → Br(g)

Question 3

First Ionisation Enthalpy

Answer 3

standard enthalpy change when one mole of electrons is removed from one mole of gaseous atoms to give one mole of gaseous ions each with a single positive charge.

e.g. Ca(g) → Ca+(g) + e-

Question 4

Mean Bond Enthalpy

Answer 4

enthalpy change when one mole of gaseous molecules each break a covalent bond to form two free radicals, averaged over a range of compounds.

e.g Br2(l) → 2Br(g)

Question 5

First Electron Affinity

Answer 5

standard enthalpy change when one mole of gaseous atoms is converted into a mole of gaseous ions, each with a single negative charge under standard conditions.

e.g. Cl(g) + e- → Cl-(g)

Question 6

Lattice Formation Enthalpy

Answer 6

standard enthalpy change when one mole of solid ionic compound is formed from its gaseous ions

e.g. Mg2+(g) + 2Br-(g) → MgBr2(s)

Question 7

Lattice Dissociation Enthalpy

Answer 7

standard enthalpy change when one mole of solid ionic compound dissociates into its gaseous ions.

e.g. MgBr2(s) → Mg2+(g) + 2Br-(g)

Question 8

Born Haber Cycle

Answer 8

Question 9

Suggest one reason why the 1st electron affinity of O2 is exothermic

Answer 9

• atomic radius is small, strong attraction between nucleus and e-
• energy is released

Question 10

2nd e- affinity trend

Answer 10

• ion already negative
• repulsion between e- and negative ion must be overcome
  (energy in)

Question 11

Trends in Lattices

Answer 11

Charge of ion:
greater charge an ion has, geater the attraction to an oppositely charged ion

Size of an ion (atomic radius):
smaller the ion. greater the attraction to an oppositely charged ion

Compare using CRAM: (2 diff compounds)
Charge on ion
Radius of ion
Attraction between ions
More exo/endo

Question 12

Explain why the lattice formation of K2O is more exothermic than K2S

Answer 12

• both O and S have a -2 charge
• O2- has a smaller ionic radius than S2-
• O2- has a stronger attraction to potassium ion
• K2O has a more exothermic lattice enthalpy

Question 13

Predicting characters with covalent character

Answer 13

• have a positive ion which is small and highly charged
• have a negative ion which is large and highly charged

Question 14

Covalent character (theoretical & experimental)

Answer 14

Theoretical:
* perfect ionic model
* type of ion= point charges, perfect spheres (not polarisable)
* purely ionic

Experimental:
* born haber
* type of ion= polarisable ions
* ionic bonding & covalent character
(stronger bonds = more exo)

Question 15

Covalent character Vs no covalent character

Answer 15

If an ionic compound has no covalent character:
ΔHLF calculated by perfect ionic model is equal to ΔHLF calculated by experimental model

If there is covalent character in an ionic compound:
- ΔHLF calculated by perfect ionic model is less exothermic than the ΔHLF calculated by experimental model.
- Experimental Born Haber model allows for covalent character and predicts stronger bonding.

Question 16

Standard Enthalpy of Solution

Answer 16

Enthalpy change when one mole of solute dissolves to form its aqueous ions

e.g. NaCl(s) + aq → Na+(aq) + Cl-(aq)

Question 17

Standard Enthalpy of Hydration

Answer 17

Enthalpy change when one mole of gaseous ions is converted into one mole of aqueous ions

e.g. Cl-(g) + aq → Cl-(aq)

Question 18

Hess’ Law

Answer 18

Question 19

How does solubility change as temperature is increased
MgCl2(s) + aq → Mg2+(aq) + 2Cl-(aq) -155kJmol-1

Answer 19

• forward reaction is exothermic
• equilibrium shifts to the left to oppose increase in temperature
• MgCl2 will be less soluble

Question 20

Explain why the enthalpy of hydration becomes less exothermic from Li+ to K+

Answer 20

ions get bigger so attraction to δ+ of water gets weaker

Question 21

Suggest why the electron affinity of Cl is an exothermic change

Answer 21

net attraction between chlorine nucleus and extra e-

Question 22

Explain why the enthalpy of hydration is more negative/ exothermic

Answer 22

Ion is smaller so has stronger attraction to δ+ of water ion
Energy is released when attraction forms

Question 23

Explain why magnesium ions are attracted to H2O molecule

Answer 23

H2O is polar
δ- on O2 attracted to positive charge on Mg2+ ion

Question 24

Entropy definition

Answer 24

measure of amount of disorder in system
units= Jk-1mol-1 (/1000 = kJk-1mol-1)

Question 25

Entropy changes

Answer 25

• gases have much higher entropy than (l) and (g)
• solid - low entropy = particles highly ordered
• gases = high entropy = particle move in disordered way

Question 26

Calculating entropy change

Answer 26

∆Sᶿ = ∑Sᶿ (products) - ∑Sᶿ (reactants)

• ∆S is positive = disorder
• ∆S is negative = order

Question 27

Feasible

Answer 27

If a reaction is feasible it can happen at a given temperature

reaction is feasible when ∆G less than or equal to 0

Question 28

Gibbs free energy equation (thermodynamics equation)

Answer 28

∆G = ∆H - T∆S

∆G = kJmol-1
∆H = kJmol-1
T = Kelvin
∆S = kJK-1mol-1

Question 29

Calculating the temperature at which a reaction becomes feasible

Answer 29

T = ∆H / ∆S

Question 30

Changes is ∆G when T changes

Answer 30

Question 31

Kinetic Factors

Answer 31

A reaction may not occur in real life.
The reaction may still have a very high activation energy such that very few particles have sufficient energy to react or it may occur at a very slow rate.

Question 32

Gibbs Free Energy and Graphs

Answer 32

y = m x + c. m=gradient (dy/dx). c=intercept
∆G = -∆S T + ∆H