Thermodynamics (P1+P3)

Question 1

Define standard enthalpy of formation

Answer 1

Enthalpy change when 1 mole of a compound is formed from its elements under standard conditions, all reactants and products in their standard states.

Na(s) + 1/2Cl2(g) –> NaCl(s)

Question 2

Define 1st ionisation enthalpy

Answer 2

Standard enthalpy change when 1 mole of electrons is removed from 1 mole of atoms to give 1 mole of gaseous ions with a single positive charge.

Ca(g) –> Ca+(g) + e-

Question 3

Define standard enthalpy of atomisation

Answer 3

Enthalpy change when 1 mole of gaseous atoms are formed from an element in its standard state.

1/2Br2(l) –> Br(g)

Question 4

Define mean bond enthalpy

Answer 4

Enthalpy change when 1 mole of gaseous molecules each break a covalent bond to form 2 free radicals over a range of compounds.

Br2(l) –> 2Br(g)

Question 5

Define 1st electron affinity

Answer 5

Standard enthalpy change when 1 mole of gaseous atoms is converted into 1 mole of gaseous ions each with a single negative charge under standard conditions.

O(g) + e- –> O-(g)

Question 6

Define lattice formation enthalpy

Answer 6

Standard enthalpy change when 1 mole of a solid ionic compound is formed from its gaseous ions.

Mg2+(g) + 2Br(g) –> MgBr2(s)

Question 7

Define lattice dissociation enthalpy

Answer 7

Enthalpy change when 1 mole of solid ionic compound dissociates into its gaseous ions.

MgBr2(s) –> Mg2+(g) +2Br-(g)

Question 8

Define standard enthalpy of hydration

Answer 8

Standard enthalpy change when 1 mole of gaseous ions in converted into 1 mole of aqueous ions.

Cl-(g) + aq –> Cl-(aq)

Question 9

Define standard enthalpy of solution

Answer 9

Standard enthalpy change when 1 mole of a solute dissolves to form its aqueous ions.

NaCl(s) + aq –> Na+(aq) + Cl-(aq)

Question 10

How does charge and ionic radius determine how exothermic a lattice will be?

Answer 10

The bigger the charge and the smaller the ion, the greater the charge density.

Question 11

The more negative the lattice formation means…

Answer 11

More stronger ionic bonds

Question 12

How to determine if a compound will have covalent character

Answer 12

• A positive ion which is small and highly charged.
• A negative ion which is large and negatively charged.

Question 13

The perfect ionic (theoretical) model suggests that…

Answer 13

Ions have point charges and are perfectly ionic.

Question 14

The born haber cycle (experimental) suggests that…

Answer 14

Ions are polarisable and have SOME covalent character.

Question 15

If an ionic compound has NO covalent character…

Answer 15

∆HLf value for perfect ionic model and experimental model would be very similar

Question 16

If an ionic compound HAS covalent character…

Answer 16

∆HLf value calculated by the perfect ionic model should be less exothermic than the
∆HLf calculated by the experimental born haber cycle

Question 17

Define entropy

Answer 17

A measure of disorder in a system.

Question 18

What happens to the entropy from a solid to a liquid to a gas?

Answer 18

Increases

Question 19

What happens to the entropy from a gas to a liquid to a solid?

Answer 19

Decreases

Question 20

What happens to the entropy from a small number of moles to a large number of moles?

Answer 20

Increases

Question 21

What happens to the entropy from a large number of moles to a small number of moles?

Answer 21

Decreases

Question 22

Equation for entropy (∆S)

Answer 22

Products - Reactants

Question 23

Equation for Gibbs free energy (∆G)

Answer 23

∆G = ∆H - T∆S

∆H = enthalpy change Kj mol-1
T = temp in K
∆S = entropy change KjK-1mol-1

key point = divide ∆S by 1000

Question 24

Equation to calculate the temperature at which the reaction becomes feasible?

Answer 24

T = ∆H / ∆S

Question 25

When does a reaction become feasible?

Answer 25

When ∆G is LESS than 0

Question 26

Equation for Gibbs free energy in terms of y = mx + c

Answer 26

∆G = ∆S T + ∆H

If positive gradient - ∆S is negative
If negative gradient - ∆S is positive

Question 27

If ∆H is negative and ∆S is positive, ∆G will always be… and feasible at…

Answer 27

• negative
• any temp

Question 28

If ∆H is positive and ∆S is negative, ∆G will always be… and NOT feasible at…

Answer 28

• positive
• at any temp

Question 29

If ∆H is negative and ∆S is negative, ∆G will always be… and feasible at…

Answer 29

• negative
• at low temp

Question 30

If ∆H is positive and ∆S is positive, ∆G will always be… and feasible at…

Answer 30

• negative
• at high temp