Thermodynamics

Question 1

Standard molar enthalpy of formation

Answer 1

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

Question 2

Standard enthalpy of atmoisation

Answer 2

The enthalpy change which accompanies the formation of one mole of gaseous atoms from the element in its standard state under standard conditions. (Endothermic)

Question 3

First ionisation energy

Answer 3

The standard enthalpy change when one mole of gaseous atoms is converted into a mole of gaseous ions each with a single positive charge. (Endothermic)

Question 4

Second ionisation energy

Answer 4

Loss of a mole of electrons from a mole of singly positively charged ions. (Endothermic)

Question 5

The first electron affinity

Answer 5

The standard enthalpy change when a mole of gaseous atoms is converted to a mole of gaseous ions, each with a single negative charge. (Exothermic)

Question 6

The second electron affinity

Answer 6

The enthalpy change when a mole of electrons is added to a mole of gaseous ions each with a single negative charge to form ions each with two negative charges. (Endothermic)

Question 7

Lattice formation enthalpy

Answer 7

The standard enthalpy change when one mole of solid ionic compound is formed from its gaseous ions. (Exothermic)

Question 8

Enthalpy of lattice dissociation

Answer 8

One mole of ionic compound separates into its gaseous ions. (Endothermic)

Question 9

Enthalpy of hydration

Answer 9

Standard enthalpy change when water molecules surround one mole of gaseous ions.(Exothermic)

Question 10

Enthalpy of solution

Answer 10

Standard enthalpy change when one mole of solute dissolves completely in sufficient solvent to form a solution in which the molecules or ions are far enough apart not to interact with each other. (Endothermic)

Question 11

Mean bond enthalpy

Answer 11

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

Question 12

Why can lattice enthalpies obtained by calculation differ from those from a Born-Haber cycle?

Answer 12

Bond has covalent character

Question 13

What factors increase polarisation?

Answer 13

1. Cation : small size, high charge

2. Anion : large size, high charge

Question 14

Describe the process of polarisation

Answer 14

1. When the small cation approaches the electron clouds of the anion, it distorts the electrons attracting them towards it
2. The distortion means that there are more electrons than expected conc. between the nuclei’s
3. This represents a degree of covalency which accounts for the lattice enthalpy discrepancy
4. The anion is said to be polarised

Question 15

Why is the anion easy to distort?

Answer 15

Large size means the electrons are far from the nucleus. (maybe double charge)

Question 16

Lattice enthalpy measures

Answer 16

strength of ionic bond

Question 17

Bond enthalpies measure

Answer 17

strength of covalent bonding

Question 18

Define Entropy

Answer 18

The randomness of a system expressed mathematically

Question 19

Units of Entropy

Answer 19

JK-1 mol -1

Question 20

Entropy =

Answer 20

Entropy of Products - Reactants

Question 21

The Gibbs free energy change

Answer 21

The enthalpy change and the entropy change which combined govern the feasibility of a chemical reaction.

Question 22

G =

Answer 22

H - TS

Question 23

What happens at G = 0?

Answer 23

The reaction is just feasible. In a closed system an equilibrium exists around this temperature in which both products and reactants are present.

Question 24

How can G be used to calculate entropy?

Answer 24

When there is a change of state eg. melting point G = 0 so enthalpy and the temperature at melting point will tell you entropy change

Question 25

What is a good way of extracting metals and why?

Answer 25

Heat them with carbon. Coke is cheap. CO2 diffuses away so there’s no problem of separating it from the metal.

Question 26

∆H(solution) =

Answer 26

Σ(∆H(hydration)) - ∆H(lattice formation)
or
Σ(∆H(hydration))+ ∆H(lattice dissociation)

Question 27

Why do many endothermic reactions proceed spontaneously under normal conditions?

Answer 27

because there is an increase in entropy. Some exothermic reactions do not proceed spontaneously because there is a decrease in entropy.