Y13 Enthalpy and entropy

Question 1

Lattice enthalpy definition

Answer 1

The enthalpy change when ONE MOLE of an ionic compound is formed from its GASEOUS ions under standard conditions

Question 2

True or false: lattice enthalpies are always positive

Answer 2

False (energy is released when the ionic bonds are made = exothermic)

Question 3

Why does LiF (-1038) have a lower enthalpy than KF (-821)?

Answer 3

(Charge is the same)
Li+ has a smaller ionic radius than K+
Li+ has a stronger attraction to F-
More energy is released when Li+ attracts F-
Lattice enthalpy is more negative

Question 4

What are the 4 marking points for a Born-Haber cycle?

Answer 4

Species must be on the lines (only 1 thing changes at a time)
State symbols
Arrows must be labelled and point in the correct direction (up = endothermic, down = exothermic)
Show calculation

Question 5

Enthalpy change of formation definition

Answer 5

Enthalpy change when one mole of a compound is formed from its constituent elements in their standard states = exothermic

Question 6

Enthalpy change of atomisation

Answer 6

Enthalpy change when 1 mole of gaseous atoms is formed from the element in its standard state e.g. Na(s) –> Na(g) = endothermic

Question 7

First electron affinity definition

Answer 7

Enthalpy change when 1 mole of gaseous atoms each gain 1 electron e.g. Cl(g) + e- –> Cl-(g) = exothermic

Question 8

First ionisation enthalpy definition

Answer 8

Enthalpy change when 1 mole of gaseous 1+ ions are formed from 1 mole of gaseous atoms e.g. Na(g) –> Na+(g) + e- = endothermic

Question 9

Why is second electron affinity endothermic instead of exothermic?

Answer 9

More energy is required to overcome the repulsion between the negatively charged e- and negative ion

Question 10

Enthalpy change of solution definition

Answer 10

Enthalpy change when ONE MOLE of a solid dissolves in a solvent under standard conditions e.g. NaCl(s) + aq –> Na+(aq) + Cl-(aq)

Question 11

When calculating △solH, how do you work out m in the q=mc△T equation?

Answer 11

Mass of water AND solid

Question 12

Is △solH endo or exothermic?

Answer 12

Can be either

Question 13

When is △solH exothermic?

Answer 13

When the magnitude of the enthalpy change of hydrations is GREATER than the magnitude of the lattice enthalpy

Question 14

When is △solH endothermic?

Answer 14

When the magnitude of the enthalpy of hydrations is LESS than the magnitude of the lattice enthalpy

Question 15

Enthalpy of hydration definition

Answer 15

Enthalpy change when GASEOUS ions dissolve in water to form ONE MOLE of AQUEOUS ions e.g. Na+(g) + aq –> Na+(aq) =exothermic

Question 16

Compare the enthalpies of hydration of Mg2+ and Na+

Answer 16

Mg2+ has a higher charge than Na+
Mg2+ have a smaller ionic radius than Na+
Mg2+ will have a stronger attraction to WATER MOLECULES
△hydH of Mg2+ would be more negative

Question 17

What is the symbol and units of entropy?

Answer 17

Symbol: S, Units: J K-1 mol-1

Question 18

Why is the entropy of water 0 at 0K?

Answer 18

Particles have no energy so no disorder

Question 19

When is △S positive and negative?

Answer 19

If a reaction leads to MORE order = NEGATIVE △S
If a reaction leads to LESS order = POSITIVE △S

Question 20

What affects entropy?

Answer 20

No. moles of gas

Question 21

2Na(s) + Cl2(g) –> 2NaCl(s) Will this change in entropy be positive or negative?

Answer 21

Entropy will decrease because the number of moles of gas decreases which increases order

Question 22

What is the equation for calculating △S?

Answer 22

△S = (sum of S of products) - (sum of S of reactants)

Question 23

What is the equation for calculating △G (feasibility)?

Answer 23

△G = △H - T△S

Question 24

What are the unit conversions when calculating △G?

Answer 24

Temperature (K): °C –> K = +273
△S (kJ K-1 mol-1): J K-1 mol-1 –> kj K-1 mol-1 = /1000

Question 25

What must be the case for a reaction to be feasible?

Answer 25

△G must be < 0 / △G must be negative

Question 26

What is the equation for calculating the minimum temperature a reaction is feasible at?

Answer 26

T = △H / △S

Question 27

When is a reaction feasible when △H is negative and △S is positive?

Answer 27

△G is always negative so the reaction is always feasible at all temperatures

Question 28

When is a reaction feasible when △H is positive and △S is negative?

Answer 28

△G is always positive so the reaction is never feasible at any temperature

Question 29

When is a reaction feasible when both △H and △S are negative?

Answer 29

Reaction is only feasible at LOW temperatures

Question 30

When is a reaction feasible when both △H and △S are positive?

Answer 30

Reaction is only feasible at HIGH temperatures

Question 31

How does an increase in temperature affect feasibility when both △H and △S are negative?

Answer 31

△G = △H - T△S
For the reaction to be feasible, △G must be negative
When both △H and △S are negative, the reaction is feasible at high temperatures
As T increases, T△S becomes more positive
Eventually T△S will have a greater magnitude than △H, leading to a negative △G value
Feasibility increases as T increases

Question 32

What are the 2 limitations of predicting feasibility using △G?

Answer 32

Reactions may have a:
1. Large activation energy
2. Slow rate