Module 5: Enthalpy and Entropy V1

Question 1

Define the term “first ionisation energy”. Include the symbol in your answer and predict whether the enthalpy change will be exothermic (negative) or endothermic (positive). Explain why.

Answer 1

ΔH sign: Positive. ✓
Energy required to break electrostatic attraction between negative electron and positive protons. ✓

Question 2

Define the term “second ionisation energy”. Include the symbol in your answer and predict whether the enthalpy change will be exothermic (negative) or endothermic (positive). Explain why.

Answer 2

ΔH sign: Positive. ✓

Energy required to break electrostatic attraction between negative electron and positive protons. ✓

Question 3

Define the term “first electron affinity”. Include the symbol in your answer and predict whether the enthalpy change will be exothermic (negative) or endothermic (positive). Explain why.

Answer 3

ΔH sign: Negative. ✓

Bond formed between negative electron and positive protons in nucleus. ✓

Question 4

Define the term “second electron affinity”. Include the symbol in your answer and predict whether the enthalpy change will be exothermic (negative) or endothermic (positive). Explain why.

Answer 4

ΔH sign: Positive. ✓

Energy required to overcome repulsion between negative ion and negative electron. ✓

Question 5

Explain why electron affinity decreases down the group.

Answer 5

Atomic radius increases. ✓

Number of shells increases, and so shielding increases. ✓

Nuclear charge increases, but this is outweighed by increase in shielding and atomic radius. ✓

Less nuclear attraction, so more difficult for larger atoms to attract electrons needed to form an ion. ✓

Remember, gaining or losing electrons is always explaining using NANCARS

Question 6

Define the term “enthalpy change of atomisation”. Include the symbol in your answer and predict whether the enthalpy change will be exothermic (negative) or endothermic (positive). Explain why.

Answer 6

ΔH sign: Positive. ✓

Energy is required to break a bond. ✓

Question 7

Define the term “lattice enthalpy”. Include the symbol in your answer and predict whether the enthalpy change will be exothermic (negative) or endothermic (positive). Explain why.

Answer 7

ΔH sign: Negative. ✓

Energy is released when forming a bond. ✓

Question 8

Define the term “enthalpy change of hydration”. Include the symbol in your answer and predict whether the enthalpy change will be exothermic (negative) or endothermic (positive). Explain why.

Answer 8

Negative. ✓

Bond formation between ions and water molecules releases energy. ✓

Question 9

Define the term “enthalpy change of solution”. Include the symbol in your answer and predict whether the enthalpy change will be exothermic (negative) or endothermic (positive). Explain why.

Answer 9

ΔH sign: Positive or negative. ✓

Depends on the balance between bond breaking for the lattice ✓ and bond making between the ions and water molecules. ✓

Question 10

Determine the type of enthalpy change for each of the following equations.

Answer 10

Question 11

Write equations and determine whether the enthalpy change is positive/ negative for the following:

a. The first electron affinity of sulfur
b. The second electron affinity of sulfur
c. The atomisation of bromine
d. The atomisation of oxygen

Answer 11

Question 12

Write equations and determine whether the enthalpy change is positive/ negative for the following:

a. The first ionisation of aluminium
b. The formation of propan-1-ol
c. The lattice enthalpy of potassium chloride
d. The lattice enthalpy of magnesium chloride
e. The lattice enthalpy of aluminium oxide

Answer 12

Question 13

Which enthalpy change(s) is/are exothermic?

1 The third electron affinity of nitrogen
2 The bond enthalpy of the C–H bond
3 The standard enthalpy change of formation of Cl2(g)

A 1, 2 and 3
B Neither
C Only 1 and 2
D Only 2 and 3

Answer 13

B, Niether are correct ✓

Question 14

Which equation represents the change that accompanies the standard enthalpy change of atomisation of bromine?

Answer 14

D ✓

Bromine is a liquid under standard states and conditions

Question 15

How does lattice enthalpy link to ionic bonding?

Answer 15

Lattice enthalpy is a measure of the strength of an ionic bond. ✓

Question 16

How does average bond enthalpies link to covalent bonding?

Answer 16

average bond enthalpy is a measure of the strength of a covalent bond. ✓

Question 17

What does the size of lattice enthalpy indicate?

Answer 17

A large exothermic value for lattice enthalpy means there are very strong electrostatic forces of attraction ✓

between the oppositely charged ions ✓

and a stronger ionic bond. ✓

Question 18

Write the equation for the lattice enthalpy of sodium chloride. Include state symbols in your answer.

Answer 18

Question 19

Explain why NaCl has a more exothermic lattice enthalpy than CsCl?

Answer 19

Na+ is a smaller ion than Cs+ ✓

Na+ ions can pack closer together and have a greater charge density ✓

Na+ ions have a stronger electrostatic force of attraction for Cl- ions ✓

NaCl has the more exothermic lattice enthalpy ✓

Question 20

Explain why magnesium chloride has a more exothermic lattice enthalpy than sodium chloride.

Answer 20

Mg2+ has a greater charge compared to Na+ ✓

Mg2+ ions can pack closer together and have a greater charge density ✓

Mg2+ ions have a stronger electrostatic force of attraction for Cl- ions ✓

MgCl2 has the more exothermic lattice enthalpy ✓

Question 21

Describe how the lattice enthalpy of MgO differs from the lattice enthalpy of BaO.

Answer 21

Mg2+ is a smaller ion than Ba2+ ✓

Mg2+ ions can pack closer together and have a greater charge density ✓

Mg2+ ions have a stronger electrostatic force of attraction for O2- ions ✓

MgO has the more exothermic lattice enthalpy ✓

Question 22

Explain why magnesium oxide has a more exothermic lattice enthalpy than sodium oxide.

Answer 22

Mg2+ has a greater ionic charge than Na+ ✓

Mg2+ ions can pack closer together and have a greater charge density ✓

Mg2+ ions have a stronger electrostatic force of attraction for O2- ions ✓

MgO has the more exothermic lattice enthalpy ✓

Question 23

Why is it difficult to predict whether the lattice enthalpy for magnesium sulfide is more or less exothermic than the lattice enthalpy of sodium oxide.

Answer 23

Mg2+ smaller and has a greater charge compared to Na+ ✓

therefore Mg2+ stronger attraction ✓

O2- smaller than S2- ✓

O2- stronger attraction ✓

Hard to predict which is more exothermic ✓

Question 24

Answer 24

Question 25

Using a Born-Haber Cycle and the data provided, calculate the lattice enthalpy

Answer 25

Question 26

Answer 26

Question 27

Draw a Born-Haber cycle for BaO.

Answer 27

Question 28

Using the following data, draw a dissolving cycle and calculate a value for the standard enthalpy change of solution of KCl

Answer 28

Question 29

Calculate a value for the standard enthalpy change of hydration for chloride ions using the cycle below

Answer 29

Question 30

Using the following data, draw a dissolving cycle and calculate a value for the standard enthalpy change of solution of aluminium chloride

Answer 30

Question 31

Explain the difference in hydration enthalpies between Cl- and Br -

Answer 31

Cl- is a smaller ion than Br - ✓ Cl- has the same ionic charge as Br - but Cl- has a greater charge density ✓ Cl- has stronger electrostatic attractions for water molecules ✓ Cl- has the more exothermic hydration enthalpy ✓

Question 32

Explain the difference in hydration enthalpies between Na+ and Mg2+

Answer 32

Mg2+ is a smaller ion than Na+ ✓ Mg2+ a greater ionic charge and greater charge density than Na+ ✓ Mg2+ has stronger electrostatic attractions for water molecules ✓ Mg2+ has the more exothermic hydration enthalpy ✓

Question 33

Use the results below to determine the enthalpy change of solution for KCl

Answer 33

Question 34

Define what is meant, entropy, *S*

Answer 34

is a measure of the degree of disorder in a system. ✓

Question 35

Explain why entropy, *S* is always positive.

Answer 35

All substances have some degree of disorder ✓ Particles are in constant motion ✓

Question 36

Explain what a +ΔS shows

Answer 36

A positive ΔS value means the system is becoming more disordered. ✓

Question 37

Explain what a -ΔS shows

Answer 37

A negative ΔS value means the system is becoming more ordered. ✓

Question 38

State whether the entropy change will be a positive or negative value for the following. Explain your answer.

Answer 38

Positive because entropy increases ✓ Liquid becomes gas and there is more disorder ✓

Question 39

State whether the entropy change will be a positive or negative value for the following. Explain your answer.

Answer 39

Negative because entropy decreases ✓ 3 moles of gas to 2 moles of gast herefore there is less moles of gas product.✓ therefore more ordered ✓

Question 40

State whether the entropy change will be a positive or negative value for the following. Explain your answer.

Answer 40

Positive because entropy increases ✓ Solid becomes aqueous, aqueous particles are more disordered than solid particles ✓

Question 41

State whether the entropy change will be a positive or negative value for the following. Explain your answer.

Answer 41

Negative because entropy decreases ✓ 4 moles of gas to 2 moles of gas therefore there is less moles of gas product. ✓ therefore more ordered ✓

Question 42

State whether the entropy change will be a positive or negative value for the following. Explain your answer.

Answer 42

Positive because entropy increases ✓ 1 moles of gas to 2 moles of gas therefore there is more moles of gas product.✓ therefore more disorder ✓

Question 43

State whether the entropy change will be a positive or negative value for the following. Explain your answer.

Answer 43

Positive because entropy increases ✓ Solid becomes gas and liquid and there is more disorder ✓

Question 44

State whether the entropy change will be a positive or negative value for the following. Explain your answer.

Answer 44

Negative because entropy decreases ✓ 2 moles of gas to 1 moles of gas therefore there is less moles of gas product.✓ therefore more ordered ✓

Question 45

State whether the entropy change will be a positive or negative value for the atomisation of iodine. Explain your answer.

Answer 45

Positive because entropy increases ✓ Solid becomes gas so there is more disorder ✓

Question 46

State whether the entropy change will be a positive or negative value for the lattice enthalpy of calcium bromide. Explain your answer.

Answer 46

Negative because entropy decreases ✓ Gases become solid, order increases ✓

Question 47

Give the units for entropy, *S* and entropy change, *ΔS*

Answer 47

Question 48

Answer 48

Question 49

Answer 49

Question 50

Answer 50

Question 51

Answer 51

Question 52

Explain why this exothermic reaction is spontaneous at low temperatures but does not occur at very high temperatures. 

Answer 52

∆H is negative (exothermic reaction) ✓ ∆S is negative (3 moles of gas gives 2 moles of gas) so T∆S is negative ✓ At higher temperatures, T∆S becomes more negative ✓ At higher temperatures, T∆S becomes more negative than ∆H and outweighs ∆H ✓ At higher temperatures, ∆G becomes more positive, and the reaction becomes unfeasible ✓

Question 53

Complete the following table

Answer 53

Question 54

Write the equation for Gibbs Free energy in terms of the equation of a line (y=mx+c) and state how you can find each term graphically.

Answer 54

Plot graph of ∆G on the y-axis against temperature on the x-axis gives a straight with gradient - ∆S. ✓ y-Intercept: ∆H. ✓ Gradient: -∆S. ✓

Question 55

Answer 55

a) No, as ∆G is positive ✓ b) 760 – 273 = 487 degrees celcius ✓ c) 160 kJ mol-1 ✓ d) -∆S = -0.25 kJ K-1 mol-1 ✓ so ∆S = 0.25 x 1000 = 250 J K-1 mol-1 ✓

Question 56

This question is about free energy changes, ∆G, enthalpy changes, ∆H, and temperature, T. The Gibbs’ equation is shown below. ∆G = ∆H − T∆S A chemist investigates a reaction to determine how ∆G varies with T. The results are shown What is significant about the gradient of the line and the values P and Q shown Explain your reasoning.

Answer 56

ΔG = −ΔST + ΔH ✓ Gradient = −ΔS ✓ P: ΔH enthalpy change ✓ Q: Temperature for a reaction to be feasible ✓

Question 57

Using the graph attached, determine the value for ∆S, ∆H and ∆G for lines A, B, C, D

Answer 57

a. ∆S = -ve ∆H = +ve ∆G = +ve. ✓ b. ∆S = -ve ∆H = -ve ∆G = +ve at high temperatures only. ✓ c. ∆S = +ve ∆H = +ve ∆G = +ve at low temperatures only. ✓ d. ∆S = +ve ∆H = -ve ∆G = -ve. ✓

Question 58

Answer 58

∆H is negative (exothermic reaction) ✓ ∆S is positive so T∆S is positive ✓ At higher temperatures, T∆S becomes more positive ✓ At higher temperatures, ∆G becomes more negative and the reaction becomes more feasible ✓ ∆G is always going to be negative. ✓

Question 59

Answer 59

∆H is negative (exothermic reaction) ✓ ∆S is negative (4 moles of gas gives 2 moles of gas) so T∆S is negative ✓ At higher temperatures, T∆S becomes more negative ✓ At higher temperatures, T∆S becomes more negative than ∆H and outweighs ∆H ✓ At higher temperatures, ∆G becomes more positive and the reaction becomes unfeasible ✓

Question 60

Answer 60

∆H is positive (endothermic reaction) ✓ ∆S is positive so T∆S is positive ✓ At higher temperatures, T∆S becomes more positive ✓ At higher temperatures, T∆S becomes more positive than ∆H and outweighs ∆H ✓ At higher temperatures, ∆G becomes more negative and the reaction becomes feasible ✓

Question 61

∆S for an endothermic reaction is negative. Will the reaction ever be feasible? Explain your answer using both low and high temperature.

Answer 61

∆H is positive (endothermic reaction) ✓ ∆S is negative so T∆S is negative ✓ At higher temperatures, T∆S becomes more negative ✓ Negative value is subtracted from positive ∆H, so ∆G will always be positive and reaction will always be not feasible. ✓

Question 62

∆S for an exothermic reaction is positive. Will the reaction ever be feasible? Explain your answer using both low and high temperature.

Answer 62

∆H is negative (exothermic reaction) ✓ ∆S is positive so T∆S is postive ✓ At higher temperatures, T∆S becomes more positive ✓ Positive value is subtracted from negative ∆H, so ∆G will always be negative and reaction will always be feasible. ✓