Chapter 2 Entropy

Question 1

The entropy (S) of a given system

Answer 1

is the number of possible arrangements of the particles and their energy in a given system
In other words, it is a measure of how disordered a system is

Question 2

When a system becomes more disordered, its entropy will

Answer 2

• increase
• An increase in entropy means that the system becomes energetically more stable
• For example, during the thermal decomposition of calcium carbonate (CaCO₃) the entropy of the system increases:

CaCO₃(s) → CaO(s) + CO₂(g)

Question 3

CaCO₃(s) → CaO(s) + CO₂(g)

• In this decomposition reaction, a gas molecule

Answer 3

• (CO₂) is formed
• The CO₂ gas molecule is more disordered than the solid reactant (CaCO₃), as it is constantly moving around
• As a result, the system has become more disordered and there is an increase in entropy

Question 4

• Another typical example of a system that becomes more disordered is when a solid is melted
• For example, melting ice to form liquid water:

H₂O(s) → H₂O(l)

Answer 4

• The water molecules in ice are in fixed positions and can only vibrate about those positions
• In the liquid state, the particles are still quite close together but are arranged more randomly, in that they can move around each other
• Water molecules in the liquid state are therefore more disordered
• Thus, for a given substance, the entropy increases when its solid form melts into a liquid

Question 5

Melting a solid will cause the particles to become more disordered resulting in a more energetically stable system

Answer 5

Question 6

• All elements have positive standard molar entropy values
• The order of entropy for the different states of matter are as follows:

Answer 6

gas > liquid > solid

• There are some exceptions such as calcium carbonate (solid) which has a higher entropy than mercury (liquid)

Question 7

Simpler substances with fewer atoms have

Answer 7

lower entropy values than complex substances with more atoms

• For example, calcium oxide (CaO) has a smaller entropy than calcium carbonate (CaCO₃)

Question 8

Harder substances have

Answer 8

lower entropy than softer substances of the same type

• For example, diamond has a smaller entropy than graphite

Question 9

• The entropy of a substance changes during a change in state
• The entropy …. when a substance melts (change from solid to liquid)

Answer 9

Increases

• Increasing the temperature of a solid causes the particles to vibrate more
• The regularly arranged lattice of particles changes into an irregular arrangement of particles
• These particles are still close to each other but can now rotate and slide over each other in the liquid
• As a result, there is an increase in disorder

Question 10

The entropy ….. when a substance boils (change from liquid to gas)

Answer 10

increases

• The particles in a gas can now freely move around and are far apart from each other
• The entropy increases significantly as the particles become very disordered

Question 11

Similarly, the entropy …… when a substance condenses (change from gas to liquid) or freezes (change from liquid to solid)

Answer 11

decreases

• The particles are brought together and get arranged in a more regular arrangement
• The ability of the particles to move decreases as the particles become more ordered
• There are fewer ways of arranging the energy so the entropy decreases

Question 12

The entropy of a substance increases when the temperature is raised as particles become more disordered

Question 13

The entropy of a substance increases when the temperature is raised as particles become more disordered

Answer 13

Question 14

The entropy also increases when a solid is

Answer 14

• dissolved in a solvent
• The solid particles are more ordered in the solid lattice as they can only slightly vibrate
• When dissolved to form a dilute solution, the entropy increases as:
  
  • The particles are more spread out
  • There is an increase in the number of ways of arranging the energy

Question 15

The crystallisation of a salt from a solution is associated with a

Answer 15

decrease in entropy

• The particles are spread out in solution but become more ordered in the solid

Question 16

Gases have higher entropy values than

Answer 16

• solids
• So, if the number of gaseous molecules in a reaction changes, there will also be a change in entropy
• The greater the number of gas molecules, the greater the number of ways of arranging them, and thus the greater the entropy

Question 17

• For example the decomposition of calcium carbonate (CaCO₃)

CaCO₃(s) → CaO(s) + CO₂(g)

• The CO₂ gas molecule is more disordered than the solid reactant (CaCO₃) as it can

Answer 17

• freely move around whereas the particles in CaCO₃ are in fixed positions in which they can only slightly vibrate
• The system has therefore become more disordered and there is an increase in entropy

Question 18

Similarly, a decrease in the number of gas molecules results in a

Answer 18

decrease in entropy causing the system to become less energetically stable

Question 19

The standard entropy change (ΔS_system^ꝋ ) for a given reaction can be calculated using the

Answer 19

• standard entropies (S^ꝋ ) of the reactants and products
• The equation to calculate the standard entropy change of a system is:

ΔS_system^ꝋ = ΣΔS_products^ꝋ - ΣΔS_reactants^ꝋ

(where Σ = sum of)

Question 20

• For example, the standard entropy change for the formation of ammonia (NH₃) from nitrogen (N₂) and hydrogen (H₂) can be calculated using this equation

Answer 20

N₂(g) + 3H₂(g) ⇋ 2NH₃(g)

ΔS_system^ꝋ = (2 x ΔS^ꝋ(NH₃)) - (ΔS^ꝋ(N₂) + 3 x ΔS^ꝋ(H₂))

Question 21

The feasibility of a reaction does not only depend on the entropy change of the reaction, but can also be affected by the

Answer 21

• enthalpy change
• Therefore, using the entropy change of a reaction only to determine the feasibility of a reaction is inaccurate

Question 22

The Gibbs free energy (G) is the energy change that takes into account

Answer 22

both the entropy change of a reaction and the enthalpy change

Question 23

• The Gibbs equation is:

Answer 23

ΔG^ꝋ = ΔH_reaction^ꝋ - TΔS_system^ꝋ

• The units of ΔG^ꝋ are in kJ mol^-1
• The units of ΔH_reaction^ꝋ are in kJ mol^-1
• The units of T are in K
• The units of ΔS_system^ꝋ are in J K^-1 mol^-1 (and must therefore be converted to kJ K^-1 mol^-1 by dividing by 1000)

Question 24

• The Gibbs equation can be used to calculate the Gibbs free energy change of a reaction

Answer 24

ΔG^ꝋ = ΔH_reaction^ꝋ - TΔS_system^ꝋ

• The equation can also be rearranged to find values of ΔH_reaction^ꝋ, ΔS_system^ꝋ or the temperature, T

Question 25

* The Gibbs equation can be used to calculate whether a reaction is **feasible** or not

Answer 25

**Δ*****G*****^ꝋ = Δ*****H_reaction*****^ꝋ - TΔ*****S_system*****^ꝋ** * When Δ*G*^ꝋ is **negative**, the reaction is **feasible** and likely to occur * When Δ*G*^ꝋis **positive**, the reaction is **not feasible** and unlikely to occur

Question 26

The **feasibility** of a reaction can be affected by the

Answer 26

* **temperature** * The Gibbs equation will be used to explain what will affect the feasibility of a reaction for exothermic and endothermic reactions

Question 27

**Exothermic reactions** * In exothermic reactions, Δ*H_reaction*^ꝋ is **negative**

Answer 27

* If the Δ*S_system*^ꝋ is **positive:** * Both the first and second term will be **negative** * Resulting in a **negative** Δ*G*^ꝋ so the reaction is **feasible** * Therefore, regardless of the temperature, an exothermic reaction with a positive Δ*S_system*^ꝋ will **always be feasible**

Question 28

**Exothermic reactions** * In exothermic reactions, Δ*H_reaction*^ꝋ is **negative** * If the Δ*S_system*^ꝋ is **positive:**

Answer 28

* Both the first and second term will be **negative** * Resulting in a **negative** Δ*G*^ꝋ so the reaction is **feasible** * Therefore, regardless of the temperature, an exothermic reaction with a positive Δ*S_system*^ꝋ will **always be feasible**

Question 29

**Exothermic reactions** * In exothermic reactions, Δ*H_reaction*^ꝋ is **negative** * If the Δ*S_system*^ꝋ is **negative:**

Answer 29

* The first term is **negative** and the second term is **positive** * At high temperatures, the -*T*Δ*S_system*^ꝋ will be very **large** and **positive** and will overcome Δ*H_reaction*^ꝋ * Therefore, at **high temperatures** Δ*G*^ꝋ is **positive** and the reaction is not feasible * The reaction is more **feasible** at low temperatures, as the second term will not be large enough to overcome Δ*H_reaction*^ꝋ resulting in a negative Δ*G*^ꝋ

Question 30

This corresponds to Le Chatelier’s principle which states that for **exothermic reactions** an increase in temperature will cause the

Answer 30

equilibrium to shift position in favour of the reactants, i.e. in the endothermic direction * In other words, for exothermic reactions, the products will **not be formed** at high temperatures * The reaction is **not feasible** at high temperatures

Question 31

***The diagram shows under which conditions exothermic reactions are feasible***

Answer 31

Question 32

**Endothermic reactions** * In endothermic reactions, Δ*H_reaction*^ꝋ is **positive** * If the Δ*S_system*^ꝋ is **negative:**

Answer 32

* * Both the first and second term will be **positive** * Resulting in a **positive** Δ*G*^ꝋ so the reaction is **not feasible** * Therefore, regardless of the temperature, endothermic with a negative Δ*S_system*^ꝋ will **never be feasible**

Question 33

**Endothermic reactions** * In endothermic reactions, Δ*H_reaction*^ꝋ is **positive** * If the Δ*S_system*^ꝋ is **positive:**

Answer 33

* The first term is **positive** and the second term is **negative** * At low temperatures, the -*T*Δ*S_system*^ꝋ will be **small** and **negative** and will not overcome the larger Δ*H_reaction*^ꝋ * Therefore, at low temperatures Δ*G*^ꝋ is **positive** and the reaction is less feasible * The reaction is **more** **feasible** at **high temperatures** as the second term will become negative enough to overcome the Δ*H_reaction*^ꝋ resulting in a negative Δ*G*^ꝋ

Question 34

This again corresponds to Le Chatelier’s principle which states that for **endothermic reactions** an increase in temperature will cause the

Answer 34

equilibrium to shift position in favour of the products * In other words, for endothermic reactions, the products will **be formed** at high temperatures * The reaction is therefore **feasible**

Question 35

***The diagram shows under which conditions endothermic reactions are feasible***

Answer 35