Entropy is a measure of the number of possible arrangements of particles and their energy in a system — a measure of how disordered or chaotic the system is. The greater the disorder, the higher the entropy.

Give two examples of physical or chemical changes that lead to an increase in entropy.

Any two from: Melting a solid to form a liquid (e.g. ice → water) Dissolving a solid in water (e.g. NH 4 NO 3 in water) Producing a gas in a reaction (e.g. CaCO 3 → CaO + CO 2 )

True or False? An increase in entropy makes a reaction more thermodynamically favourable.

True. A positive entropy change (ΔS > 0) contributes to a more negative value of ΔG (ΔG = ΔH − TΔS), making a reaction more thermodynamically favourable.

What symbol and units are used for entropy ?

Entropy is given the symbol S , with standard entropy change denoted Δ S ꝋ . Units: J K -1 mol -1

When a solid melts to form a liquid, the particles change from fixed positions to a more random arrangement, so entropy .......... and the system becomes .......... stable.

When a solid melts to form a liquid, the particles change from fixed positions to a more random arrangement, so entropy increases and the system becomes more stable.

Why do endothermic reactions occur if products end up in a higher energy state?

Endothermic reactions occur because enthalpy change (Δ H ꝋ ) alone does not determine whether a reaction is feasible . The increase in entropy (disorder) of the system provides the driving force — a more disordered state is energetically more stable.

True or False? Producing a gas in a reaction always decreases entropy.

False. Producing a gas increases entropy. Gas molecules are much more disordered than solids or liquids, so the system becomes more chaotic and entropy rises.

In the decomposition CaCO 3 (s) → CaO (s) + CO 2 (g), entropy .......... because a .......... is produced, increasing disorder.

In the decomposition CaCO 3 (s) → CaO (s) + CO 2 (g), entropy increases because a gas is produced, increasing disorder.

Why is it difficult to predict whether dissolving an ionic solid is endothermic or exothermic?

Two competing processes occur simultaneously: Bond breaking between ions increases disorder and absorbs energy Bond forming between ions and solvent molecules decreases disorder and releases energy The overall enthalpy change depends on the relative magnitudes of these two processes.

What is the total entropy change of a reaction?

The total entropy change is the sum of the entropy change of the system and the entropy change of the surroundings: Δ S ꝋ total = Δ S ꝋ sys + Δ S ꝋ surr

What are the units for standard entropy change, Δ S ꝋ ?

The units for standard entropy change, Δ S ꝋ , are J K -1 mol -1 . Note: entropy changes are measured in joules (not kilojoules), as they are an order of magnitude smaller than enthalpy changes.

What equation is used to calculate the entropy change of the system ?

The entropy change of the system is calculated using: Δ S ꝋ system = ΣΔ S ꝋ products − ΣΔ S ꝋ reactants where Σ means "sum of".

True or False? The standard entropy of an element in its standard state is zero.

False. Unlike standard enthalpy of formation, the standard entropy of elements is not zero. Entropy values for elements can be found in data books.

For the reaction 2Mg (s) + O 2 (g) → 2MgO (s), Δ S ꝋ system = ΣΔ S ꝋ .......... − ΣΔ S ꝋ ..........

For the reaction 2Mg (s) + O 2 (g) → 2MgO (s), Δ S ꝋ system = ΣΔ S ꝋ products − ΣΔ S ꝋ reactants

What equation gives the entropy change of the surroundings ?

The entropy change of the surroundings is calculated using: Δ S ꝋ surr = −Δ H / T where Δ H is the enthalpy change (in J mol -1 ) and T is the absolute temperature in kelvin (K).

True or False? A given energy transfer to surroundings at a low temperature produces a greater entropy change than the same transfer at a high temperature.

True. Δ S ꝋ surr = −Δ H / T , so at lower T the same energy input produces a larger Δ S ꝋ surr . The surroundings are more sensitive to energy input when they are cold.

Why can a reaction with a negative Δ S ꝋ sys still be spontaneous?

A reaction can still be spontaneous if the entropy change of the surroundings (Δ S ꝋ surr ) is sufficiently positive to make the total entropy change positive: Δ S ꝋ total = Δ S ꝋ sys + Δ S ꝋ surr > 0 For example, an exothermic reaction releases energy to the surroundings, greatly increasing their disorder.

To use the equation Δ S ꝋ surr = −Δ H / T , the enthalpy change must be converted from kJ mol -1 to .......... mol -1 by multiplying by .......... .

To use the equation Δ S ꝋ surr = −Δ H / T , the enthalpy change must be converted from kJ mol -1 to J mol -1 by multiplying by 1000 .

What is Gibbs free energy ?

Gibbs free energy ( G ) is a thermodynamic quantity that combines enthalpy and entropy to determine whether a reaction is feasible (spontaneous). A reaction is feasible when Δ G ꝋ ≤ 0.

True or False? A reaction is feasible when Δ G ꝋ is greater than zero.

False. A reaction is feasible (spontaneous) when Δ G ꝋ is equal to or less than zero (Δ G ꝋ ≤ 0). A positive Δ G ꝋ means the reaction is not spontaneous.

Before substituting Δ S ꝋ (in J K -1 mol -1 ) into the Gibbs equation, you must divide by .......... to convert it to .......... K -1 mol -1 .

Before substituting Δ S ꝋ (in J K -1 mol -1 ) into the Gibbs equation, you must divide by 1000 to convert it to kJ K -1 mol -1 .

What two factors determine whether a chemical reaction is feasible?

The feasibility of a reaction is determined by: The enthalpy change (Δ H ꝋ ) The entropy change (Δ S ꝋ ) These are combined in the Gibbs equation: Δ G ꝋ = Δ H ꝋ − T Δ S ꝋ

State the Gibbs equation .

The Gibbs equation is: Δ G ꝋ = Δ H ꝋ − T Δ S ꝋ where Δ G ꝋ is in kJ mol -1 , Δ H ꝋ is in kJ mol -1 , T is in K, and Δ S ꝋ is in J K -1 mol -1 (converted to kJ K -1 mol -1 by dividing by 1000 before substituting).

How can the Gibbs equation be rearranged to find the temperature at which a reaction becomes feasible?

At the crossover point Δ G ꝋ = 0, so: 0 = Δ H ꝋ − T Δ S ꝋ T = Δ H ꝋ ÷ Δ S ꝋ (Δ S ꝋ must be in kJ K -1 mol -1 to match units with Δ H ꝋ )

True or False? A reaction with positive Δ H ꝋ and positive Δ S ꝋ can become feasible at high temperatures.

True. In the Gibbs equation Δ G ꝋ = Δ H ꝋ − T Δ S ꝋ , raising T increases the T Δ S ꝋ term. At sufficiently high T , this outweighs the positive Δ H ꝋ , making Δ G ꝋ negative and the reaction feasible .

For a reaction with Δ H ꝋ = +135 kJ mol -1 and Δ S ꝋ = +344 J K -1 mol -1 , calculate Δ G ꝋ at 298 K.

Convert Δ S ꝋ : 344 ÷ 1000 = 0.344 kJ K -1 mol -1 Δ G ꝋ = Δ H ꝋ − T Δ S ꝋ = +135 − (298 × 0.344) = +32.5 kJ mol -1 Δ G ꝋ is positive, so the reaction is not feasible at 298 K.

When Δ H ꝋ is negative and Δ S ꝋ is positive, Δ G ꝋ is always .......... and the reaction is feasible at .......... temperatures.

When Δ H ꝋ is negative and Δ S ꝋ is positive, Δ G ꝋ is always negative and the reaction is feasible at all temperatures.

What is meant by a feasible reaction ?

Feasibility describes whether a reaction is thermodynamically spontaneous. A reaction is feasible when Δ G ꝋ ≤ 0. A feasible reaction will occur spontaneously, though it may be very slow if the activation energy is large.

On a graph of Δ G ꝋ against temperature, what do the gradient and y-intercept represent?

For the straight-line form Δ G ꝋ = −Δ S ꝋ × T + Δ H ꝋ : Gradient = −Δ S ꝋ y-intercept = Δ H ꝋ x-intercept = temperature at which the reaction ceases to be feasible (Δ G ꝋ = 0)

True or False? A feasible reaction always proceeds at an observable rate.

False. Feasibility (Δ G ꝋ ≤ 0) only indicates thermodynamic spontaneity. A reaction may have a large activation energy ( E a ), making it kinetically very slow even though it is thermodynamically feasible.

State the van’t Hoff equation linking Δ G ꝋ to the equilibrium constant.

The van’t Hoff equation is: Δ G ꝋ = − RT ln K where R = 8.31 J K -1 mol -1 , T is temperature in kelvin (K), and K is the equilibrium constant.

A negative Δ G ꝋ means the forward reaction is .......... and the equilibrium constant K is .......... than 1.

A negative Δ G ꝋ means the forward reaction is favoured and the equilibrium constant K is greater than 1.

Why does the Haber process operate above the temperature at which Δ G ꝋ = 0 for ammonia synthesis?

Above the crossover temperature, Δ G ꝋ for ammonia synthesis becomes positive (less feasible), but the Haber process uses this higher temperature to ensure an acceptable rate of reaction . Additionally, use of a catalyst lowers E a , and continuous removal of ammonia shifts the equilibrium towards products.

True or False? Δ G ꝋ can be used to predict feasibility under non-standard conditions.

False. Δ G ꝋ applies only under standard conditions . Under non-standard conditions, Δ G must be calculated separately using actual concentrations, pressures, and temperature.

The Gibbs equation can be written as Δ G ꝋ = .......... × T + Δ H ꝋ , which has the form y = mx + c.

The Gibbs equation can be written as Δ G ꝋ = −Δ S ꝋ × T + Δ H ꝋ , which has the form y = mx + c.

Give an example of a reaction that is thermodynamically feasible but kinetically very slow, and explain why.

The decomposition of hydrogen peroxide at 298 K: H 2 O 2 (l) → H 2 O (l) + ½O 2 (g) Δ G = −117 kJ mol -1 Despite a negative Δ G , the reaction has a very large activation energy and proceeds too slowly without a catalyst such as MnO 2 .

A LevelChemistryEdexcelFlashcards5. Advanced Physical ChemistryEnergetics II: Entropy

Energetics II: Entropy (Edexcel A Level Chemistry): Flashcards

Exam code: 9CHO

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FrontIntroducing Entropy
What is entropy?
FrontIntroducing Entropy
Give two examples of physical or chemical changes that lead to an increase in entropy.
BackIntroducing Entropy
Any two from:
Melting a solid to form a liquid (e.g. ice → water)
Dissolving a solid in water (e.g. NH₄NO₃ in water)
Producing a gas in a reaction (e.g. CaCO₃ → CaO + CO₂)

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Cards in this collection (36)

What is entropy?
Entropy is a measure of the number of possible arrangements of particles and their energy in a system — a measure of how disordered or chaotic the system is.
The greater the disorder, the higher the entropy.
Give two examples of physical or chemical changes that lead to an increase in entropy.
Any two from:
1. Melting a solid to form a liquid (e.g. ice → water)
2. Dissolving a solid in water (e.g. NH₄NO₃ in water)
3. Producing a gas in a reaction (e.g. CaCO₃ → CaO + CO₂)
True or False?
An increase in entropy makes a reaction more thermodynamically favourable.
True.
A positive entropy change (ΔS > 0) contributes to a more negative value of ΔG (ΔG = ΔH − TΔS), making a reaction more thermodynamically favourable.
What symbol and units are used for entropy?
Entropy is given the symbol S, with standard entropy change denoted ΔS^ꝋ.
Units: J K^-1 mol^-1
When a solid melts to form a liquid, the particles change from fixed positions to a more random arrangement, so entropy .......... and the system becomes .......... stable.
When a solid melts to form a liquid, the particles change from fixed positions to a more random arrangement, so entropy increases and the system becomes more stable.
Why do endothermic reactions occur if products end up in a higher energy state?
Endothermic reactions occur because enthalpy change (ΔH^ꝋ) alone does not determine whether a reaction is feasible.
The increase in entropy (disorder) of the system provides the driving force — a more disordered state is energetically more stable.
True or False?
Producing a gas in a reaction always decreases entropy.
False.
Producing a gas increases entropy. Gas molecules are much more disordered than solids or liquids, so the system becomes more chaotic and entropy rises.
In the decomposition CaCO₃ (s) → CaO (s) + CO₂ (g), entropy .......... because a .......... is produced, increasing disorder.
In the decomposition CaCO₃ (s) → CaO (s) + CO₂ (g), entropy increases because a gas is produced, increasing disorder.
Why is it difficult to predict whether dissolving an ionic solid is endothermic or exothermic?
Two competing processes occur simultaneously:
- Bond breaking between ions increases disorder and absorbs energy
- Bond forming between ions and solvent molecules decreases disorder and releases energy
The overall enthalpy change depends on the relative magnitudes of these two processes.
What is the total entropy change of a reaction?
The total entropy change is the sum of the entropy change of the system and the entropy change of the surroundings:
ΔS^ꝋ _total = ΔS^ꝋ _sys + ΔS^ꝋ _surr
What are the units for standard entropy change, ΔS^ꝋ?
The units for standard entropy change, ΔS^ꝋ, are J K^-1 mol^-1.
Note: entropy changes are measured in joules (not kilojoules), as they are an order of magnitude smaller than enthalpy changes.
What equation is used to calculate the entropy change of the system?
The entropy change of the system is calculated using:
ΔS^ꝋ _system = ΣΔS^ꝋ _products − ΣΔS^ꝋ _reactants
where Σ means "sum of".
True or False?
The standard entropy of an element in its standard state is zero.
False.
Unlike standard enthalpy of formation, the standard entropy of elements is not zero. Entropy values for elements can be found in data books.
For the reaction 2Mg (s) + O₂ (g) → 2MgO (s), ΔS^ꝋ _system = ΣΔS^ꝋ .......... − ΣΔS^ꝋ ..........
For the reaction 2Mg (s) + O₂ (g) → 2MgO (s), ΔS^ꝋ _system = ΣΔS^ꝋ products − ΣΔS^ꝋ reactants
What equation gives the entropy change of the surroundings?
The entropy change of the surroundings is calculated using:
ΔS^ꝋ _surr = −ΔH / T
where ΔH is the enthalpy change (in J mol^-1) and T is the absolute temperature in kelvin (K).
True or False?
A given energy transfer to surroundings at a low temperature produces a greater entropy change than the same transfer at a high temperature.
True.
ΔS^ꝋ _surr = −ΔH / T, so at lower T the same energy input produces a larger ΔS^ꝋ _surr. The surroundings are more sensitive to energy input when they are cold.
Why can a reaction with a negative ΔS^ꝋ _sys still be spontaneous?
A reaction can still be spontaneous if the entropy change of the surroundings (ΔS^ꝋ _surr) is sufficiently positive to make the total entropy change positive:
ΔS^ꝋ _total = ΔS^ꝋ _sys + ΔS^ꝋ _surr > 0
For example, an exothermic reaction releases energy to the surroundings, greatly increasing their disorder.
To use the equation ΔS^ꝋ _surr = −ΔH / T, the enthalpy change must be converted from kJ mol^-1 to .......... mol^-1 by multiplying by .......... .
To use the equation ΔS^ꝋ _surr = −ΔH / T, the enthalpy change must be converted from kJ mol^-1 to J mol^-1 by multiplying by 1000.
What is Gibbs free energy?
Gibbs free energy (G) is a thermodynamic quantity that combines enthalpy and entropy to determine whether a reaction is feasible (spontaneous).
A reaction is feasible when ΔG^ꝋ ≤ 0.
True or False?
A reaction is feasible when ΔG^ꝋ is greater than zero.
False.
A reaction is feasible (spontaneous) when ΔG^ꝋ is equal to or less than zero (ΔG^ꝋ ≤ 0). A positive ΔG^ꝋ means the reaction is not spontaneous.
Before substituting ΔS^ꝋ (in J K^-1 mol^-1) into the Gibbs equation, you must divide by .......... to convert it to .......... K^-1 mol^-1.
Before substituting ΔS^ꝋ (in J K^-1 mol^-1) into the Gibbs equation, you must divide by 1000 to convert it to kJ K^-1 mol^-1.
What two factors determine whether a chemical reaction is feasible?
The feasibility of a reaction is determined by:
1. The enthalpy change (ΔH^ꝋ)
2. The entropy change (ΔS^ꝋ)
These are combined in the Gibbs equation: ΔG^ꝋ = ΔH^ꝋ − TΔS^ꝋ
State the Gibbs equation.
The Gibbs equation is:
ΔG^ꝋ = ΔH^ꝋ − TΔS^ꝋ
where ΔG^ꝋ is in kJ mol^-1, ΔH^ꝋ is in kJ mol^-1, T is in K, and ΔS^ꝋ is in J K^-1 mol^-1 (converted to kJ K^-1 mol^-1 by dividing by 1000 before substituting).
How can the Gibbs equation be rearranged to find the temperature at which a reaction becomes feasible?
At the crossover point ΔG^ꝋ = 0, so:
0 = ΔH^ꝋ − TΔS^ꝋ
T = ΔH^ꝋ ÷ ΔS^ꝋ
(ΔS^ꝋ must be in kJ K^-1 mol^-1 to match units with ΔH^ꝋ)
True or False?
A reaction with positive ΔH^ꝋ and positive ΔS^ꝋ can become feasible at high temperatures.
True.
In the Gibbs equation ΔG^ꝋ = ΔH^ꝋ − TΔS^ꝋ, raising T increases the TΔS^ꝋ term. At sufficiently high T, this outweighs the positive ΔH^ꝋ, making ΔG^ꝋ negative and the reaction feasible.
For a reaction with ΔH^ꝋ = +135 kJ mol^-1 and ΔS^ꝋ = +344 J K^-1 mol^-1, calculate ΔG^ꝋ at 298 K.
Convert ΔS^ꝋ: 344 ÷ 1000 = 0.344 kJ K^-1 mol^-1
ΔG^ꝋ = ΔH^ꝋ − TΔS^ꝋ
= +135 − (298 × 0.344)
= +32.5 kJ mol^-1
ΔG^ꝋ is positive, so the reaction is not feasible at 298 K.
When ΔH^ꝋ is negative and ΔS^ꝋ is positive, ΔG^ꝋ is always .......... and the reaction is feasible at .......... temperatures.
When ΔH^ꝋ is negative and ΔS^ꝋ is positive, ΔG^ꝋ is always negative and the reaction is feasible at all temperatures.
What is meant by a feasible reaction?
Feasibility describes whether a reaction is thermodynamically spontaneous.
A reaction is feasible when ΔG^ꝋ ≤ 0. A feasible reaction will occur spontaneously, though it may be very slow if the activation energy is large.
On a graph of ΔG^ꝋ against temperature, what do the gradient and y-intercept represent?
For the straight-line form ΔG^ꝋ = −ΔS^ꝋ × T + ΔH^ꝋ:
- Gradient = −ΔS^ꝋ
- y-intercept = ΔH^ꝋ
- x-intercept = temperature at which the reaction ceases to be feasible (ΔG^ꝋ = 0)
True or False?
A feasible reaction always proceeds at an observable rate.
False.
Feasibility (ΔG^ꝋ ≤ 0) only indicates thermodynamic spontaneity. A reaction may have a large activation energy (E_a), making it kinetically very slow even though it is thermodynamically feasible.
State the van’t Hoff equation linking ΔG^ꝋ to the equilibrium constant.
The van’t Hoff equation is:
ΔG^ꝋ = −RT ln K
where R = 8.31 J K^-1 mol^-1, T is temperature in kelvin (K), and K is the equilibrium constant.
A negative ΔG^ꝋ means the forward reaction is .......... and the equilibrium constant K is .......... than 1.
A negative ΔG^ꝋ means the forward reaction is favoured and the equilibrium constant K is greater than 1.
Why does the Haber process operate above the temperature at which ΔG^ꝋ = 0 for ammonia synthesis?
Above the crossover temperature, ΔG^ꝋ for ammonia synthesis becomes positive (less feasible), but the Haber process uses this higher temperature to ensure an acceptable rate of reaction.
Additionally, use of a catalyst lowers E_a, and continuous removal of ammonia shifts the equilibrium towards products.
True or False?
ΔG^ꝋ can be used to predict feasibility under non-standard conditions.
False.
ΔG^ꝋ applies only under standard conditions. Under non-standard conditions, ΔG must be calculated separately using actual concentrations, pressures, and temperature.
The Gibbs equation can be written as ΔG^ꝋ = .......... × T + ΔH^ꝋ, which has the form y = mx + c.
The Gibbs equation can be written as ΔG^ꝋ = −ΔS^ꝋ × T + ΔH^ꝋ, which has the form y = mx + c.
Give an example of a reaction that is thermodynamically feasible but kinetically very slow, and explain why.
The decomposition of hydrogen peroxide at 298 K:
H₂O₂ (l) → H₂O (l) + ½O₂ (g) ΔG = −117 kJ mol^-1
Despite a negative ΔG, the reaction has a very large activation energy and proceeds too slowly without a catalyst such as MnO₂.

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