O defg: Entropy

Question 1

Define entropy.

Answer 1

A measure of the number of ways of arranging molecules and their associated energy quanta.

Question 2

When may endothermic reactions be spontaneous?

Answer 2

When entropy increases during the reaction.

• Endothermic interactions happen spontaneously since energy isn’t evenly distributed
• All reactions are somewhat reversible
• But if products are messier and more spread apart than reactants, they won’t react - so it’s one-way

Question 3

How are the following represented symbolically?

• General entropy change
• The entropy change for a given reaction

Answer 3

• ΔS
• ΔS_sys

Question 4

Entropy values are quoted at standard conditions, these being…

Answer 4

298 K, 100 kPa

Question 5

What is the unit of entropy?

Answer 5

J K^-1 mol^-1

Not kJ

Question 6

What is the sign of ΔS_sys when:

• products are more entropic than reactants?
• products are less entropic than reactants?

Answer 6

• Entropy increases: positive
• Entropy decreases: negative

Question 7

What factors increase the entropy of a collection of molecules?

Answer 7

• Greater volume (molecules more spread out)
• Greater total energy
• More molecules sharing the energy
• More complex molecules (more atoms = more energy levels)

Question 8

What state of matter has the lowest entropy?

Answer 8

Solid

Question 9

For each of the following changes, state whether the entropy of the system would increase, decrease or remain the same.

• Oil mixes with petrol
• A suspension of oil in water separates into 2 layers
• Car exhaust gases are adsorbed onto the surface of a catalyst in a catalytic converter

Answer 9

• Increase
• Decrease
• Decrease

Question 10

State and explain which substance in the following pairs would have the higher standard molar entropy.

• Separate samples of copper and zinc / sample of brass
• Liquid pentane / liquid octane

Answer 10

• Brass; mixtures have higher entropies than the pure substances
• Octane; more complex molecules have higher entropies

Question 11

Answer 11

C

• Hexane and pentane are liquids, so have lower entropy
• But pentane is less complex than hexane, so has lower entropy
• Propane and butane are gases, so have higher entropy
• But propane is less complex than butane, so has lower entropy

Question 12

Answer 12

• Entropies increase for the first four alkanes as the molecules contain increasing numbers of atoms. The number of energy levels increases so there are more ways of distributing energy
• Pentane is a liquid and butane a gas so butane has the higher entropy

Question 13

Write the equation used to calculate the total entropy change of a process.

Answer 13

ΔS_tot = ΔS_sys + ΔS_surr

ΔS_tot = ΔS_sys - ΔH/T

Total entropy change = entropy change of system + entropy change of surroundings

Question 14

In mathematical terms, what is required for an endothermic reaction to be feasible?

Answer 14

ΔS_tot is positive

ΔS_sys + ΔS_surr > 0

Question 15

How do you work out ΔS_sys?

Answer 15

S_products - S_reactants

Question 16

How do you work out ΔS_surr?

Answer 16

ΔS_surr = -ΔH / T

ΔH is the enthalpy change of the system. Therefore, the enthalpy change of the surroundings is -ΔH.

Question 17

Prove that water will not freeze at 10^oC.

ΔS_sys = -22.0 J K^-1 mol^-1

ΔH = -6.010 kJ mol^-1

Answer 17

ΔS_tot = ΔS_sys + ΔS_surr

ΔH = -6010 J

T = 273 + 10 = 283 K

ΔS_surr = -ΔH/T = - (-6010) / 283 = +21.2 J K^-1 mol^-1

ΔS_tot = -22.0 + 21.2 = -0.80 J K^-1 mol^-1

ΔS_tot < 0 so entropy decrease; the water will not freeze

Question 18

Will calcium carbonate decompose to produce carbon dioxide at 1000 ^oC?

ΔS_sys = +159 J K^-1 mol^-1

ΔH = +179 kJ mol^-1

Answer 18

ΔS_tot = ΔS_sys + ΔS_surr

ΔH = +179,000 J

T = 273 + 1000 = 1273 K

ΔS_surr = -ΔH/T = -(+179,000) / 1273 = -140.6 J K^-1 mol^-1

ΔS_tot = 159 - 140.6 = +18 J K^-1 mol^-1

ΔS_tot > 0 so entropy increases; it decomposes

Question 19

What is the requirement of entropy at equilibrium?

Answer 19

ΔS_tot = 0

Should be equally easy to go forwards + backwards

Question 20

What must the temperature be for water and ice to exist together in equilibrium? Give your answer in ^oC.

ΔS_sys = +22.0 J K^-1 mol^-1

ΔH = +6.01 kJ mol^-1

Answer 20

ΔS_tot = 0 (equilibrium)

ΔS_sys + ΔH_surr = 0

ΔS_sys - ΔH/T = 0

ΔH = 6010 J

22.0 - 6010/T = 0

T = 6010/22.0 = 273 K = 0 ^oC

Question 21

Why may a reaction be feasible but not spontaneous?

Answer 21

• Sponteneity depends on enthalpy as well as entropy
• Kinetic factors affect rate: a reaction may not be spontaneous if E_a is large (i.e. so slow it is said not to occur)

Question 22

Fill in the table by stating:

• Whether a reaction in each scanario would be feasible
• The conditions required for each reaction to occur spontaneously

Answer 22

Question 23

Answer 23

Question 24

Answer 24

1) Entropy is a measure of the number of ways of arranging molecules and their associated energy quanta.
2) ΔS_sys is often positive because ions move from a regular lattice into a solution with more ways of arranging molecules and their associated energy quanta.

Ca²⁺ has high charge density, so ion-dipole interactions are stronger (solvation shells more regular) than bonding in ionic lattice. Lattice has higher entropy than solution, so ΔS_sys is negative.

Question 25

2NO(g) + O₂(g) → 2NO₂(g) Δ_rH = -114 kJ mol^-1

Give the sign of ΔS_sys for this reaction, with a reason. Use this to explain whether the reaction becomes more or less feasible at higher temperatures. (3)

Answer 25

• ΔS_sys must be positive for reaction to be feasible
• ΔS_tot = ΔS_sys -ΔH/T. ΔH is negative so -ΔH/T is positive. As T increases, -ΔH/T decreases, so ΔH_tot decreases. Becomes less feasible at higher temperatures

Question 26

An experiment was carried out to determine the standard enthalpy change of solution for a compound which readily dissolves at room temperature. During the dissolving process a decrease in temperature was recorded.

What can be assumed from these results?

1. The compound dissolves more readily at lower temperatures.
2. The dissolving process has a positive Δ_surrS.
3. The dissolving process has a positive Δ_sysS.
4. The process has a high activation energy.

Answer 26

3

• 1 incorrect since it is endothermic, so should say “higher temperatures”*
• 2 incorrect since endothermic → Δ is H positive → Δ_surrS = -ΔH/T is negative*
• 3 correct since endothermic, so entropy must increase for it to be feasible*
• 4 incorrect since readily dissolves at room temperature*