Chapter 23 (Part 2) Entropy & Gibbs

Question 1

Entropy.

Answer 1

• Measure of disorder / randomness of a system OR
• The number of possible arrangements of the particles and their energy in a given system

Question 2

Units of entropy.

Answer 2

ΔS = J / k / mol

Question 3

Conditions for standard molar entropy, S°.

Answer 3

• Pressure: 1 atm / 1 x 10^5 Pa
• Temperature: 298 K
• Each substance involved in the reaction is in its normal physical state (solid, liquid or gas) at 1 atm & 298 K
• Units: S° = J / K / mol

Question 4

Standard molar entropy.

Answer 4

The entropy of one mole of substance in its standard states.

Question 5

Factors affecting entropy.

Answer 5

• Physical state
• Temperature
• Number of gaseous molecules

Question 6

What happens to entropy in a reaction when physical state changes from solid to liquid to gas?

Answer 6

• Gases have higher entropy than liquids & liquids higher than solids.
• For similar types of substances, harder substances have a lower entropy value (softer compounds = more disorder).

Question 7

Entropy changes in exothermic reactions.

Answer 7

Exothermic:
- Energy released to surroundings
- Increases number of ways of arranging energy (particles) in the system
- Increase in ΔS (entropy)
- Spontaneous

Question 8

Entropy changes in endothermic reactions.

Answer 8

Endothermic:
- Energy absorbed from surroundings
- Decreases number of ways of arranging energy in the system (because more particles in a smaller area)
- Decrease in ΔS (entropy)
- Not spontaneous

Question 9

Entropy when temperature increases?

Answer 9

• Temperature increases
• KE increases
• ΔS (entropy) increases

Question 10

Entropy when number of gaseous molecules increases?

Answer 10

• More molecules / atoms
• More ways of arranging
• Higher ΔS (entropy)

Question 11

Entropy changes in reactions - ΔSsystem.
Why does entropy change?

Answer 11

• Magnitude of entropy changes in a reaction
• Because entropy of reactants are different from the product
• Entropy can increase or decrease

Question 12

CaCO3 (s) –> CaO (s) + CO2 (g)
Entropy change? Explain.

Answer 12

• Increase in entropy
• Because gas is produced
• Reactant CaCO3 is a solid (low entropy)

Question 13

2N2O5 (g) –> 4NO2 (g) + O2(g)
Entropy change? Explain.

Answer 13

• Increase in ΔSsystem (entropy of the system)
• Greater number of molecules in products (5 moles) than reactants (2 moles)

Question 14

Formula for calculating ΔSsystem.

Answer 14

• ΔSsystem = ΔS°products - ΔS°reactants
• ΔS is scalar: +ve = increase, -ve = decrease

Question 15

Purpose of Gibbs free energy change

Answer 15

To determine whether a reaction is likely to be spontaneous

Question 16

What is standard molar Gibbs free energy of formation, ΔG°f ?

Answer 16

The free energy change that accompanies the formation of one mole of compound from its elements in their standard state.

Question 17

Formula for ΔG.

Answer 17

• ΔG = -TΔStotal
• ΔG = ΔHr - TΔSsytem
• For ΔS, divide by 1000.
• T in Kelvins (K)

Question 18

Units for ΔG°f.

Answer 18

Unit = kJ / mol

Question 19

When is a reaction spontaneous?

Answer 19

When ΔG < 0

Question 20

Formula which includes ΔG° & ΔE°cell.

Answer 20

Since both predict feasibility of a reaction, there is a relation between this 2:
ΔG° = -nΔE°cellf

• ΔG°: Gibbs free energy in J / mol
• n = no. of electrons transferred in the cell reaction
• F = Faraday constant
• ΔE°cell = standard cell potential

Question 21

Predict spontaneity using sign of ΔG:
ΔH = -ve
ΔS = +ve

Answer 21

• Spontaneous at any temperature because
• ΔG = ΔH - TΔSsystem
  = (-ve) - T(+ve) < 0

Question 22

Predict spontaneity using sign of ΔG:
ΔH = -ve
ΔS = -ve

Answer 22

• Spontaneous at low temperature because
• ΔG = ΔH - TΔSsystem
  = (-ve) - T(-ve)
• For ΔG to be < 0, T has to be low

Question 23

Predict spontaneity using sign of ΔG:
ΔH = +ve
ΔS = +ve

Answer 23

• Spontaneous at high temperature because
• ΔG = ΔH - TΔSsystem
  = (+ve) - T(+ve)
• For ΔG to be < 0, T has to be high

Question 24

Predict spontaneity using sign of ΔG:
ΔH = +ve
ΔS = -ve

Answer 24

• Not spontaneous / feasible at any temperature
• ΔG = ΔH - TΔSsystem
  = (+ve) - T(-ve)
• ΔG is always positive