Chapter 10 - Spontaneous Change

Question 1

Entropy (S), microscopic

Answer 1

S = klnΩ
S: entropy
k: Boltzmann constant, 1.38066e-23 J/K = R/Na
ln: natural log
Ω: number of microstates (omega)

Describes entropy on a microscopic level

Question 2

Entropy (S), macroscopic

Answer 2

∆S = qrev/T
qrev: q reversible (because q isn’t a state function)

Question 3

∆G < 0

Answer 3

Spontaneous

Question 4

∆G > 0

Answer 4

Non-spontaneous

Question 5

∆G = 0

Answer 5

Equilibrium

Question 6

What is ∆G equal to (no K)?

Answer 6

∆G = ∆H - T∆S
∆G: Gibbs free energy
∆H: enthalpy
∆S: change in entropy
Can add ° and it’s the same

Question 7

∆G for Q

Answer 7

∆G = ∆G° + RTlnQ
∆G°: ∆G of formation, ∆G standard
R: 8.3145 J/molK (might need to convert to kJ)
ln: natural log
Q: reaction quotient

Question 8

∆G° for K

Answer 8

∆G° = -RTlnK
R: 8.3145 J/molK
ln: natural log
K: equilibrium constant

Question 9

Calculate change in entropy of surroundings

Answer 9

∆Ssurr = -∆Hsys/T
∆Ssurr: change in entropy of surroundings
∆Hsys: change in enthalpy of system
T: temperature in K

Question 10

When ∆Suniv is > 0, is the reaction product-favored or reactant-favored?

Answer 10

Product-favored

Question 11

When ∆Suniv is < 0, is the reaction product-favored or reactant-favored?

Answer 11

Reactant-favored

Question 12

How is the enthalpy term affected when ∆H° is greater or less than 0?

Answer 12

∆H° < 0, enthalpy is favorable
∆H° > 0, enthalpy is unfavorable

Question 13

How is the entropy term affected by ∆S° is greater or less than 0?

Answer 13

∆S° < 0, entropy is unfavorable
∆S° > 0, entropy is favorable