Entropy and Free Energy

The Effect of Temperature on Spontaneity

• Endothermic rxn: heat flow from surrounding to system

• Exothermic rxn: heat flow from system to surrounding

• Heat flow to and from the system affects the entropy of the surrounding

(NOTE: SPONTANEITY deals with whether a reaction is spontaneous or non-spontaneous)

• Heat flow will affect the entropy of the surrounding DEPENDING ON THE TEMPERATURE at which the process happens

• The significance of heat flow to and from the surrounding will be GREATER at lower temperature, i.e., the entropy of the surrounding will change significantly if the temperature (at which the reaction happens) is lower.

Characteristics of Entropy, S

1. The sign of ΔSsurr depends on the direction of the heat flow

If the temperature is constant,

a. An exothermic reaction will release heat to the surrounding which will increase molecular disorder of the surrounding thus increasing the entropy of the surrounding, giving a positive ΔSsurr

b. An endothermic reaction will absorb heat from the surrounding which will decrease the molecular disorder of the surrounding thus decreasing the entropy of the surrounding, giving a negative ΔSsurr

Δssystem ΔSsurrounding Δsuniverse = Δssystem +ΔSsurrounding

Is the process spontaneous?

+ + + YES

- - - NO (but if you reverse process it will be

spontaneous)

+ - ? YES but only if the Δssystem >> Δssurrounding

- + ? YES but only if Δssurrounding >> ΔSsystem

Entropy and Spontaneity

ΔSuniverse has to be POSITIVE for a process to be spontaneous

2. Magnitude of ΔSsurrounding depends on temperature

Δssurrounding is directly proportional to the heat flow but inversely proportional to the temperature

Δssurrounding can be expressed in terms of the enthalpy of the system, ΔHsystem. If we consider a process occurring at CONSTANT PRESSURE,

ΔHsystem = qp (Note:recall calorimetry)

Ssurrounding

=heat flow

temperature

If we consider a condition at constant pressure and constant temperature (in Kelvin),

(NOTE: The minus sign is needed because we are looking at the entropy change of the surrounding but we are using the enthalpy change of the system)

For an EXOTHERMIC REACTION, ΔHsystem is negative so the ΔSsurrounding is positive, i.e., heat flows from system to surrounding (negative ΔHsystem ) which will increase the entropy of the surrounding (positive Δssurrounding)

Ssurrounding

= -

system

T

Ssurrounding

= --

system)

T

For an ENDOTHERMIC REACTION, ΔHsystem is positive so the ΔSsurrounding is negative, i.e., heat flows from surrounding to system (positive ΔHsystem ) which will decrease the entropy of the surrounding (negative Δssurrounding)

Ssurrounding

= -

system)

T

e.g. In metallurgy of antimony, the pure metal is recoveredvia different reactions, depending on the composition ofthe ore. For example, iron is used to reduce the antimony insulfide ores:

(a) Sb2S3(s) + 3Fe(s) ---> 2Sb(s) + 3FeS(s) ΔH= -125 kJ

Carbon is used as the reducing agent for the oxide ores:

(b) Sb2O6(s) + 6C(s) ---> 4Sb(s) + 6CO(g) ΔH= 778 kJ

Calculate Δssurrounding for each reactions at 25oC at 1 atm.

Solution:

1. Use

2. Convert 25oC to K, T= 25 + 237 = 298 K

3. For reaction (a), ΔSsurrounding = 419 J/K, i.e., the reaction is exothermic so heat flows to the surrounding which increases the entropy of the surrounding.

4. For reaction (b), ΔSsurrounding = -2.61 x 10 3 J/K, i.e., the reaction is endothermic so heat flows to the system which decreases the entropy of the surrounding.

Ssurrounding

= -

system)

T

Calculating for entropy

The entropy of a perfect crystal at 0 K is zero. This is an ideal value which is taken as a standard but never actually observed.

The entropy (disorder)of a non-ideal substance increases with temperature because of the increase in the random vibrational motion.

Since the S is zero for a perfect crystal at 0K, the standard entropy value, So, for substance at a particular temperature can be calculated by knowing its temperature dependence.

The standard entropy value So represents the increase in entropy when a substance is heated from 0 K to 298 K at 1 atm.

Entropy is a state function of the system like enthalpy.

The entropy change (So) for a given chemical reaction can be calculated from the standard entropy values (So) of the products and reactants

Soreaction = S nproduct So

product - S nreactant Soreactant

Calculate the So for the reduction of Aluminum oxide by hydrogen gas: Al2O3(s) + 3H2(g) ---> 2Al(s) + 3H2O(g)

Consider the following standard entropy values:

Solution:

1. Use Soreaction = S nproduct So

product - S nreactant Soreactant

2. Soreaction = [2 mol Al (28 J/K-mol)+3 mol H2O (189

J/K-mol)] – [1 mol Al2O3 (51 J/K-mol)+ 3 mol H2 (131 J/K-mol)] = 179 J/K

Substance So (J/K-mol)

Al2O3(s) 51

H2(g) 131

Substance So (J/K-mol)

Al(s) 28

H2O(g) 189

Summary• Entropy, S, is a measure of randomness. • The second law of thermodynamics states that in any

spontaneous process, there is ALWAYS an increase in the entropy of the universe.

• Suniverse = Ssystem+Ssurrounding

• Suniverse is positive, there is an increase in the entropy of the universe, and the process is spontaneous. Suniverse is negative, there’s a decrease in the entropy of the universe, the REVERSE process is spontaneous.

• standard entropy value, So, for a substance can be known• The change in the entropy of a chemical reaction can be

calculated from So of products and reactants in their standard statesSo

reaction = S nproduct Soproduct - S nreactant So

reactant

Free Energy, GFree energy, G, is another thermodynamic function

that is related to spontaneity and the temperature dependence of spontaneity. It is defined as :

G = H – TS (Note: H is enthalpy, T is the Kelvin temperature & S is the entropy)

At constant temperature T, the change in the free energy of the system is

ΔGsystem = ΔHsystem -T ΔSsystem

We can manipulate this equation to see how it relates to spontaneity.

If we divide the equation by the temperature T,

then multiply both sides by -1

Gsystem

= H

system - T S

system

T T

Gsystem

T

=H

system

T

- Ssystem

Gsystem

T

=H

system

T

+ Ssystem- -

Gsystem

T

= Ssurrounding+ S

system = S

universe-

The equation above indicates that a process carried out at constant temperature & pressure will be SPONTANEOUSonly if ΔG is NEGATIVE, i.e., a process at constant T & P is spontaneous in the direction which decreases the free energy.

For a chemical reaction, spontaneity will depend on the temperature and which state function will control the reaction based on the relationship given by

ΔGsystem = ΔHsystem -T ΔSsystem

Gsystem

T

= Suniverse

at constant T & P-

∆G° = ∆H° – T∆S°

Because 0 ≤ H ≥ 0 and 0 ≤ S ≥ 0, there are four possibilities for G.

H° S° G° Reaction spontaneity

– + – Spontaneous at all T’s.

– – Temp dependent Spont at low Temp.

+ + Temp dependent Spont at high Temp.

+ – + NON Spont at all T’s.

Free energy and chemical reactionsStandard free energy change,Go, is the change in

the free energy if the reactants in their standard states are converted to the products in their standard states.

e.g. N2 (g) + 3H2(g) 2NH3(g) Go = -33.3 kJ

Change in the free energy when 1 mole of nitrogen gas at 1 atm reacts with 3 moles of hydrogen gas at 1 atm to produce 2 moles gaseuous ammonia at 1 atm.

Standard free energy change,Go is not measured like enthalpy or entropy. It is calculated from other measured quantities. We can use the relationship ΔGo = ΔHo -T ΔSo to calculate for the standard free energy

e.g.

C(s) + O2 (g) ---> CO2 (g) ΔHo=-393.5 kJ ΔSo= 305 J/K

Calculate Go at 25oC.

Solution:

1. Convert temperature to K: T= 25+273= 298 K

2. ΔGo = ΔHo -T ΔSo=(-3.935x105 J)-[(298K)(305 J/K)] = -3.944x105J

or -394.4 kJ/mol CO2 formed.

e.g. Consider the following:

2SO2(g) + O2(g) ---> 2SO3(g)

The above reaction was carried out at 25oC at 1 atm. Calculate ΔHo, ΔSo & ΔGo using the following data:

Substance ΔHof (kJ/mol) So (J/K-mol)

SO2 (g) -297 248

SO3 (g) -396 257

O2(g) 0 205

Solution: 2SO2(g) + O2(g) ---> 2SO3(g)

1. ΔHo = S nproduct ΔHof product - S nreactant ΔHo

f reactant

ΔHo = [2mol SO3(g) x(-396 kJ/mol)]-[2mol SO2(g)x (-297 kJ/mol)]+[1mol O2(g)x(0 kJ/mol)]

ΔHo = -792 kJ + 594 kJ = -198 k

2. ΔSo = S nproduct Soproduct - S nreactant So

reactant

ΔSo = [2mol SO3(g) x(257 J/K-mol)]-[(2mol SO2(g)x 248 J/K-mol)+(1mol O2(g)x 205 J/K-mol)]

= 514 J/K - 701 J/K = -187 J/K

3. ΔGo = ΔHo -T ΔSo

= -198 kJ – ( 298 K)(-187 J/K)( )

= -142 kJ

1 kJ1000 J

Bond energy and enthalpy• Chemical reactions involves breaking and forming of

bonds. We can use bond energy values to calculate for the approximate energies for reactions.

Consider the reaction below:

H2(g) + F2(g) ---> 2HF(g)

This reaction involves BREAKING 1 H-H bond and 1 F-F bond and FORMING 2 H-F bonds.

Bond BREAKING requires energy to be ADDED to the system (endothermic process, positive sign)

Bond FORMING requires energy to be RELEASED to the surrounding (exothermic process, negative sign)

We can then express the change in energy of the reaction based on bonds broken (positive sign) and bonds formed (negative sign)

ΔH = Sn x D(bonds broken) - Sn x D(bonds formed)

where

D = represents the bond energy per mole of bonds; always a POSITIVE VALUE given in kJ/mol

n = moles of a particular type of bond

e.g. Consider the formation of HF:H2(g) + F2(g) ---> 2HF(g)

Solution:ΔH = Sn x D(bonds broken) - Sn x D(bonds formed)

= [1mol H-H (432 kJ/mol) + 1mol F-F (154 kJ/mol)]-[2 mol H-F (565 kJ/mol)]= -544 kJ

When 1 mol H2(g) reacts with 1 mol F2(g) to form 2 mol HF(g), 544 kJ of heat should be released.

Bond Average Bond energy (kJ/mol)

H-H 432

F-F 154

H-F 565

Using bond energies, calculate the ΔH for the reaction of methane with chlorine and flourine to give Freon-12.

CH4(g) + 2Cl2(g) +2F2(g) ---> CF2Cl2(g) +2HF(g) + 2HCl (g)

Solution:1. Identify bonds broken and formed.Bonds broken:1 mol CH4 has 4 C-H bonds, so there are 4 mol C-H bond broken

per 1 mol of CH4

2 mol Cl2 has 2 mol Cl-Cl broken2 mol F2 has 2 mol F-F broken

Bonds formed:1 mol CF2Cl2 has 2 C-F bond & 2 C-Cl bond, so there are 2 mol C-F

bond formed & 2 mol C-Cl bond formed per 1 mol of CF2Cl22 mol H-F bond formed2 mol H-Cl bond formed

2. Calculate energy involved in bond breaking and bond formation using bond energies in kJ/mol.

CH4(g) + 2Cl2(g) +2F2(g) ---> CF2Cl2(g) +2HF(g) + 2HCl (g)

Bonds broken:

CH4: 4 mol (413 kJ/mol) = 1652 kJ

2Cl2: 2 mol (239 kJ/mol) = 478 kJ

2F2: 2 mol (154 kJ/mol) = 308 kJ

TOTAL ENERGY for bonds broken: 2438 kJ

Bond Bond energy

(kJ)/mol

Bond Bond energy

(kJ/mol)

Bond Bond energy

(kJ/mol)

Bond Bondenergy

(kJ/mol)

C-H 413 C-F 485 C-Cl 339 Cl-Cl 239

F-F 154 H-F 565 H-Cl 427

CH4(g) + 2Cl2(g) +2F2(g) ---> CF2Cl2(g) +2HF(g) + 2HCl (g)

Bonds formed:CF2Cl2: 2 mol C-F (485 kJ/mol)= 970 kJ

: 2 mol C-Cl (339 kJ/mol)= 678 kJ2HF(g) : 2 mol H-F ( 565 kJ/mol)= 1130 kJ2HCl(g) : 2 mol H-Cl (427 kJ/mol)= 854 kJTOTAL ENERGY for bonds formed: 3632 kJ

3. Calculate for the ΔH for the formation of Freon-12.ΔH = Sn x D(bonds broken) - Sn x D(bonds formed)ΔH = 2438 kJ – 3632 kJ = -1194 kJ

There is 1194 kJ of energy released per mole of CF2Cl2(g) formed.