Unit 9 Chp 19 B and L notes.ppt

Entropy,

Free

Energy,

and

Equilibrium

Unit 9

Thermodynamics

• Thermodynamics provides evidence for the driving

forces in processes and reactions.

• 2 main driving forces:

Enthalpy (ΔH) and Entropy (ΔS)

• “Spontaneous” reactions occur naturally but may be

very slow. Spontaneity has

nothing

to do with

speed. (Spontaneous also known as

“Thermodynamically favorable”)

Thermodynamically favorable

Physical & Chemical

Processes

• A waterfall runs downhill

• A lump of sugar dissolves in a cup of coffee

• At 1 atm, water freezes below 0oC and ice melts above 0oC

• Heat flows from a hotter object to a colder object • A gas expands in an evacuated bulb

• Iron exposed to oxygen and water forms rust

Favorable

Non Favorable Thermodynamically

favorable process:

Favorable

Not Favorable

Does a decrease in enthalpy (-H0) mean a reaction

is thermodynamically favorable?

CH₄ (g) + 2O₂ (g) CO₂ (g) + 2H₂O (l) H0 = -890.4 kJ

H+ _(aq) + OH- _(aq) H

2O (l) H0 = -56.2 kJ

H₂O (s) H₂O (l) H0 = 6.01 kJ

NH₄NO₃ (s) NH₄+_(aq) + NO

3- (aq) H0 = 25 kJ

H₂O

Favorable reactions

Need more than just enthalpy (_{to determine}

Entropy (



S

)

Entropy (S) is a measure

of the

randomness or

disorder

of a system.

•Units: J/K

•Higher



S = greater

disorder

•Nature favors disorder!

•+



S favored = more

disorder

For any substance, the solid state is more ordered than the liquid state and the liquid state is more

ordered than gas state

S

<

S

<<

S

H₂O (s) H₂O (l) _{S > 0}

Predicting changes in entropy

(



S)

1. Gases have most

entropy

2. Mixtures have

more entropy than

pure substances or

elements

3. Side of the

chemical equation

with more moles gas

is the favored side.

Least

How does the entropy of a system change for each of the following processes?

(a) Condensing water vapor

Randomness decreases Entropy decreases (-_{S )}

(b) Forming sucrose crystals from a supersaturated solution

Randomness decreases Entropy decreases (-S₎

(c) Heating hydrogen gas from 600C to 800C

Randomness increases Entropy increases (+S₎

(d) Subliming dry ice

How does the entropy of a system change for each of

the following processes?

H₂(g) + Cl₂(g)  2HCl(aq)

Randomness decreases Entropy decreases (-S )

NaCl(s) + H₂O(l)  H₂O(l) + NaCl(aq)

Randomness increases Entropy increases (+S)

N₂(g) + 3H₂(g)  2NH₃(g)

Randomness decreases Entropy decreases (-S )

I₂(s)  I₂(g)

All processes listed lead to an

increase in entropy (_{S > 0)}

Why?

Models and

Entropy

The diagram above represents the gas-phase reaction

of NO

(g) to form N

O

(g) at a certain temperature.

Based on the diagram, predict and explain the sign of

the entropy change for the reaction, ΔS°rxn ?

Predicting Molar Entropy Values, S°

1. Unlike enthalpies of formation, standard molar

entropies of elements not zero!

2. Standard molar entropies of gases greater than

liquids and solids

Ex. S° H

O

(l)

= 69.9 J/K, S° H

O

(g)

= 188.8 J/K

3. Standard molar entropies increase with increasing

molar mass

Ex. S° H

(g)

= 130.6 J/K S° O

(g)

= 205.0 J/K

4. Standard molar entropies increase with an increasing

number of atoms in the formula

Gibbs Free Energy:

For a constant-temperature

process:



G =



H

sys

-T



S

sys

_{G < 0 The reaction is thermodynamically favored in the}

forward direction. (G is negative, - G)

_{G > 0 The reaction is not thermodynamically favored as}

written. (G is positive, + G)

Calculate the standard free energy change for the

formation of NO

(g)

from N

(g)

and O

(g)

at 298K:

N

(g)

+ O

(g)



2 NO

(g)

ΔH° = 180.7 kJ and ΔS° = 24.7 J/K



G =



H

-T



S

Answer: 173.3 kJ

N₂(g) + 3 H₂(g)  2 NH₃(g)

Assume that ΔH° and ΔS° do not change with temperature. ΔH° = -92.38 kJ and ΔS° = -198.3 J/K

A) Predict the ΔG° for the reaction with increasing temperature. B) Calculate the ΔG° at 25°C and 500°C.

@ 25

°C ΔG° = -33.3 kJ

Thermodynamically Favored

@ 500°C ΔG° = 61 kJ

Gibbs can also be used to find T

N

(g)

+ 3 H

(g)



2 NH

(g)

ΔH° = -92.38 kJ and ΔS° = -198.3 J/K

Find the maximum temperature at which

this reaction can be spontaneous.

(*Set ΔG = 0 and solve for temperature)





H,

H,





S,

S,





G and Spontaneity

G and Spontaneity

Value of H Value of

_S Value of _G Thermodynamically _{favorability.}

Negative Positive Negative Favored

Positive Negative Positive Not Favored

Negative Negative ??? Favored if the absolute value of

H is greater than the absolute value of TS

(low temperature) _{H > T}_S

Positive Positive ??? Favored if the absolute value of TS is greater than the absolute value of H

(high temperature)

H < TS





G =

G =





H - T

H - T





S

S

H is enthalpy, T is Kelvin temperature

When solid ammonium chloride, NH

Cl(

s

), is added

to water at 25

C, it dissolves and the temperature of

the solution decreases. Which of the following is true

for the values of

ΔH

and

ΔS

for the dissolving

aA + bB cC + dD

G0

rxn = [cG0f (C) + dG0f (D)] - [aG0f (A) + bG0f (B)]

G0

rxn = nG0f (products) - mG0f (reactants)

Can use

Hess’ Law







G0 of any element in its stable form is zero, same as

H0. NOT true for f S0.

f f

_S0

rxn= [ cS0(C) + dS0(D) ] - [aS0(A) + bS0(B)]

S0

rxn =  nS0(products)-  mS0(reactants)

2C₆H₆ (l) + 15O₂ (g) 12CO₂ (g) + 6H₂O (l)

G0

rxn = nG0f (products) - mG0f (reactants)

What is the standard free-energy change for the following reaction at 25 0C?

G0

rxn = [12G0f (CO2)+6G0f (H2O)] - [ 2Gf0 (C6H6)]

G0

rxn = [ 12x–394.4 + 6x–237.2 ] – [ 2x124.5 ] = -6405 kJ

Is the reaction spontaneous at 25 0C?

G0 = -6405 kJ < 0

spontaneous

G0 C

6H6(l) = 124.5 kJ G0 H2O(l) = -237.2 kJ

G0 CO

Hess’ Law Examples

C₂H₄(g) + 6 F₂(g) 2 CF₄(g) + 4 HF(g)

H₂(g) + F₂(g) 2 HF(g) H = - 537 kJ C(s) + 2 F₂(g) CF₄(g) H = - 680 kJ

2 C(s) + 2 H₂(g) C₂H₄(g) H = +52.3 kJ

From the enthalpies of reaction listed below, calculate _H

for the reaction of ethylene with F₂:

What is the standard entropy change for the following reaction at 25oC? 2CO _(g) + O

2 (g) 2CO2 (g)

So(CO) = 197.9 J/K_•mol

So(O

2) = 205.0 J/K•mol

So(CO

2) = 213.6 J/K•mol

So

rxn= 2 x S0(CO2) – [2 x S0(CO) + S0 (O2)]

So

One more Hess’ Law Example:

Calculate the enthalpy change for the rxn:

2 C

(s)

+ H

(g)



C

H

(g)

Given the following thermochemical equations:

C

H

(g)

+ 5/2 O

(g)



2 CO

(g)

+ H

O

(l)



H = -1299.5 kJ

C

(s)

+ O

(g)



CO

(g)



H = -393.5 kJ

H

(g)

+ ½ O

(g)



H

O

(l)



H = -285.8 kJ

Free Energy and Equilibrium

Free Energy and Equilibrium

•

Equilibrium point occurs at the lowest

value of free energy available to the

reaction system

•

At equilibrium,



G

= 0



G

K



G

= 0

K = 1



G

< 0

K > 1



G

> 0

K < 1

*This table is

looking at reactions written in the

Free Energy Versus Extent of Reaction

G0 < 0 (-)

*move forward Favorable

G0 > 0 (+)

G0 =  RT lnK

18.6 R is the gas constant (8.314 J/K•mol) *unit conversion needed!

T is the absolute temperature (K)

18.6 Free Energy Versus Extent of Reaction

G0 < 0 (-)

K > 0

*move forward spontaneously

G0 > 0

K < 0

Standard Conditions?



G

°

vs.



G

•

For a gas

: std state is pure gas at 1 atm pressure

•

For a liquid or solid

: std state is the pure substance

in its most stable form present at 1 atm pressure and

at the temp of interest (usually room temp, 25°C)

•

For a solution

: std state for a substance in solution

is a concentration of 1 M

*In general: 25°C and 1 atm

Use tabulated free energy of formation values

to calculate the equilibrium constant for the

reaction at 298 K.



G

N

O

(g)

= 99.8 kJ/mol



G

NO

2

(g) =

51.3 kJ/mol

What else can we calculate?

Calculate the equilibrium concentration of

ammonia at 25°C given the following information:

N

(g)

+ 3 H

(g)



2 NH

(g)

ΔG

= -33.32 kJ

[N

] = 0.020

M

(Integrating mC

Δ

T and stoich: Review)

Magnesium metal reacts with hydrochloric acid:

Mg

(s)

+ 2HCl

(aq)



MgCl

(aq)

+ H

(g)

In an experiment, 0.158 g Mg metal is

combined with 100.0 mL HCl in a coffee cup

calorimeter. The Mg fully reacts. The

temperature rises from 25.6°C to 32.8°C.

Find

ΔH in kJ/mol

. (Use 1.00 g/mL as density of

soln and C

= 4.18J/g°C as the specific heat.)

Gibbs Free Energy and Chemical Equilibrium

for Nonstandard conditions

G = G0 + RT lnQ

R is the gas constant (8.314 J/K•mol) *unit conversion needed!

T is the absolute temperature (K)

Q is the reaction quotient (at non-equilibrium conditions)

At Equilibrium

G = 0 Q = K

0 = G0 + RT lnK

G0 =  RT lnK

Calculate G at 298 K for a mixture of 1.0 atm N₂, 3.0 atm H₂ and 0.50 atm NH₃: N₂(g) + 3 H₂(g)  2 NH₃(g).

1. Find



G°

using Hess’ Law

G_f° N₂ = 0 G_f° H₂ = 0 G_f° NH₃ = -16.66 kJ/mol

2. Solve for Q

3. Plug into the equation:



G

=



G

+

RT

ln

Q

Q = 9.3 x 10



G =

-44.9 kJ/mol

*make sense of answer:



G

is 1.0 atm for all gases at

standard conditions,



G

becomes more negative as

pressures of reactants inc. Larger



G

indicates rxn

The synthesis of NH₃ is represented by the equation above. Based on the equilibrium constant, K, and ΔH°rxn given above, which of the following can best be used to justify that the reaction is thermodynamically favorable at 298K and constant pressure?

A.ΔG°=−RTlnK>0 because K>>1 B.ΔG°=−RTlnK<0 because K>>1

C.ΔG°=ΔH°−TΔS°<0 because ΔH°<0 and ΔS°<0 D.ΔG°=ΔH°−TΔS°>0 because ΔH°<0 and ΔS°<0

The formation of HCl(g) from its atoms is represented by the

equation above. Which of the following best explains why the

reaction is thermodynamically favored?

A.ΔG>0 because energy is released as the bond between the H and Cl atoms forms, and entropy increases because the number of gaseous product particles is less than the number of gaseous reactant particles.

B.ΔG>0 because energy is absorbed as the bond between the H and Cl atoms forms, and entropy decreases because the number of gaseous product particles is less than the number of gaseous reactant particles.

C.ΔG<0 because although energy is absorbed as the bond between

the H and Cl atoms forms, entropy increases because the number of gaseous product particles is less than the number of gaseous reactant particles.

D.ΔG<0 because although entropy decreases because the number of gaseous product particles is less than the number of gaseous reactant particles, energy is released as the bond between the H and Cl atoms forms.

Based on the chemical equation and ΔG° given above, which

of the following justifies the claim that HA(aq) is a weak acid?

A.Because ΔG°>>0, Ka>>1 , and HA completely dissociates.

B.Because ΔG°>>0, Ka>>1, and HA almost completely

dissociates.

C.Because ΔG°>>0, Ka<<1, and HA only partially dissociates.

D.Because ΔG°>>0, Ka<<1, and HA does not dissociate.