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H 2 O(l) --> H 2 O(s) Normal freezing point of H 2 O = 273.15K PowerPoint Presentation
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H 2 O(l) --> H 2 O(s) Normal freezing point of H 2 O = 273.15K - PowerPoint PPT Presentation


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The Gibbs Free Energy Function and Phase Transitions. D H. D S=. -6007 J. =. = -21.99 J/K. T. 273.15K. H 2 O(l) --> H 2 O(s) Normal freezing point of H 2 O = 273.15K

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slide1

The Gibbs Free Energy Function and Phase Transitions

DH

DS=

-6007 J

=

= -21.99 J/K

T

273.15K

H2O(l) --> H2O(s)

Normal freezing point of H2O = 273.15K

The change in enthalpy is the enthalpy of freezing - enthalpy change associated when one mole of liquid water freezes at 1atm and 273.15K

DH = -6007J

Entropy change

DG = DH - TDS = -6007J - (273.15K)(-21.99J/K) = 0J

=> at 273.15K and 1 atm, liquid and solid water are at equilibrium

slide2

As water is cooled by 10K below 273.15K to 263.15K

Assume that DH and DS do not change with T

DG = DH - TDS = -6007J - (263.15K)(-21.99J/K) = -220J

Since DG < 0 water spontaneously freezes

At 10K above the freezing point, 283.15K

DG = DH - TDS = -6007J - (283.15K)(-21.99J/K) = 219J

DG > 0 water does not spontaneously freeze at 283.15K, 10K above the normal freezing point of water

Above 273.15K, the reverse process is spontaneous - ice melts to liquid water.

slide3

Freezing not spontaneous; melting spontaneous

freezing spontaneous

Plot of DH and TDS vs T for the freezing of water. At 273.15 K the system is at equilibrium, DG = 0

H2O(l) --> H2O(s)

slide4

For phase transitions, at the transition temperature, the system is at equilibrium between the two phases.

Above or below the transition temperature, the phase that is the more stable phase is determined by thermodynamics - DG of the phase transition.

slide5

Reaction Free Energy

DGr = SnGm(products) -SnGm(reactants)

Standard Reaction Free Energy

DGor = SnGom (products) -SnGom(reactants)

Difference in free energy of the pure products in their standard states and pure reactants in their standard states.

slide6

1/2N2 (g) + 3/2 H2 (g) --> NH3(g) DGfo = -16 kJ

Standard Free-Energy of Formation

The standard free energy of formation, DGfo, of a substance is the standard reaction free energy per mole for the formation of a compound from its elements in their most stable form

O2 (g) --> O2 (g) DGfo = 0

slide7

DGfo is an indication of a compounds stability relative to its elements

Thermodynamically stable compound DGfo < 0

Thermodynamically unstable compound DGfo > 0

slide8

Standard Reaction Free Energy

For a reaction: aA + bB --> cC + dD

DGro = c DGfo (C) + d DGfo(D) - a DGfo(A) -b DGfo(B)

The standard free energy of formation of elements in their most stable form at 298.15K is zero.

slide9

Since the reactants N2(g) and H2(g) are in their standard state at 298 K their standard free energies of formation are defined to be zero.

DGro = 2DGfo[NH3(g)] - DGfo[N2(g)] - 3DGfo[H2(g)]

= 2 x -16.66 = -33.32 kJ/mol

For the reverse reaction: 2NH3(g) --> N2(g) + 3H2 (g)

DGro = +33.32 kJ/mol

Problem: Calculate the standard free-energy change for the following reaction at 298K:

N2(g) + 3H2 (g) --> 2NH3(g)

Given that DGfo[NH3(g)] = -16.66 kJ/mol.

What is the DGo for the reverse reaction?

slide10

C3H8(g) + 5O2(g) --> 3CO2 (g) + 4H2O(l) DHo = - 2220 kJ

a) Without using the information from the tables of thermodynamic quantities predict whether DGo for this reaction will be less negative or more negative than DHo.

b) Use data from App. D to calculate DGro for this reaction.

Whether DGro is more negative or less negative than DHo depends on the sign of DSo since DGro = DHo - TDSo

The reaction involves 6 moles of gaseous reactants and produces 3 moles of gaseous product

=> DSo < 0 (since fewer moles of gaseous products than reactants)

Since DSo < 0 and DHo < 0 and DGro = DHo - TDSo

=> DGro is less negative than DHo

slide11

DGro = 3DGfo(CO2 (g))+4DGfo(H2O(l)-DGfo(C3H8(g))-5DGfo(O2(g))

= 3(-394.4) + 4(-237.13) - (-23.47) - 5(0)

= -2108 kJ

slide12

T =

DHo

DSo

Effect of temperature on DGo

Values of DGro calculated using the tabulated values of DGfo apply only at 298.15K.

For other temperatures the relation:

DGro = DHo - T DSo

can be used, assuming DHo and DSo do not vary significantly with temperature.

When DGro = 0,

Use this expression to determine temperature above or below which reaction becomes spontaneous.

slide15

a) CCl4(l) CCl4(g)

Problem: The normal boiling point is the temperature at which a pure liquid is in equilibrium with its vapor at 1 atm.

a) write the chemical equation that defines the normal boiling point of liquid CCl4

b) what is the value of DGo at equilibrium?

c) Use thermodynamic data to estimate the boiling point of CCl4.

b) At equilibrium DG = 0. In any equilibrium for a normal boiling point, both the liquid and gas are in their standard states. Hence, for this process DG is DGo.

slide16

32.6 kJ

=

= 343 K

DHo

DHo

T =

T =

0.095 kJ/K

DSo

DSo

c)

where for this reaction T is the boiling point

Note: To determine the boiling point accurately we need the values of DHo and DSo for the vaporization process at the boiling point of CCl4

DHo = (1mol)(-106.7 kJ/mol) - (1mol)(-139.3kJ/mol) = 32.6 kJ

DSo = (1mol)(309.4 J/mol-K) -(1mol)(214.4 J/mol-K) = 95.0 J/K

slide17

The Gibbs Function and the Equilibrium Constant

DG = DGro + R T ln Q

Q - reaction quotient

Under standard conditions the concentrations of all reactants and products are set to 1

(standard pressure of gases = 1 atm

Standard pressure of solutions = 1 M)

Under standard conditions, ln Q = 0

=> DG = DGro

slide18

Q

DG = R T ln

K

At equilibrium DG = 0 and Q = K

At equilibrium

DGro = - R T ln K or K = e- DGro /RT

DGro < 0 => K > 1

DGro > 0 => K < 1

DG = DGro + R T ln Q = - RT ln K + RT ln Q

when Q = K => DG = 0; equilibrium

slide19

DGforward < 0

DGforward > 0

If the reaction mixture has too much N2 or H2 relative to NH3, Q < K, and reaction moves spontaneously in the direction to form NH3 till equilibrium is established; Q = K. If there is too much NH3 in the mixture, Q > K, decomposition of NH3 is spontaneous, till equilibrium is established; Q = K

slide20

-DGr

- DHo

DSo

ln K =

=

+

R T

R T

R

Temperature Dependence of K

The temperature dependence of K can be used to determine DHo and DSo of a reaction

slide21

- DHo

- DHo

DSo

DSo

ln K1 =

ln K2 =

+

+

R T1

R T2

R

R

1

T2

K2

DHo

1

ln K2 - ln K1 =

ln

=

(

-

)

K1

R

T1

If the equilibrium constant of a reaction is known at one temperature and the value of DHo is known, then the equilibrium constant at another temperature can be calculated

If DHo is negative (exothermic), an increase in T reduces K; if it is positive (endothermic), an increase in T increases K

slide22

Liquid Gas K = Pgas

1

T2

p2

DHvap

1

ln

=

(

-

)

p1

R

T1

Clausius-Clapeyron equation

For the equilibrium between a pure liquid and its vapor, the equilibrium constant is simply the equilibrium vapor pressure (since pure liquids do not appear in the equilibrium expression)

Hence

The Clausius-Clapeyron equation indicates the variation of vapor pressure of a liquid with temperature

slide23

Driving Non-spontaneous Reactions

The magnitude of DG is a useful tool for evaluating reactions.

For a spontaneous process at constant pressure and temperature, the change in free energy for a process equals the maximum amount of work that can be done by the system on its surroundings wmax = DGr

For non-spontaneous reactions (DGr > 0), the magnitude of DGr is a measure of the minimum amount of work that must be done on the system to cause the process to occur.

A reaction for which DGr is large and negative (like the combustion of gasoline), is much more capable of doing work on the surroundings than a reaction for which DGr is small and negative (like the melting of ice).

slide24

Many chemical reactions, including a large number that are central to biological systems, are non-spontaneous.

For example: Cu can be extracted from the mineral chalcolite which contains Cu2S.

The decomposition of Cu2S is non-spontaneous

Cu2S --> 2Cu(s) + S(s) DGro = +86.2 kJ

Since this reaction is not spontaneous Cu cannot be directly obtained from the above reaction.

Instead, work needs to be done on this reaction, and this is done by coupling this non-spontaneous reaction with a spontaneous reaction so that the overall reaction is spontaneous.

slide25

Consider the following reaction:

S(s) + O2(g) --> SO2 (g) DGro = -300.4kJ

Coupling this reaction with the extraction of Cu from Cu2S

Cu2S --> 2Cu(s) + S(s) DGro = +86.2 kJ

S(s) + O2(g) --> SO2 (g) DGro = -300.4kJ

Net: Cu2S(s) + O2 (g) --> 2Cu(s) + SO2(g) DGro = -214.2kJ

slide26

Biological systems employ the same principle by using spontaneous reactions to drive non-spontaneous reactions.

Many biochemical reactions are not spontaneous.

These reactions are made to occur by coupling them with spontaneous reactions which release heat.

The metabolism of food is the usual source of free energy needed to do the work to maintain biological systems.

C6H12O6(s) + 6O2 (g) --> 6CO2 (g) + 6H2O(l)

DHo = -2803 kJ

slide27

A means of transporting the energy released by glucose metabolism to the reactions that require energy is needed.

The free energy released by the metabolism of glucose is used to convert lower-energy ADP (adenine diphosphate) to higher energy ATP (adenine triphosphate).

The energy stored in the ATP molecule is then available to a biochemical reaction that requires energy and in the process ATP is converted to ADP.

slide28

C6H12O6

Free energy released by ATP converts simple molecules to more complex molecules

ATP

Free energy released by oxidation of glucose converts ADP to ATP

ADP

6CO2 (g) + 6H2O(l)