cm1502 chapter 8 part ii

8/12/2019 CM1502 Chapter 8 Part II

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CM1502

Chapter 8

Part II


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Third Law and Absolute Entropies

All elements exists in three different states of aggregation.

Solid phases are most stable forms at low temperature. Solid Liquid Gas Determination of entropy of O 2 at 298.15 K is described here.

dq rev = Cp dT


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To obtain the numerical value for S m(T), S m(0K) must also be known.

The entropy can beobtained as a function oftemperature bynumerically integrating

the area under the curveand adding the entropychanges associated withthe phase changes at thetransition temperatures.


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S gas >> S liq > S solid

S increases withmolecular size

S(JK -1mol -1)

H2(g) 130.7

CH 4(g) 186.3

C 3H8(g) 270.3


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CM 1502 5

Free Energy

Entropy is the fundamental consequence of the second law ofthermodynamics. There are two other functions which utilise entropy intheir derivation and thus can also be used to predict spontaneity.

The two functions are the Helmholtz free energy (A) (constant V and T)

and the Gibbs free energy (G) (constant P and T) .

G can be used:- to predict if a process will be spontaneous or not other than using S in

S universe = S system + S surroundings 0 - has been employed extensively in connection with the study ofequilibrium and the direction of chemical changes.


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CM 1502 6

Gsystem

= Hsystem

T. Ssystem

G = n G (products) - n G (reactants)

G = H T.SG is also a state function as are H, T and S.For a change occurring at constant T and P,

G = n G (products) - n G (reactants)

= ( n H (products) - n H (reactants)) T.( n S (products) n S (reactants))

G = H T.S

Gibbs Free Energy

*

Josiah Willard Gibbs(1839 1903)


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7

G system = H system T.S system (1)

For spontaneous process, S total = S sys + S surr 0 and S surr = - Hsys /T=> S total = S system Hsystem /T 0 (2)

Combine (1) and (2) by multiplying (2) x -T=> -T. S total = -T. S system + Hsystem 0

(multiply by -ve, so inequality change direction)=> G system 0 for spontaneous process (opposite to S total 0)

G 0 => spontaneous process, G > 0 => non-spontaneous process.

Relationship between G and S

From here, will not specify that properties are referring to system.*


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CM 1502 8

Reaction Spontaneity and the Signs of H 0, S 0, and G 0

H0 S 0 G 0 Description

- + -

+ - +

+ + + or -

- - + or -

Spontaneous at all T

Non-spontaneous at all T

Spontaneous at higher T (G 0 -ve);Non-spontaneous at low er T (G 0 +ve)

G = H T.S

Depends on TSpontaneous at lower T (G 0 -ve);Non-spontaneous at higher T (G 0 +ve)

*

The standard free energy change of a reaction ( G ) can be readilycalculated from:

- using H and S for the reaction in equation: G = H T.S


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CM 1502 10

Standard Conditions

G = RT ln Q RT ln K = RT In Q + G =>

Under standard conditions (if T not stated, then T = 298 K), all specieshave concentrations of 1 M (1 atm for gases), so Q = 1. Equation reducesto:

G = -RT ln K

K = exp(- G/RT)

G = RT ln 1 RT ln K

G = RT ln Q/K = RT ln Q RT ln K

G =

G + RT In Q

Eqn 2

Eqn 3


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CM 1502 11

Eqn 1: G andLe Chateliers Principle (LCP)

1. Q < K, Q/K < 1, G = RT In (Q/K) < 0 => forward reaction isspontaneous. LCP also predicts that forward reaction is favored due toexcess reactants.

2. Q > K, Q/K > 1, G = RT In (Q/K) > 0 => forward reaction is non-spontaneous. LCP also predicts that reverse reaction is favored due toexcess products.

3. Q = K, Q/K = 1, G = 0 => equilibrium. LCP also predicts that systemis at equilibrium.

G = RT ln Q/K = RT ln Q RT ln K


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CM 1502 12

F ORWA RDRE A

C T I ON

RE V E R

S E RE A

C T I

ON

Eqn 2: Relationship between G and K

G(kJ) K Significance

200

100

50

10

9x10 -36

3x10 -18

2x10 -9

2x10 -2

Essentially no forward reaction; reversereaction FAVOURED goes to completionREACTANTS DOMINATE AT EQUILIBRIUM.

1

0

-1

7x10 -1

1

1.5

Forward and reverse reactionsproceed to same extent

-10 5x101

-50

-100

-200

6x10 8

3x10 17

1x10 35

Forward reaction FAVOURED goes tocompletion; essentially no reversereactionPRODUCTS DOMINATE AT EQUILIBRIUM.

G = -RT In K

K < 1

K > 1

G > 0 non-spont.

G < 0 spont.


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CM 1502 13

Eqn 3: G and Equilibrium

General reaction: aA + bB cC + dD

G = G + RT In

or

Recap:

Q is the reaction quotient = =

cf. Equilibrium constant, K =

ba

dc

[B][A][D][C]

bB

a A

dD

cC

aaaa

activities

eqb

eqa

eqd

eqc

[B][A][D][C]

ba

dc

[B][A][D][C]

G = G + RT In Q

*


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CM 1502 14

K dependence on T

From experimental data, if K is measured as T changes and plot In K vs.1/T, the equation of the straight line: In K = - H/RT + constant

Justification from previous equations:

G = H T.S and

G = -RT In K

=> - RT In K = H T.S (x -1/RT)In K = - H/RT + S /R (cf y = mx + c)

If plot In K vs. 1/T, then get a straight line with slope - H/R andintercept S /R .

The above equations can be used to calculate K, using standard H and S values.


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CM 1502 15

If have two (not very accurate!) data points, K 1 and T 1; and K 2 and T 2=> In K 1 = - H/RT 1 + S /R (a)

and In K 2 = - H/RT 2 + S /R (b)

(b) (a) In (K 2/K1) = (- H/R)(1/T 2 1/T 1)

CHECK: Slope = Rise/Run -H/R = (InK 2 InK 1)/(1/T 2 1/T 1)

Vant Hoff equation calculates K when the T is changed. (K dependsonly on T).

van't Hoff equation

*


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CM 1502 16

Using the thermodynamic data given below, calculate the following forthe reaction:

H2O (l) H + (aq) +OH - (aq)

Calculate the following using the data in the table:(i) Kw at 25C(ii) K

w at 30C

Hf / kJ mol -1 G f / kJ mol -1

H2O(l) -285.84 -237.15

H+(aq) 0 0

OH-(aq) -229.94 -157.30

Answer

at 25 C


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CM 1502 17

(i) G = n G f (products) n G f (reactants)= [0 + (-157.3)] (-237.15)= 79.85 kJ

G = -RT In KKw = exp(- G/RT)

= exp [-(79.85*1000)/(8.314*298)] [R = 8.314 J K-1 mol -1]= 1.007 x 10 -14

(ii) H = n Hf (products) n H

f (reactants)

= (0 + -229.94) (-285.84)= 55.9 kJ

In (K 2/K1) = (- H/R)(1/T 2 1/T 1)

In (K w 303K /1.007 x 10-14

) = [-(55.9*1000)/8.314] (1/303 1/298)In (K w 303K /1.007 x 10 -14 ) = 0.3723(Kw 303K /1.007 x 10 -14 ) = exp 0.3723

Kw 303K = 1.451 x (1.007 x 10 -14 )Kw 303K = 1.46 x 10 -14

*

Endothermic

cm1502 chapter 8 part ii

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