The van’t Hoff plot II - Clear Cognitiongenchem.clearcognition.org/GeneralChemistryslideshows/...2...

The van’t Hoff plot II

Constructing the van’t Hoff plot from ThermodynamicData

1

©copyright by James B. Condon (2013) all rights reserved.

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In a previous slideshow the Clausius-Calpyron equation was usedto create a van’t Hoff plot of water vapor pressure. This equation,derivable from Thermodynamics is:

It was noted that the abscissa (x) was the quantity 1/T and theordinate (y) was lnP. Thus, the slope was –∆H

v.

For the case of vapor pressure, the ∆H > 0 always. This is not ingeneral the case for reactions and the equilibrium values forreactions at two different temperatures may not be available, atleast initially. Therefore it is useful to be able to calculate thevan’t Hoff plot given only Thermodynamic data. Conversely, froman experimental van’t Hoff plot one can obtain theThermodynamic data.

ln = –‰ ÷P

1

P2

∆Hv

R‰ ÷ ‰ ÷ 1 1

T2 T

1

3


Starting with the relationship between thermodynamics and theequilibrium constant:

and the relationship of Gibbs’ free energy and the enthalpy andentropy:

∆Gº = ∆H

º – T∆Sº

Taking the ln of the first equation and substituting ∆Hº – T∆Sº for

∆Gº yields:

Thus, –∆Hº/R is the slope and ∆S

º/R is the intercept.

lnK = + ∆S

º

R‰ ÷ ‰ ÷–∆H

º

R

1

T

K = e–∆G

º

RT

4


Usually this equation is used by designating two temperatures, amaximum and minimum, and a calculation of the two values oflnK.

First, a look at the parametric arrangement.

lnK = + ∆S

º

R‰ ÷ ‰ ÷–∆H

º

R

1

T

5

lnK2 = ! +

∆Hº 1 ∆S

º

R T2 R ‰ ÷‰ ÷

lnK1 = ! +

∆Hº 1 ∆S

º

R T1 R ‰ ÷‰ ÷

!


This equation could be used between two temperature to yield anequation similar to the Clausius-Calpyron equation. Todemonstrate this use two temperatures, T

1 and T

2 corresponding to

two values of K, K1 and K

2. Substrating equation 1 from 2:

ln = !∆H

º 1 1

R T2 T

1 ‰ ÷‰ ÷K

1

K2‰ ÷

For the reaction: (l) º (g), this becomes the Clausius-Calpyron equation.

lnK = + ∆S

º

R‰ ÷ ‰ ÷–∆H

º

R

1

T

6


Here is an example:

Draw a van’t Hoff plot for the reaction between 100EC and 500EC:

N2(g) + 3H

2(g) ! 2NH

3(g)

The thermodynamic quantities for the reaction are required.

∆Hº = 2 ∆H

º(NH3) – ∆H

º(N2) – 3 ∆H

º(H2)

∆Hº = 2(–46.11 kJ mol–1) – 0 – 3(0) = –92.22 kJ mol–1

∆Sº = 2 Sº(NH

3) – Sº(N

2) – 3 Sº(H

2)

∆Sº = 2(192.3) – (191.1) – 3(130.6) J mol–1 K–1 = –198.3 J mol–1 K–1

rxn

rxn

Usually this equation is used by designating two temperatures, amaximum and minimum, and a calculation of the two values oflnK.

rxn f f f

rxn

lnK = + ∆S

º

R‰ ÷ ‰ ÷–∆H

º

R

1

T

7


van’t Hoff plot for: N2 + 3H

2 ! 2NH

3

Using the information that

∆Hº = –92.22 kJ mol–1 and ∆S

º = –198.3 J mol–1 K–1

T1 = 100EC / 373 K and T

2 = 500EC / 773 K

the values for lnK and 1/T are calculated*:

rxn rxn

* Remember one must convert kJ to J. It has been done here without explicitly stating it . Youshould know that by now.

Now calculate the values for:1/T

1 = ?

1/T2 = ?

ln(K1) = ?

ln(K2) = ?

When finished, use the right arrow, !, to compare answers.

lnK = + ∆S

º

R‰ ÷ ‰ ÷–∆H

º

R

1

T

8


van’t Hoff plot for: N2 + 3H

2 ! 2NH

3

Using the information that

∆Hº = –92.22 kJ mol–1 and ∆S

º = –198.3 J mol–1 K–1

T1 = 100EC / 373 K and T

2 = 500EC / 773 K

the values for lnK and 1/T are calculated*:

rxn rxn

lnK2 = +

+92.2 × 103 J mol–1 1 –198.3 J mol–1 K–1

8.314 J mol–1 K–1 773 K 8.314 J mol–1 K–1‰ ÷‰ ÷lnK

2 = –9.50

lnK1 = +

+92.2 × 103 J mol–1 1 –198.3 J mol–1 K–1

8.314 J mol–1 K–1 373 K 8.314 J mol–1 K–1‰ ÷‰ ÷lnK

1 = 5.89

1T

2

= 1.29 × 10–3 K–1 1T

1

= 2.68 × 10–3 K–1 and

lnK = + ∆S

º

R‰ ÷ ‰ ÷–∆H

º

R

1

T

9


Using the following information, a van’t Hoff plot may be formed:

Starting witha blank grid

1T

2

= 1.29 × 10–3 K–1 1T

1

= 2.68 × 10–3 K–1 and

lnK2 = –9.50lnK

1 = 5.89 and

lnK = + ∆S

º

R‰ ÷ ‰ ÷–∆H

º

R

1

T

10

N2 + 3H

2 ! 2NH

3



Provide a title

lnK = + ∆S

º

R‰ ÷ ‰ ÷–∆H

º

R

1

T

1T

2

= 1.29 × 10–3 K–1 1T

1

= 2.68 × 10–3 K–1 and

lnK2 = –9.50lnK

1 = 5.89 and

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Label theabscissa (x)axis.

Notice that itis beingscaled by1000 and theunits areprovidedbehind a “/”.

1/T × 1000 /K-1

lnK = + ∆S

º

R‰ ÷ ‰ ÷–∆H

º

R

1

T

1T

2

= 1.29 × 10–3 K–1 1T

1

= 2.68 × 10–3 K–1 and

lnK2 = –9.50lnK

1 = 5.89 and

N2 + 3H

2 ! 2NH

3

12



Label theordinate (y)axis.

Notice thatthe quantitythat is a ln isunit-less.

lnK = + ∆S

º

R‰ ÷ ‰ ÷–∆H

º

R

1

T

1T

2

= 1.29 × 10–3 K–1 1T

1

= 2.68 × 10–3 K–1 and

lnK2 = –9.50lnK

1 = 5.89 and

1/T × 1000 /K-1

N2 + 3H

2 ! 2NH

3

13



1.2 1.6 1.8 2.0 2.2 2.4 2.6 2.8 3.01.41.0

Write in thenumberlables on the“x” axis as alinear scale.

lnK = + ∆S

º

R‰ ÷ ‰ ÷–∆H

º

R

1

T

1T

2

= 1.29 × 10–3 K–1 1T

1

= 2.68 × 10–3 K–1 and

lnK2 = –9.50lnK

1 = 5.89 and

1/T × 1000 /K-1

N2 + 3H

2 ! 2NH

3

14



-2

-6

-4

-10

0

2

4

6

8

10

-8

Write in thenumberlables on the“y” axis as alinear scale.

lnK = + ∆S

º

R‰ ÷ ‰ ÷–∆H

º

R

1

T

1T

2

= 1.29 × 10–3 K–1 1T

1

= 2.68 × 10–3 K–1 and

lnK2 = –9.50lnK

1 = 5.89 and

1/T × 1000 /K-1

N2 + 3H

2 ! 2NH

3

1.2 1.6 1.8 2.0 2.2 2.4 2.6 2.8 3.01.41.0

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Draw in theline betweenthe twotemperature.

lnK = + ∆S

º

R‰ ÷ ‰ ÷–∆H

º

R

1

T

1T

2

= 1.29 × 10–3 K–1 1T

1

= 2.68 × 10–3 K–1 and

lnK2 = –9.50lnK

1 = 5.89 and

-2

-6

-4

-10

0

2

4

6

8

10

-8

1/T × 1000 /K-1

N2 + 3H

2 ! 2NH

3

1.2 1.6 1.8 2.0 2.2 2.4 2.6 2.8 3.01.41.0


Constructing the van’t Hoff plot from ThermodynamicData

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