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Temperature and pressure

dependence of K

Chemistry 201

NC State University

Lecture 10

Factors that affect K

• How does the free energy relate to the

standard state?

• How does temperature change affect

energetics and equilibrium?

• What is the pressure dependence of K?

Text : Sections 4.7, 5.1

Relationship of free energy

to standard state The free energy change for a reaction, DG,

is composed of individual free energies Gn

that correspond to each reactant. We can

write the free energy of each component as

where an is the activity of component n and

DGno is the standard free energy. Activity

means concentration, but it is used for “real”

solutions. See the next slide for comparison.

Definition of activity

Activity is a concentration and has units of

molarity. We use activity to account for the

fact that there may an effective or non-ideal

concentration due to solvent effects. You

May also think of the free energy as follows:

where [n] is the effective concentration (activity).

Combining free energies to make a

free energy change

The molar free energy change of a reaction

is given by the difference between the free

energy of the products minus the reactants.

For the hypothetical reaction

we have

Combining free energies to make a

free energy change (cont’d)

Each free energy is related to the standard

free energy for that species so we have

We can rearrange this to

and finally

The standard state

Thus, the standard free energy change is

the difference in free energy of each species

at its standard state,

And the reaction quotient is the ratio of each

concentration to the standard concentration,

which is 1 molar (so it does not appear).

Defining the reaction quotient

The reaction quotient is unitless. The units of the

individual activities (concentrations) are cancelled

by the units of the standard state.

N2O4(g) 2 NO2(g)

DHo = +58.20 kJ DSo = +176.6 J/K

At what temp. does DGo = 0?

(K = 1)

Temperature dependence of K

N2O4(g) 2 NO2(g)

DHo = +58.20 kJ DSo = +176.6 J/K

What is K at 425 K?


N2O4(g) 2 NO2(g)

DHo = +58.20 kJ DSo = +176.6 J/K

What is K at 425 K?


Remember to convert DHo to units of J (multiply by 1000)

H2(g) + 1/2 O2 (g) H2O (g)

DHo = -241.82 kJ DSo = -44.38 J/K

At what T, if any, does K = 1?


H2(g) + 1/2 O2 (g) H2O (g)

DHo = -241.82 kJ DSo = -44.38 J/K

At what T, if any, does K = 1?


The temperature corresponds to DGo = 0

H2(g) + 1/2 O2 (g) H2O (g)

DHo = -241.82 kJ DSo = -44.38 J/K

Evaluate K at 1500 K.


The temperature dependence of DGo

As we have shown previously, DG, will decrease until it

reaches 0. Then we have reached equilibrium. The

equilibrium condition is

DGo = -RT ln K

Next we consider the fact that we can use the temperature

dependence of the free energy to obtain information about

the enthalpy.

DHo - TDSo = -RT ln K

If we assume that DHo and DSo are independent of

temperature, then we can obtain the values of K at two

temperatures as follows,

DHo – T1DSo = -RT1 ln K1

DHo – T2DSo = -RT2 ln K2

The temperature dependence of DGo

Then we can divide each equation by its respective

temperature to obtain,

DHo /T1 – DSo = - R ln K1

DHo /T2 – DSo = - R ln K2

We subtract temperature T2 from T1.

DHo (1/T1 – 1/T2) = - R ln(K1/K2)

ln(K2/K1) = -DHo /R(1/T2 – 1/T1)

This equation says that if we plot ln(K) vs 1/T, we obtain a

line, and the slope of that line is -DHo /R.

2 NO2 (g) 2 NO (g) + O2 (g)

T1 at 190 K K1 = 18.4

T2 at 200 K K2 = 681

find DHo and DSo

Using equilibrium data to obtain DHo and DSo

Solution: starting with the equation

Solve for DHo

Substitute in the given values

Using equilibrium data to obtain DHo and DSo

Van’t Hoff plots

The standard method for obtaining the

reaction enthalpy is a plot of ln K vs. 1/T

Slope = -DHo/R

Note: DHo > 0

Van’t Hoff plot for drug binding

A practical example of the application of the

van’t Hoff equation can be found in drug binding.

The equilibrium constant for drug binding to an

active site can be measured by fluorescence,

NMR, etc. at various temperatures. Then one

may plot ln K vs. 1/T and fit the result to a line.

In most cases the binding will be exothermic to

that DHo < 0 and then slope of the line will be

positive rather than negative as shown in the

previous slide.

Van’t Hoff plot for drug binding

In this example, the slope is positive because the enthalpy

of binding is negative (i.e. binding is exothermic).

Slope = -DHo/R

Note: DHo < 0

Example: preventing inflammation by binding to prostaglandin synthase 2

COX-2 crystal structure

DHo and DSo can be measured

using ln(K) as a function 1/T.

Kinetic and Thermodynamic Assessment of Binding of Serotonin Transporter

Inhibitors. J Pharm Exp Tech (2008) vol. 327, pp. 991-1000

Example: preventing depression serotonin transport inhibitors

DHo and DSo can be measured using ln(K) as a function 1/T.

This article shows specific differences in the enthalpy of

binding of drugs based on analysis of so-called van’t Hoff

plots (i.e. plots of ln(K) vs. 1/T).

Kinetic and Thermodynamic Assessment of Binding of Serotonin Transporter

Inhibitors. J Pharm Exp Tech (2008) vol. 327, pp. 991-1000

Basic conclusion: fluvoxamine binds exclusively based on entropic driving force

Entropically driven binding is relatively rare. Usually the

entropy of binding is unfavorable since a flexible drug

molecule (large W) will be forced to adopt a fixed

conformation (small W or W = 1) upon binding to a protein.

Ligand binding in myoglobin Ligand binding to myoglobin is described as a chemical equilibrium

between a bound state MbCO, a dissociated state Mb:CO and a

solvent state Mb + CO.

Mb + CO Mb:CO MbCO

If we ignore the intermediate state the binding can be described by

An overall equilibrium process.

Mb + CO MbCO

with equilibrium constant K. We have:

The fraction bound is:

K =[MbCO]

[Mb][CO]

f =[MbCO]

[Mb] + [MbCO]

We use CO in

Many studies.

The same equations

Hold for O2.

Ligand binding curve The fraction bound f can be related to the binding constant K:

This type of binding curve is plotted below.

K =[MbCO]

[Mb][CO]

f =[MbCO]

[Mb]+[MbCO]=

K[Mb][CO]

[Mb] + K[Mb][CO]=

K[CO]

1 + K[CO]

Pressure dependence of species

We can see from the gas phase form of the equilibrium

constant that pressure of species depend on pressure.

For the general gas phase reaction,

we can write the equilibrium constant as

And the free energy is

From Dalton’s law

Pressure dependence of species

If we substitute these mole fractions and total pressure into

the equilibrium constant we have

Which depends on the total pressure unless z – c – d = 0.

This expression shows that, in general, the free energy

depends on the total pressure. This means that for the

fixed pressure may affect the proportion of products

to reactants.

Equilibrium of smog formation

Why is the dissociation greater at low pressure?

Can the trend be explained in simple terms?

N2O4(g) 2 NO2(g)



Yes, this is an example of Le Chatelier’s principle.

N2O4(g) 2 NO2(g)



Yes, this is an example of Le Chatelier’s principle.

Can the trend be explained quantitatively?

N2O4(g) 2 NO2(g)

The Haber-Bosch Process

The Haber-Bosch process

Involves both high temperature and pressure.

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



Why high pressure?

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



Why high pressure?

Forces the equilibrium to the right. This is an

example of Le Chatelier’s principle.

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



Why high pressure?



Why high temperature?

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



Why high pressure?



Why high temperature?

Although this is an exothermic reaction, it also has a

large barrier. The process uses a catalyst. But, what

does the temperature do to the equilibrium?

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


The enthalpy change of the reaction is equal to the

enthalpy of formation of NH3. DfHo = -45.9 kJ/mol

The entropy can be calculated from tabulated

absolute entropies.

1/2 N2 (g) + 3/2 H2 (g) NH3(g)

Mol So NH3 192.77 N2 153.3 H2 114.7


Under standard conditions

At high pressure there is a shift in the free energy for

the process is given by:

The entropy DrxnSo is obtained from the tables as well.

We note that the standard (tabulated) conditions

correspond to Q = 1 and therefore RTlnQ = 0.

Suppose we increase the pressure of N2 and H2

to 500 atm, while NH3 is maintained at 1 atm.

What happens to the free energy of reaction?


1/2 N2 (g) + 3/2 H2 (g) NH3(g)


1/2 N2 (g) + 3/2 H2 (g) NH3(g)

The shift to high pressure will shift the reaction further

towards products. This is clear intuitively from

Le Chatelier’s principle, but it is quantified using the

equilibrium expressions we have learned.

Since the free energy change is negative, this might

seem unnecessary. However, the Haber-Bosch

Process is challenging since the reaction must occur

at high temperature as well. We will see how all of the

factors combine in future lectures.

Hydrogenation: use of pressure

Skills

• Calculate how DGo changes with

temperature and pressure

• Calculate how K changes with

temperature and pressure

• Calculate DHo from the temperature

dependence of K

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