Download Calculating Enthalpy Changes

Chemistry 201
Lecture 10
Temperature and pressure
dependence of K
NC State University
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.
Temperature dependence of K
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
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
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
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
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?
Temperature dependence of K
N2O4(g)  2 NO2(g)
DHo = +58.20 kJ
DSo = +176.6 J/K
What is K at 425 K?
Temperature dependence of K
N2O4(g)  2 NO2(g)
DHo = +58.20 kJ
DSo = +176.6 J/K
What is K at 425 K?
Temperature dependence of 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)
Temperature dependence of K
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?
Temperature dependence of K
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
Temperature dependence of K
H2(g) + 1/2 O2 (g)  H2O (g)
DHo = -241.82 kJ
DSo = -44.38 J/K
Evaluate K at 1500 K.
Temperature dependence of K
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.
Using equilibrium data to obtain
DHo and DSo
2 NO2 (g)  2 NO (g) + O2 (g)
T1 at 190 K
T2 at 200 K
K1 = 18.4
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
Van’t Hoff plots
Slope = -DHo/R
Note: DHo > 0
The standard method for obtaining the
reaction enthalpy is a plot of ln K vs. 1/T
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
Slope = -DHo/R
Note: DHo < 0
In this example, the slope is positive because the enthalpy
of binding is negative (i.e. binding is exothermic).
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.
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.
Kinetic and Thermodynamic Assessment of Binding of Serotonin Transporter
Inhibitors. J Pharm Exp Tech (2008) vol. 327, pp. 991-1000
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
with equilibrium constant K. We have:
MbCO
[MbCO]
K=
[Mb][CO]
The fraction bound is:
[MbCO]
f=
[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:
[MbCO]
K=
[Mb][CO]
[MbCO]
K[Mb][CO]
K[CO]
f=
=
=
[Mb]+[MbCO] [Mb] + K[Mb][CO] 1 + K[CO]
This type of binding curve is plotted below.
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?
N2O4(g)  2 NO2(g)
Can the trend be explained in simple terms?
Why is the dissociation greater at low pressure?
N2O4(g)  2 NO2(g)
Can the trend be explained in simple terms?
Yes, this is an example of Le Chatelier’s principle.
Why is the dissociation greater at low pressure?
N2O4(g)  2 NO2(g)
Can the trend be explained in simple terms?
Yes, this is an example of Le Chatelier’s principle.
Can the trend be explained quantitatively?
Why is the dissociation greater at low pressure?
N2O4(g)  2 NO2(g)
Can the trend be explained in simple terms?
Yes, this is an example of Le Chatelier’s principle.
Can the trend be explained quantitatively?
The Haber-Bosch Process
The Haber-Bosch process
N2 (g) + 3 H2 (g)  2 NH3(g)
Involves both high temperature and pressure.
The Haber-Bosch process
N2 (g) + 3 H2 (g)  2 NH3(g)
Involves both high temperature and pressure.
Why high pressure?
The Haber-Bosch process
N2 (g) + 3 H2 (g)  2 NH3(g)
Involves both high temperature and pressure.
Why high pressure?
Forces the equilibrium to the right. This is an
example of Le Chatelier’s principle.
The Haber-Bosch process
N2 (g) + 3 H2 (g)  2 NH3(g)
Involves both high temperature and pressure.
Why high pressure?
Forces the equilibrium to the right. This is an
example of Le Chatelier’s principle.
Why high temperature?
The Haber-Bosch process
N2 (g) + 3 H2 (g)  2 NH3(g)
Involves both high temperature and pressure.
Why high pressure?
Forces the equilibrium to the right. This is an
example of Le Chatelier’s principle.
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?
The Haber-Bosch process
1/2 N2 (g) + 3/2 H2 (g)  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.
Mol
NH3
N2
H2
So
192.77
153.3
114.7
The Haber-Bosch Process
Under standard conditions
The Haber-Bosch Process
1/2 N2 (g) + 3/2 H2 (g)  NH3(g)
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?
The Haber-Bosch Process
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