Lecture 18' Chemical Reactions Ch' 5

1
Lecture 18. Chemical Reactions (Ch. 5)
In chemical reactions, the products of reaction
are intermixed with the reacting substances
(reactants). Thus, the process is governed by two
factors (a) the energy change (V,Tconst) or
enthalpy change (T,Pconst), and (b) the entropy
change
For a reaction to be energetically favorable, the
Gibbs energy for products should be lower than
the Gibbs energy for reactants.
2
Chemical Reactions
- processes of molecular transformations that
involve at least one of the following changes
the number of atoms in a molecule, the type of
atoms, their mutual positioning in a molecule
(isomers), or their charge.
The other work (electrical, chemical, etc.)
performed on a system at T const and P const
in a reversible process is equal to the change in
the Gibbs free energy of the system
reac- tants
products
?
reaction coordinate
the enthalpy released in the reaction at
P,Tconst
During this reaction, some bonds should be broken
and other bonds (with a more negative potential
energy) should be formed. This process is
characterized by a potential barrier thus, the
Boltzmann factor! For the direct reaction, the
barrier is , for the reverse one -
.
The reactions are characterized by
directionality (which free energy is lower,
reactants or products), energy release, and rate.
3
Directionality of Chemical Reactions
The directionality of a chemical reaction at
fixed P,T is governed by the Gibbs free energy
minimum principle. Two factors are at play the
entropy and the enthalpy. Since the change in G
is equal to the maximum useful work which can
be accomplished by the reaction, then ?G lt 0
indicates that the reaction will proceed
spontaneously.
- clearly, S increases. Also, the energy of the
products of the reaction is lower than the
reactant (the energy is released in the TNT
explosion) the reaction is shifted strongly
toward the products.
- though S increases, the equilibrium is shifted
to the left (?H gtT?S) at 300K
- though S decreases, the equilibrium is shifted
to the right because the decrease of ?H
overpower the increase of -T?S
4
Examples
Several channels of the reaction between CO and
H2 at 300K
Several channels of the reaction between CO and
H2 at 600K
at this T, CH3OH and CH3COOH will spontaneously
dissociate
These estimates tell us nothing about the
reaction rate ! (the process of transformation of
diamond into graphite also corresponds to
negative ?G -2.9 kJ/mol, but our experience
tells us that this process is extremely slow).
5
Problem
Molar values of ?H and ?S for the reaction of
dissolving of NH4Cl in water at standard
conditions (P1 bar, T298 K) are 34.7 kJ/mol and
0.167 kJ/(Kmol), respectively. (1) Does the
reaction proceed spontaneously under these
conditions? (2) How does the entropy of the
environment and the Universe change in this
reversible process?
?G lt0, thus the reaction proceeds spontaneously
The reversible energy (heat) for this process
6
Oxidation of Methane
Consider the reaction of oxidation of methane
For this reaction ?H -164 kJ/mol, ?S -162
J/mol?K.

• Find the temperature range where this reaction
  proceeds spontaneously.
• Calculate the energy transferred to the
  environment as heat at standard conditions (T
  298 K, P 1 bar) assuming that this process is
  reversible.
• Calculate the change in the entropy of
  environment, ?Senv, at standard conditions
  assuming that this process is reversible. What is
  the total entropy change for the Universe (the
  system environment) if the process is
  reversible?

(a) For this reaction to proceed spontaneously,
?G must be negative
(b) The energy transferred to the environment as
heat
(c)
For reversible processes
7
Glucose Oxidation
Mammals get the energy necessary for their
functioning as a result of slow oxidation of
glucose
At standard conditions, for this reaction ?H
-2808 kJ/mol, ?S 182.4 J/mol?K.
Thus, at T298K, this reaction will proceed
spontaneously.
If this process proceeds as reversible at
P,Tconst, the maximum other work (chemical,
electrical, etc.) done by the system is
This work exceeds the energy released by the
system (?H -2808 kJ/mol). Clearly, some energy
should come from the environment. For reversible
processes
Thus, the environment transfers to the system
54.4 kJ/mol as the thermal energy (heat). The
system transforms into work both the energy
released in the system (?H) and the heat received
from the environment (qenv). For all reactions
that are characterized by enthalpy decrease and
entropy increase, Wmax exceeds ?H. Interestingly
that the Nature selected this process as a source
of work it not only releases a great deal of
energy, but also pumps the energy out of the
environment!
8
The Rates of Chemical Reactions
reac- tants
The rates of both direct and reverse rections
are governed by their Boltzmann factors (the
activation over the potential barrier). Thus,
each rate is an exponential function of T.
Catalysts reduce the height of an activation
barrier.
products
reaction coordinate
The rate is proportional to the probability of
collisions between the molecules (concentration
of reactants).
Chemical equilibrium dynamical equilibrium, the
state in which a reaction proceeds at the same
rate as its inverse reaction.
association
equilibrium
rate
dissociation
t
9
Chemical Equilibrium
This plot shows that despite the fact that a
particular reaction could be energetically
favorable, it hardly ever go to comletion. At any
non-zero T, there is a finite concentration of
reactants.
association
equilibrium
rate
dissociation
t
This can be understood using the concept of the
minimization of the Gibbs free energy in
equilibrium. The reason for the incompleteness
of reactions is intermixing of reactants and
products. Without mixing, the scenario would be
straightforward the final equilibrium state
would be reached after transforming 100 of
reactants into products. However, because the
products are intermixed with the reactants,
breaking just a few products apart into the
reactant molecules would increase significantly
the entropy (remember, there are infinite slopes
of G(x) at x 0,1), and that shifts the
equilibrium towards x lt 1.
GA
no mixing
GB
ideal mixing
x ?
reactants
products of reaction
chemical equilibrium (strongly shifted to the
right, but still there is a finite concentration
of reactants)
10
Chemical Equilibrium (cont.)
Lets consider a general chemical equation
stoichiometric coefficients
reactants
products
The sign of coefficients ai is different for the
reactants and for the products
The numbers of different kinds of molecules, Ni,
cannot change independently of each other the
equation of chemical reaction must be satisfied
dNi must be proportional to the numbers of
molecules appearing in the balanced chemical
equation
Here ? is a constant of proportionality, dNigt0
(dNilt0) for molecules formed (disappeared) in the
reaction.
In equilibrium, at fixed T and P, The Gibbs free
energy is at minimum
- the general condition for chemical equilibrium
Combining with the expression for dNi
The chemical potentials ?i are functions of T, P,
and all Ni. Hence this condition implies that in
equilibrium, there is a definite connection
between the mean numbers of molecules of each
kind. In principle, the statistical physics
allows one to calculate the chemical potentials
?i and thus to deduce explicitly the connection
between the numbers Ni.
11
Chemical Equilibrium between Ideal Gases
Lets consider the reaction that occurs in the
gas phase, and assume that each reactant/product
can be treated as an ideal gas. For this case, we
know ? ?(T,P).
Example transformation of the nitrogen in air
into a form that can be used by plants
ammonia
The chemical potential of an ideal gas
?0 represents the chemical potential of a gas in
its standard state, when its partial pressure
is P P 0 (usually P 0 1 bar).
In equilibrium
x NA

• the tabulated change in G
• for this reaction at P0 1 bar

12
The Law of Mass Action
the equilibrium constant K
In general, for a reaction
- all pressures are normalized by the standard
pressure P0
The product of the concentration of the reaction
partners with all concentrations always taken to
the power of their stoichiometric factors, equals
a constant K which has a numerical value that
depends on the temperature and pressure. In
particular, - the exponential temperature
dependence of the equilibrium constant K is due
to the Boltzmann factor
Generalization of this law for the concentrations
of the reaction partners in equilibrium (not
necessarily in the gas phase) is known as the law
of mass action (Guldberg-Waage, 1864)
13
Ammonia Synthesis
At T 298K and P 1 bar, ?G -32.9 kJ for
production of two moles of ammonia
Thus, the equilibrium is strongly shifted to the
right, favoring the production of ammonia from
nitrogen and hydrogen.
The calculation of the equilibrium constant K is
only the first step in evaluating the reaction
(e.g., its usefulness for applications). However,
the value of K tells us nothing about the rate of
the reaction. For this particular reaction, at
the temperatures below 7000C, the rate is
negligible (remember, the rapture of N-N and H-H
bonds is an activation process). To increase the
rate, either a high temperature or a good
catalist is required.
Haber Process, developed into an industrial
process by C. Bosch - a major chemical
breakthrough at the beginning of the 20th century
(1909) T 5000C, P 250 bar, plus a catalist
(!!!). At this temperature, K 6.910-5 (the
drop of K can be calculated using vant Hoffs
equation and ?H0 -46 kJ, see Pr. 5.86). To shift
the reaction to the right (higher concentration
of the product), a very high pressure is needed.
14
Example of Application of the Law of Mass Action
Lets look at a simple reaction
Notice that we have the same of moles on both
sides of the reaction equation. We start with
n0H2  and n0CO2  moles of the reacting gases and
define as the yield y the number of moles of H2O
that the reaction will produce at equilibrium
equilibrium concentrations of H2 and CO2
The mass action law requires
This is a quadratic equation with respect to y,
the solution is straightforward but messy. What
kind of starting concentrations will give us
maximum yield? To find out, we have to solve the
equation dy/dn0H2 0. The result
- maximum yield is achieved if you mix just the
right amounts of the starting stuff. This result
is always true, even for more complicated
reactions.
15
The Temperature Dependence of K (vant Hoff Eq.)
Its important to know how the equilibrium
concentrations are affected by temperature (Pr.
5.85). We also need this result for solving
Problems 5.86 and 5.89.
Lets find the partial derivative of lnK with
respect to T at Pconst
For either the reactants or the products,
vant Hoffs equation
?H 0 is the enthalpy change of the reaction. If
?H 0 is positive (if the reaction requires the
absorption of heat), then higher T shifts the
reaction to the right (favor higher
concentrations of the products). For the
exothermic reactions, the shift will be to the
left (higher concentration of the reactants).
16
Chemical Equilibrium in Dilute Solutions
Water dissociation
GA
Under ordinary conditions, the equilbrium is
strongly shifted to the left, but still there is
a finite concentration of ions H and OH-
dissolved in water. In equilibrium
no mixing
GB
ideal mixing
Assuming the solution is very dilute
0
1
x ?
HOH-
H2O
the shift is exaggerated ?xeq 110-7
where ?0 are the chemical potentials for the
substance in its standard state pure liquid
for the solvent, 1 molal for the solutes. This
differs from the reactions in the gas phase,
where the standard state corresponds to the
partial pressure 1 bar.
In the final equation, partial pressures are
replaced with the molalities. Most importantly,
the water concentration vanishes from the left
side its standard concentration remains 1
because the number of dissociated molecules is
tiny.
The 7 is called the pH of pure water.
17
Chemical Equilibrium between Gas and Its Dilute
Solution
The same technique can be applied to the
equilibrium between molecules in the gas phase
and the same molecules dissolved in a solvent.
Example oxygen dissolved in water.
The Gibbs free energy change ?G0 for this
reaction is for one mole of O2 dissolved in 1
kG of water at P 1bar and T 298 K.
In equilibrium
For O2 gas
For O2 dissolved in water
Henrys law (the amount of dissolved gas is
proportional to the partial pressure of this gas)
For O2 in water
PO2 0.2 bar, msolute 1.310-3 x 0.2 2.6
10-4 ? equivalent of 6.6 cm3 of O2 gas at normal
conditions (1 mol at P1 bar 25 liters).