Download Lecture 18. Chemical Equilibrium (Ch. 5)

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,T=const) or enthalpy change (T,P=const), and (b) the entropy change:
G = U + PV − TS
For a reaction to be energetically favorable, the Gibbs energy for products should be
lower than the Gibbs energy for reactants.
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.
r
ΔH
H
reactants
s
ΔH
ΔH
the enthalpy
released
in the reaction
at P,T=const
products
reaction “coordinate”
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:
δ W = − PdV + δ Wother → dG = (δ Wother )T , P
CH4 + 2O 2 → CO2 + 2H2O
H
H-C-H
H
O=O
O=O
O=C=O
H-O-H
H-O-H
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
r
s
Δ H , for the reverse one - Δ H .
The reactions are characterized by “directionality” (which free energy is lower,
reactants or products), energy release, and rate.
“Directionality” of Chemical Reactions
ΔG = ΔH − TΔS
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 ΔG<0 indicates that the reaction will proceed spontaneously.
ΔH
ΔS
ΔG = ΔH-TΔS
-
+
-
reaction will go spontaneously
+
-
+
reaction needs input of energy to go
-
-
+
+
The sign of ΔG
depends on the
relative values of ΔH,
ΔS and T
If reaction is spontaneous, the
reaction
is
entropy-driven:
the
increase in disorder is sufficient to
drive the reaction despite the fact that
it absorbs heat from its surroundings.
2C3H5N3O9 → 6CO 2 + 5H2O + 3N2 + 0.5O 2
- 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.
H2 → H + H - though S increases, the equilibrium is shifted to the left (ΔH >TΔS) at 300K
2H2 + O 2 → 2H2O
- though S decreases, the equilibrium is shifted to the right
because the decrease of ΔH “overpower” the increase of -TΔS
Several channels of the reaction between CO and H2 at 300K
reaction
ΔH
kJ/mol
-T ΔS
kJ/mol
ΔG
kJ/mol
CO+H2→C+H2O
-130
+39
-91
CO+2H2→CH3OH
-85
+65
-20
2CO+2H2→CH3COOH
-257
+147
-110
Examples:
Several channels of the reaction between CO and H2 at 600K
reaction
ΔH
kJ/mol
-T ΔS
kJ/mol
ΔG
kJ/mol
CO+H2→C+H2O
-130
+78
-52
CO+2H2→CH3OH
-85
+130
45
2CO+2H2→CH3COOH
-257
+394
137
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 ΔG = -2.9
kJ/mol, but our experience tells us that this process is extremely slow).
Exchange with the Environment
The difference between the internal energies (V,T –const) or enthalpies (P,T – const)
of reactants and products represent the “heat” of reaction:
(δQ )T , P = ΔH
< 0 – exothermic, > 0 - endothermic
If an exothermic reaction proceeds spontaneously, it means that (a) it is entropydriven, and (b) the system gets some energy necessary for the reaction from its
environment (heat bath). The latter process happens whenever the entropy of the
ptoducts is greater than that for the reactants.
“Universe” = a nonisolated system + its environment
For reversible processes,
ΔS"Universe" = 0
environment
system
ΔSUniverse = ΔS system + ΔS env = 0
ΔS in the system
ΔS in the environment
If the entropy of a system is increased in the process of
chemical reaction, the entropy of the environment must
decrease. The associated with this process heat transfer from
the environment to the system in a reversible process:
δ q = TΔS env < 0
Thus, an entropy increase “pumps” some energy out of the environment into the system.
Problem
Molar values of ΔH and ΔS for the reaction of dissolving of NH4Cl in water at
standard conditions (P=1 bar, T=298 K) are 34.7 kJ/mol and 0.167 kJ/(K·mol),
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?
ΔG = ΔH − TΔS = 34.7 − 298 × 0.167 = −15.1 J/mol < 0
ΔG <0, thus the reaction proceeds spontaneously (despite its endothermic character).
High temperatures favor the spontaneity of endothermic processes.
The “reversible” energy (heat) for this process:
revers
qsyst
= TΔS syst = 298 × 0.167 = 49.8kJ / mol
ΔS env =
qsys
qenv
49.8 kJ / mol
=−
=−
= −0.167 kJ /( K × mol )
298 K
T
T
ΔSUniverse = ΔS syst + ΔS env. = 0.167 kJ /( K × mol ) − 0.167 kJ /( K × mol ) = 0
Oxidation of Methane
Consider the reaction of oxidation of methane:
CH4 + 0.5 O 2 → CH3OH
For this reaction ΔH = -164 kJ/mol, ΔS = -162 J/mol×K.
(a)
(b)
(c)
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, ΔSenv, 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, ΔG must be negative:
ΔG = ΔH − TΔS < 0
T<
164 kJ / mol
≈ 103 K
162 J / (mol × K )
(b) The energy transferred to the environment as heat:
revers
qsyst
= ΔH − ΔG = TΔS = 298 K × (− 162 J / K × mol ) = −48.28 kJ / mol
qenv = − qsyst = 48.28kJ / mol
qenv 48.28 kJ / mol
Δ
S
=
=
= 162 J / K × mol
env
(c)
T
298 K
ΔS total = ΔS system + ΔS env = 0
For reversible processes:
Glucose Oxidation
Mammals get the energy necessary for their
functioning as a result of slow oxidation of glucose :
C 6 H12 O 6 + 6O 2 → 6CO 2 + 6H 2 O
At standard conditions, for this reaction ΔH = -2808 kJ/mol, ΔS = 182.4 J/mol×K.
ΔG = ΔH − TΔS = −2808 × 103 − 298 × 182.4 = −2862.4 × 103 J/mol < 0
Thus, at T=298K, this reaction will proceed spontaneously.
If this process proceeds as reversible at P,T=const, the maximum “other” work
(chemical, electrical, etc.) done by the system is:
3
Wmax = − ΔG = 2862 × 10 J/mol
This work exceeds the energy released by the system (ΔH = -2808 kJ/mol). Clearly,
some energy should come from the environment. For reversible processes:
qenv = TΔS env = 298 × (− 182.4 ) J / mol = −54.4 kJ / mol
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 (ΔH) and the
heat received from the environment (qenv). For all reactions that are characterized by
enthalpy decrease and entropy increase, Wmax exceeds ΔH. 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!
r
ΔH
H
reactants
s
ΔH
ΔH
products
reaction “coordinate”
The Rates of Chemical Reactions
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.
The rate is proportional to the probability of collisions between the molecules
r
(concentration of reactants).
⎛
⎝ k BT ⎠
s
⎛ ΔH ⎞
⎟
= [conc.produ cts]× exp⎜⎜ −
⎟
⎝ k BT ⎠
association
rate
Chemical equilibrium = dynamical equilibrium,
the state in which a reaction proceeds at the same
rate as its inverse reaction.
⎞
[conc.reactants]× exp⎜⎜ − ΔH ⎟⎟
equilibrium
dissociation
t
Chemical Equilibrium
association
rate
equilibrium
dissociation
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.
t
ΔG = ΔH − TΔS
GA
no
m
ixin
g
ideal mixing
reactants
x→
GB
products
of reaction
chemical equilibrium (strongly
shifted to the right, but still there is
a finite concentration of reactants)
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 < 1.
Let’s consider a general chemical equation:
Chemical Equilibrium (cont.)
m
∑a B
aR1 R1 + aR2 R2 + a R3 R3 .... ↔ a P1 P1 + a P2 P2 + a P3 P3 ....
reactants
i =1
i
i
=0
stoichiometric coefficients
products
The sign of coefficients ai is different for the reactants and for the products:
H2 + Cl2 = 2HCl
⇒ H2 + Cl2 − 2HCl = 0
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:
dN i = λai
Here λ is a constant of proportionality, dNi>0 (dNi<0)
for molecules formed (disappeared) in the reaction.
dN HCl : dN H 2 : dN Cl2 = 2 : −1 : −1
In equilibrium, at fixed T and P, The Gibbs free energy is at minimum: dG =
m
∑ μ dN
i =1
m
Combining with the expression for dNi :
2 μ HCl − μ H 2 − μCl2 = 0
μ HCl
(
1
= μ H 2 + μCl2
2
)
∑a μ = 0
i =1
i
i
i
i
=0
- the general condition
for chemical equilibrium
The chemical potentials μi 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
μi and thus to deduce explicitly the connection between the numbers Ni.
Chemical Equilibrium between Ideal Gases
Let’s 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 μ = μ(T,P).
Example: transformation of the nitrogen in air into a form that can be used by plants
N 2 + 3H 2 ↔ 2NH 3
ammonia
The chemical potential of an ideal gas:
∂μ
∂ ⎛G⎞
V k BT
=
=
=
⎜ ⎟
∂P ∂P ⎝ N ⎠T , N N
P
μ (T , P ) = μ (T , P 0 ) + k BT ln⎜
⎛ P ⎞
⎛ P ⎞
0
(
)
=
+
μ
T
k
T
ln
⎜ 0⎟
B
0 ⎟
P
⎝ ⎠
⎝P ⎠
μ0 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:
⎛ PN
k BT ln⎜⎜ 02
⎝P
x NA
μ
0
N2
⎛ PN 2
+ k BT ln⎜⎜ 0
⎝P
⎞
⎛P
⎟⎟ + 3k BT ln⎜⎜ H02
⎠
⎝P
( )
( )( )
⎛ P P 3
N2
H2
RT ln ⎜
2
⎜ P0 2 P
NH
3
⎝
⎛ PH 2
⎞
0
⎟⎟ + 3μ H 2 + 3k BT ln⎜⎜ 0
⎝P
⎠
⎛P
⎞
⎟⎟ − 2k BT ln⎜⎜ NH0 3
⎠
⎝ P
⎛ PNH 3
⎞
0
⎟⎟ = 2 μ NH 3 + 2k BT ln⎜⎜ 0
⎠
⎝ P
( )
( )( )
⎛ P P 3 ⎞
⎞
⎟ = 2 μ 0 − 3μ 0 − μ 0
⎟⎟ = k BT ln⎜ N 22 H 2
NH 3
H2
N2
⎜ P0 P 2 ⎟
⎠
NH 3
⎝
⎠
⎞
⎟ = N 2 μ 0 − 3 μ 0 − μ 0 = Δ G 0 - the tabulated change in G
A
NH 3
H2
N2
for this reaction at P0 = 1 bar
⎟
⎠
(
⎞
⎟⎟
⎠
)
The Law of Mass Action
(P ) (P )
P (P )
( )
( )( )
⎛ P P 3 ⎞
N2
H2
⎟ = ΔG 0
RT ln⎜
⎜ P0 2 P 2 ⎟
NH 3
⎝
⎠
0 2
2
NH 3
3
H2
N2
(P ) (P )
P (P )
0 2
⎛ ΔG 0 ⎞
⎟⎟
= exp⎜⎜ −
⎝ RT ⎠
N2
2
NH 3
3
H2
=K
the equilibrium constant K
m
In general, for a reaction
∑a B
i =1
i
i
=0
⎛ ΔG 0 ⎞
⎟⎟ = K P 0 , T
P P P ... = exp⎜⎜ −
⎝ RT ⎠
a1
1
a2
2
a3
3
(
(
)
ΔG 0
ln P P P ... = −
RT
)
a1
1
a2
2
a3
3
- 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:
0
⎛ ΔG
K = exp⎜⎜ −
⎝ RT
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):
⎞
⎛ ΔU ⎞
⎛ ΔS ⎞
⎟⎟ = exp⎜ −
exp
⎟
⎜−
⎟
R
RT
⎠
⎝
⎠
⎝
⎠
⎛ ΔG 0 ⎞
⎟⎟ = K
N N N ... = exp⎜⎜ −
RT
⎝
⎠
a1
1
a2
2
a3
3
Ammonia Synthesis
N 2 + 3H 2 ↔ 2NH 3
(P ) (P )
P (P )
0 2
N2
2
NH 3
3
H2
⎛ ΔG ⎞
= exp⎜ −
⎟
RT
⎝
⎠
At T = 298K and P = 1 bar, ΔG = -32.9 kJ for
production of two moles of ammonia
⎛ 32.9 × 103 J ⎞
⎛ ΔG ⎞
⎟⎟ = 5.9 ⋅105
K = exp⎜ −
⎟ = exp⎜⎜
⎝ RT ⎠
⎝ (8.3 J/K )(298 K ) ⎠
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 = 6.9·10-5 (the drop
of K can be calculated using van’t Hoff’s equation and ΔH0 =-46 kJ, see Pr. 5.86). To
shift the reaction “to the right” (higher concentration of the product), a very high
pressure is needed.
Example of Application of the Law of Mass Action
Let’s look at a simple reaction
H 2 + CO 2 ↔ H 2 O + CO
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:
nH 2O = y
0
nCO = y nH 2 = nH0 2 − y nCO 2 = nCO
−y
2
0
0
0
n
=
n
=
n
+
n
∑
H2
CO 2
equilibrium concentrations of H2 and CO2
The mass action law requires:
(n
0
H2
y2
=K
0
− y nCO 2 − y
)(
)
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:
nH0 2 = n 0 / 2
0
0
0
nCO
=
n
/
2
=
n
H
2
2
- 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.
The Temperature Dependence of K (van’t Hoff Eq.)
It’s 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.
⎛ ΔG 0 ⎞
⎟⎟
K = exp⎜⎜ −
RT
⎝
⎠
Let’s find the partial derivative of lnK with respect to T at P=const
∂ΔG 0
0
T
−
Δ
G
0
⎛
⎞
∂
(ln K ) = ∂ ⎜⎜ − ΔG ⎟⎟ = − 1 ∂T 2
∂T
∂T ⎝ RT ⎠
R
T
For either the reactants or the products,
(
)
∂G 0
= −S
∂T
∂
1 − TΔS 0 − ΔH 0 − TΔS 0
ΔH 0
(ln K ) = −
=
2
∂T
R
T
RT 2
∂ΔG 0
= −ΔS 0
∂T
∂
ΔH 0 van’t Hoff’s
(ln K ) = 2 equation
∂T
RT
ΔH 0 is the enthalpy change of the reaction. If ΔH 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).
Tf
Tf
ΔH 0
∫T d (ln K ) = T∫ RT 2 dT
i
i
ΔH 0 ⎛⎜ 1 1 ⎞⎟
ln K (T f ) − ln K (Ti ) =
−
R ⎜⎝ Ti T f ⎟⎠
Chemical Equilibrium in Dilute Solutions
Water dissociation:
GA
no
m
x→
0
H++OH-
ΔG 0 = 79.9 kJ
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.
ixin
g
ideal mixing
H2O ↔ H+ + OH-
GB
μ H + μ OH = μ H O
In equilibrium:
+
−
2
Assuming the solution is very dilute:
1
0
(T , P ) − k BT
μsolvent = μsolvent
H2O
the shift is exaggerated:
Δxeq ~ 1·10-7
μsolute
N solute
0
(T , P )
≈ μsolvent
N solvent
= μ 0 (T , P ) + k BT ln msolute
0
μ H0 O = μ H0 + k BT ln mH + μ HO
+ k BT ln mHO
+
2
+
-
-
where μ0 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.
(
RT ln (mH + mHO - ) = − N A μ
0
H+
+μ
0
HO -
−μ
0
H 2O
) = − ΔG
0
mH + mHO -
⎛ ΔG 0 ⎞
⎟⎟
= exp⎜⎜ −
⎝ RT ⎠
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.
mH + = mHO - = 1 ⋅10 −7
The 7 is called the pH of pure water.
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.
O 2 (gas ) ↔ O 2 (aqueous )
ΔG 0 = 16.4 kJ
The Gibbs free energy change ΔG0 for this “reaction” is for one mole of O2 dissolved
in 1 kG of water at P = 1bar and T = 298 K.
μO
In equilibrium:
μ
0
gas
2 ( gas )
(T , P ) = μO ( aqueous ) (T , P )
2
For O2 gas:
μ gas (T , P ) = μ 0 (T , P 0 ) + k BT ln⎜
For O2 dissolved in water:
0
(
μsolute = μsolute
⎛ P ⎞
0
+ k BT ln⎜ 0 ⎟ = μsolute
+ k BT ln msolute
P
⎝ ⎠
⎛ ΔG 0 ⎞
msolute
⎟⎟
= exp⎜⎜ −
0
P/P
⎝ RT ⎠
⎛ P ⎞
0 ⎟
P
⎝ ⎠
0
T , P + k BT ln msolute
)
⎛ P / P0 ⎞
0
0
⎟⎟ = N A μ solute
RT ln⎜⎜
− μ gas
= ΔG 0
⎝ msolute ⎠
(
)
Henry’s law (the amount of dissolved gas is proportional to
the partial pressure of this gas)
⎛ ΔG 0 ⎞
⎛
16.4 ⋅103 J ⎞
⎟⎟ = exp⎜⎜ −
⎟⎟ ≈ 1.3 ⋅10 −3
For O2 in water: exp⎜⎜ −
⎝ RT ⎠
⎝ (8.3 J/K )(298 K ) ⎠
PO2 =0.2 bar, msolute = 1.3·10-3 x 0.2 = 2.6 ·10-4 ⇒ equivalent of 6.6 cm3 of O2 gas at
normal conditions (1 mol at P=1 bar ~25 liters).