Download PHYSICAL CHEMISTRY ERT 108 Semester II 2010

PHYSICAL CHEMISTRY
ERT 108
Semester II 2010/2011
Huzairy Hassan
School of Bioprocess Engineering
UniMAP
Reaction Equilibrium in
Ideal Gas Mixtures
Introduction
Calculate the equilibrium composition for an
ideal-gas reaction from the initial
composition, the temperature and pressure,
and
.
To apply the equilibrium condition
to
an ideal-gas reaction, we need to relate the
chemical potential of a component of an
ideal gas mixture to observable properties.
Chemical Potentials in an Ideal
Gas Mixture
Chemical Potential of a Pure Ideal Gas, µ
- µ for a pure gas depends on T and P only.
- Since for reaction equilibrium  mostly
studied at constant T while the partial pressure
of the reacting gases vary.
 variation of µ with pressure.
The Gibbs equation for dG for a fixed amount of
substance is:
and division by the number of moles of the pure ideal
gas gives;
Since the chemical potential µ of a pure substance
equals
(Eq 4.86), for constant T, this equation
becomes;
If the gas undergoes an isothermal change of state from
pressure P1 and P2, integration gives;
)
Figure 6.1 plots
at fixed T.
For a pure ideal gas;
and Hm is
independent of pressure.
So, the pressure dependence
of µ in the figure is due to the
change of Sm with P.
In the zero pressure, S
becomes infinite, and
µ goes to - .
Chemical Potentials in an Ideal Gas Mixture
Definition;
- An ideal gas mixture is a gas mixture having the
following properties:
(1) the equation of state
is obeyed for
all temperatures, pressures, and compositions (ntot
is the total number of moles of gas)
(2) if the mixture is separated from pure gas i (i is
any one of the mixture’s component) by a thermally
conducting rigid membrane permeable to gas i only,
then at equilibrium the partial pressure
of gas
i in the mixture is equal to the pressure of the puregas-i system.
The standard state of component i of an ideal gas
mixture at temperature T is defined to be pure ideal
gas i at T and pressure P⁰ ≡ 1 bar.
Let
be the chemical potential of gas i in the
mixture, and let
be the chemical potential
of the pure gas in equilibrium with the mixture
through the membrane.
For phase equilibrium,
From condition (2),
at equilibrium.
Therefore for ideal gas mixture;
(Eq. 6.3)
From Equations 6.2 and 6.3;
Is the fundamental
thermodynamic equation
for an ideal gas mixture.
Where
is the partial pressure of gas i in the mixture, and
is the chemical potential of pure ideal gas i at the
standard pressure of 1 bar and at the same temperature T as
the mixture.
Ideal-Gas Reaction Equilibrium
Now, we specialize to the case where all reactants
and products are ideal gasses.
For the ideal-gas reaction:
(Eq. 6.14)
(Eq. 6.15)
is the standard equilibrium constant (or the standard pressure
equilibrium constant) for the ideal-gas reaction.
For example, for the ideal-gas reaction;
From Eq. 6.7;
⨳
The standard equilibrium constant
(Eq. 6.18)
is dimensionless.
Example 6.1 - Finding
equilibrium composition.
and ΔGº from the
Solution:
Since 0.711 mmol of CS2 was formed, so, for H2 (the same);
4(0.711 mmol) = 2.84 mmol of H2 was formed.
For CH4 (0.711 mmol reacted), 5.48 mmol – 1(0.711) mmol =
4.77 mmol CH4 present at equilibrium.
For H2S, 11.02 mmol – 2(0.711) mmol = 9.60 mmol H2S present
at equilibrium.
To find
, we need partial pressure . We have
P = 762 torr and xi are the mole fractions.
and
At equilibrium;
we have total moles = 9.60 + 4.77 + 2.84 + 0.711 = 17.92
and partial pressure,
For the
;
⨳
Temperature Dependence of the
Equilibrium Constant
Differentiation of
Use of
with respect to T;
gives;
(pressure fixed at 1 bar)
From Equation 6.32 & 6.33;
Where
is the reaction’s standard entropy change, then;
Since ΔGº = ΔHº - T ΔSº, then;
van’t Hoff equation
Multiplication of Eq. 6.36 by dT and integration from T1 to
T2 gives;
*
Example 6.2
Find
Change of
with T
at 600 K for
(a) Using the approximation that ΔHº is
independent of T;
(b) Using the approximation that
is
independent of T
Solution:
(a)From Appendix data, for NO2(g) and N2O4(g);
= 57.20 kJ/mol
= 4730 J/mol
From
ΔG⁰ = - RT ln
=
= e – ((4730 J/mol) / (- 8.314 J/mol K x 298 K))
= 0.148
From Eq. 6.39;
⨳
(b)
is assumed independent of T, then;
Appendix data gives
, then
⨳
Figure 6.6
(a) – plots ΔHº, ΔSº, ΔGº and R ln Kºp vs. T
(b) - plots ln Kºp vs. 1/T
Example 6.3
ΔH⁰ from K⁰P versus T data
Use Figure 6.6b to estimate ΔH⁰ for
for temperatures in the range 300K to 500K.
Solution:
Since only an estimation, consider the line is straight.
We have 2 points;
T-1 = 0.0040 K-1 , ln Kºp = 20.0
T-1 = 0.0022 K-1 , ln Kºp = 0
Hence, the slope is;
(20 – 0) / (0.004 K-1 - 0.0022 K-1)
= 1.1 x 104 K
The slope of plot ln Kºp vs. K-1 is -ΔHº / R, so
ΔH⁰ = -R X slope = -(1.987 cal/mol K)(1.1 x 104 K)
= -22 kcal/mol
⨳
(in agreement with Figure 6.6a)
Ideal–gas Equilibrium Calculations
Aims- find the equilibrium composition of an
ideal-gas reaction mixture and
- Equilibrium composition is related to the initial
composition by a single variable, the equilibrium extent
of reaction
;
where ni,0 is the number of moles of substance i present at the start
of reaction.
- ξ measures how much reaction has occurred.
The specific steps to find the equilibrium composition of an
ideal-gas reaction mixture are as follows;
Example 6.4 Equilibrium composition at fixed T and P
Find the equilibrium composition.
Solution:
Step 1
Use Appendix data to get
;
=[2
(NO2, g)] - [
(N2O4, g)]
= 2(51.31) – 97.89 = 4.73 kJ/mol
Step 2
Step 3
Step 4
Step 5
Step 6
Thank you