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CBE 450 Chemical Reactor Fundamentals Fall, 2009 Homework Assignment #
1. Review of Mass Balances of Reacting Systems
Natural gas (assume pure methane) is being burned in air (assume 0.79 mol% N 2 and 0.21 mol%
O 2 ). Assume complete combustion, which means that all of the natural gas is burned and is
converted into carbon dioxide and water. Assume that the oxygen is provided in 100% excess.
The flow rate of natural gas is 100 mol/s. The furnace operates at steady state.
(a) Draw a diagram with all input and output streams. Indicate known and unknown flowrates.
Indicate known and unknown compositions.
(b) Determine the flowrate of air.
(c) Determine the flowrate and composition of the output gas using atomic balances.
(d) Determine the flowrate and composition of the output gas using molecular balances.
Solution
(a) Draw a diagram with all input and output streams. Indicate known and unknown flowrates.
Indicate known and unknown compositions.
(b) Determine the flowrate of air.
The balanced reaction is
CH 4 + 2O 2 CO 2 + 2H 2 O
The stoichiometrically required amount of O 2 is two moles of O 2 for every mole of CH 4. Since
the oxygen is provided in 100% excess, you actually need twice as much, or four moles of O 2 for
every mole of CH 4. Since you have 100 mol/s of CH 4 , you need 400 mol/s of O 2. To determine
the total air flowrate,
AxO 2 A 0. 21 400 mol / s
A mol / s 1905 mol / s
- 21
(c) Determine the flowrate and composition of the output gas using atomic balances.
We have five unknowns, the flowrate and four compositions of the stack gas. Therefore, we
need five equations. The first four equations are atom balances of the form:
accumulation = in – out + generation
The accumulation term is zero because the furnace is operating at steady state. The generation
term is zero because atoms are neither created or destroyed in this furnace. Therefore, the atom
balance is
in = out
N: 2 AxN (^) 2 2 SzN 2
C: FyCH (^) 4 SzCO 2
H: 4 Fy (^) CH (^) 4 2 SzH 2 O
O: 2 AxO (^) 2 2 SzO 2 2 SzCO 2 SzH 2 O
The fifth equation is the sum of the mole fractions is unity
sum of the mole fractions is unity: 1 zO (^) 2 zCO 2 zH 2 O zN 2
Now solve.
in = out
N: SzN 2 AxN 2 1505
C: SzCO 2 FyCH 4 100
N 2 : 0 AxN (^) 2 SzN 2
CO 2 : 0 0 SzCO 2 ( 1 ) 100
H 2 O: 0 0 SzH 2 O ( 2 ) 100
O 2 : 0 AxO 2 SzO 2 ( 2 ) 100
CH 4 : 0 FyCH 4 0 ( 1 ) 100
We see that the methane balance was actually used to determine the generation term.
The fifth equation is the sum of the mole fractions is unity
sum of the mole fractions is unity: 2 2 2 2
1 zO zCO zHO zN
Now solve.
N 2 : SzN 2 AxN 2 1505
CO 2 : SzCO 2 100
H 2 O: SzH 2 O 200
O 2 : SzO 2 AxO 2 200 200
We can multiply the mole fractions sum by S and write
S SzO 2 SzCO 2 SzH 2 O SzN 2 200 100 200 1505 2005 mol / s
The compositions are thus
N:
2 ^
S
z (^) N
C:
2 ^
S
z (^) CO
H:
2 ^
S
z (^) HO
O:
2 ^
S
z (^) O
This is the same result as was obtained from the atomic balances, as it must be.
2. Molecular Description of Reaction Equilibrium
Consider the distribution of para-xylene, meta-xylene and ortho-xylene.
(a) For an ideal gas at a given temperature, what factors at the molecular level determine the
equilibrium distribution of components?
(b) At a given temperature, will the mole faction of the component with the lowest enthalpy be
greater or less than the mole fraction of the other components (all other things being equal)?
Why?
(c) At a given temperature, will the mole faction of the component with the lowest entropy be
greater or less than the mole fraction of the other components (all other things being equal)?
Why?
(d) At a given temperature, will the mole faction of the component with the lowest free energy be
greater or less than the mole fraction of the other components (all other things being equal)?
Why?
(e) As the temperature is increased, will the mole fraction of the component with the lowest
enthalpy increase or decrease (all other things being equal)? Why?
(f) As the temperature is increased, will the mole fraction of the component with the lowest
entropy increase or decrease (all other things being equal)? Why?
Solution:
(a) For an ideal gas at a given temperature, what factors at the molecular level determine the
equilibrium distribution of components?
At the molecular level, the enthalpy and the entropy can be broken into contributions from
translation of the center of the mass of the molecule rotation about the center of mass of the molecule vibration of bonds within the molecule internal rotation of internal rotors like the methyl groups in xylene
The total enthalpy and entropy (and thus the free energy) are related to these contribution. Thus
the equilibrium distribution is based on these contributions.
(b) At a given temperature, will the mole faction of the component with the lowest enthalpy be
greater or less than the mole fraction of the other components (all other things being equal)?
Why?
RT
H
R
S
RT
H T S
RT
G
K
N
N
eq
12 12 12 12 12 , 12 2
(^1) exp exp exp exp
A low free energy state is favored at equilibrium.
(e) As the temperature is increased, will the mole fraction of the component with the lowest
enthalpy increase or decrease (all other things being equal)? Why?
If state one has the lowest enthalpy, then H (^) 1 2 0 and S 1 (^) 2 0 , then the derivative of this
distribution with respect to temperature is
exp exp 2 0 12 12 12
2
1
RT
H
RT
H
R
S
N
N
T
where we neglected the temperature dependence of the enthalpies of formation. So as the
temperature increases the relative amount of N 1 decreases. In other words, the low energy state
is more favored at low temperature.
(f) As the temperature is increased, will the mole fraction of the component with the lowest
entropy increase or decrease (all other things being equal)? Why?
If state one has the lowest entropy, then S 1 (^) 2 0 and H 1 (^) 2 0 , then the derivative of this
distribution with respect to temperature is
exp exp 2 0 12 12 12
2
1
RT
H
RT
H
R
S
N
N
T
where we neglected the temperature dependence of the enthalpies of formation. So as the
temperature increases the relative amount of N 1 remains unchanged. In other words, the entropic
contribution to the distribution of components does not have a temperature dependence.
3. Review Continuum Description of Reaction Equilibrium
Consider a batch reactor initially containing nitrogen and hydrogen gases. The volume of the
batch reactor is 1 m 3
. The initial pressure is 1 atm. The temperature is kept constant at 500 K.
The initial mole fractions are 0.25 N 2 and 0.75 H 2. The relevant reaction is
N 2 3 H 2 2 NH 3
The reference enthalpies of formation are at a reference temperature of 298.15 K
H (^) f , N 2 0 kcal / mol
H (^) f , H 2 0 kcal / mol
H (^) f , NH 3 10. 96 kcal / mol
The reference Gibbs free energies of formation are at a reference temperature of 298.15 K
G (^) f , N 2 0 kcal / mol
G (^) f , H 2 0 kcal / mol
G (^) f , NH 3 3. 903 kcal / mol
In this project, you may assume that the heat capacities are constant, given by
C (^) p , N 2 7. 03 cal / mol / K
C (^) p , H 2 6. 92 cal / mol / K
C (^) p , NH 3 9. 31 cal / mol / K
(a) What are the heats of formation at 500 K for each component?
^ ^
T
T
fi fi ref pi ref
H (^) , T H , T C , dT
For constant heat capacities
H f , i T Hf , i T ref Cp , i T Tref
H f N T K 500 298 1. 42 kcal / mol
, (^2)
H f H T K 500 298 1. 40 kcal / mol
, 2 ^500 ^0.^0
H f N T K 500 298 9. 08 kcal / mol
, (^2)
(b) What are the free energies of formation at 500 K for each component. (You may assume that
the heat of formation is constant in this part (b) calculation only.)
T
T
fi fi ref ref
fi ref
dT T
H
G T T
T
T
G T 2
, , ,
For constant heats of formation,
ref
fi ref fi ref
f i T T
G T TH
T
T
G T
, , ,
24. 4 i
N No i
The number of each species is
N j No , j j
NN 2 No , N 2 N 2 6. 09 0. 5
NH 2 No , H 2 H 2 18. 3 1. 5
NNH 3 No , NH 3 NH 3 0. 0 1. 0
The reference pressure is 101325 Pa.
The final pressure is (using the ideal gas law)
V
N RT
V
NRT
p
i
o i ^
Substituting these into equations the equilibrium coefficient
1 / 2 2
3 / 2 2
3 1 / 2 2
3 / 2 2
3
1
H N
NH
H N
NH
ref
eq x x
x
x x
x
p
p K
3 / 2 1 / 2
1
3 / 2 1 / 2
1
ref ref
eq p
p
p
p K
Solving this single, non-linear algebraic equation, I find
and mole fractions
N
N
x
oj j j
,^ ^
xN
xH
xNH
