Chapter 14 :- Chemical and Phase Equilibrium¶
Example 14.1 Page no-688
In [10]:
# Given:-
# The reaction is CO + .5O2 ----> CO2
# Part(a)
T = 298.0 # in kelvin
Rbar = 8.314 # universal gas constant in SI units
# From table A-25
hfbarCO2 = -393520.0 # in kj/kmol
hfbarCO = -110530.0 # in kj/kmol
hfbarO2 = 0 # in kj/kmol
deltahbarCO2 = 0 # in kj/kmol
deltahbarCO = 0 # in kj/kmol
deltahbarO2 = 0 # in kj/kmol
sbarCO2 = 213.69 # in kj/kmol.K
sbarCO = 197.54 # in kj/kmol.K
sbarO2 = 205.03 # in kj/kmol.K
# From table A-27
logKtable = 45.066
# Calculations
import math
deltaG = (hfbarCO2-hfbarCO-.5*hfbarO2) + (deltahbarCO2-deltahbarCO-.5*deltahbarO2) - T*(sbarCO2-sbarCO-.5*sbarO2)
lnK = -deltaG/(Rbar*T)
logK = (1/math.log(10))*lnK
# Results
print '-> Part(a) the value of equilibrium constant expressed as log10K is: '
print logK
print '-> The value of equilibrium constant expressed as log10K from table A-27 is: '
print logKtable
# Part(b)
T = 2000.0 # in kelvin
# From table A-23
hfbarCO2 = -393520.0 # in kj/kmol
hfbarCO = -110530.0 # in kj/kmol
hfbarO2 = 0 # in kj/kmol
deltahbarCO2 = 100804-9364 # in kj/kmol
deltahbarCO = 65408 - 8669 # in kj/kmol
deltahbarO2 = 67881 - 8682 # in kj/kmol
sbarCO2 = 309.210 # in kj/kmol.K
sbarCO = 258.6 # in kj/kmol.K
sbarO2 = 268.655 # in kj/kmol.K
# Calculations
deltaG = (hfbarCO2-hfbarCO-.5*hfbarO2) + (deltahbarCO2-deltahbarCO-.5*deltahbarO2) - T*(sbarCO2-sbarCO-.5*sbarO2)
lnK = -deltaG/(Rbar*T)
logK = (1/math.log(10))*lnK
# From table A-27
logKtable = 2.884
# Results
print '-> Part(b) the value of equilibrium constant expressed as log10K is: '
print logK
print '-> The value of equilibrium constant expressed as log10K from table A-27 is: '
print logKtable
Example 14.2 Page no-690
In [11]:
# Given:-
# Applying conservation of mass, the overall balanced chemical reaction equation is
# CO + .5O2 -------> zCO + (z/2)O2 + (1-z)CO2
# At 2500 K, Table A-27 gives
log10K = -1.44
# Part(a)
p = 1.0 # in atm
# Calculations
import math
K = (10.0)**(log10K) # equilibrium constant
# Solving equation K = (z/(1-z))*(2/(2 + z))^.5 *(p/1)^.5 gives
z = 0.129
yCO = 2.0*z/(2.0 + z)
yO2 = z/(2.0 + z)
yCO2 = 2.0*(1.0 - z)/(2.0 + z)
# Results
print '-> Part(a) mole fraction of CO is: ',round(yCO,3)
print '-> Mole fraction of O2 is: ',round(yO2,3)
print '-> Mole fraction of CO2 is: ',round(yCO2,3)
# Part(b)
p = 10.0 # in atm
# Solving equation K = (z/(1-z))*(2/(2 + z))^.5 *(p/1)^.5 gives
z = 0.062
yCO = 2.0*z/(2.0 + z)
yO2 = z/(2.0 + z)
yCO2 = 2.0*(1.0 - z)/(2.0 + z)
# Results
print '-> Part(b) mole fraction of CO is: ',round(yCO,3)
print '-> Mole fraction of O2 is: ',round(yO2,3)
print '-> Mole fraction of CO2 is: ',round(yCO2,3)
Example 14.3 Page no-691
In [12]:
# Given:-
yCO = 0.298
p = 1 # in atm
pref = 1 # in atm
# With this value of K, table A-27 gives
T = 2881
# Calculations
# Solving yCO = 2z/(2 + z)
z = 2*yCO/(2 - yCO)
K = (z/(1-z))*(z/(2 + z))**.5*(p/pref)**.5
# Result
print '-> The temperature T of the mixture in kelvin is: ',T
Example 14.4 Page no-692
In [13]:
# Given:-
# For a complete reaction of CO with the theoretical amount of air
# CO + .5 O2 + 1.88N2 -----> CO2 + 1.88N2
# Accordingly, the reaction of CO with the theoretical amount of air to form CO2, CO, O2, and N2 is
# CO + .5O2 + 1.88N2 ---> zCO + z/2 O2 + (1-z)CO2 + 1.88N2
K = 0.0363 # equilibrium constant the solution to Example 14.2
p =1.0 # in atm
pref = 1.0 # in atm
# Calculations
# Solving K = (z*z^.5/(1-z))*((p/pref)*2/(5.76+z))^.5 gives
z = 0.175
yCO = 2.0*z/(5.76 + z)
yO2 = z/(5.76 + z)
yCO2 = 2.0*(1.0-z)/(5.76 + z)
yN2 = 3.76/(5.76 + z)
# Results
print '-> The mole fraction of CO is: ',round(yCO,3)
print '-> The mole fraction of O2 is: ',round(yO2,3)
print '-> The mole fraction of CO2 is: ',round(yCO2,3)
print '-> The mole fraction of N2 is: ',round(yN2,3)
Example 14.5 Page no-692
In [14]:
# Given:-
# Applying the conservation of mass principle, the overall dissociation reaction is described by
# CO2 ----> zCO2 + (1-z)CO + ((1-z)/2)O2
p = 1.0 # in atm
pref = 1.0 # in atm
# At 3200 K, Table A-27 gives
log10k = -.189
# Solving k = ((1-z)/2)*((1-z)/(3-z))^.5 gives
z = 0.422
# Calculations
k = 10**log10k
# From tables A-25 and A-23
hfbarCO2 = -393520.0 # in kj/kmol
deltahbarCO2 = 174695-9364 # in kj/kmol
hfbarCO = -110530.0 # in kj/kmol
deltahbarCO = 109667-8669 # in kj/kmol
hfbarO2 = 0 # in kj/kmol
deltahbarO2 = 114809-8682 # in kj/kmol
hfbarCO2r = -393520.0 # in kj/kmol
deltahbarCO2r = 0 # in kj/kmol
Qcvdot = 0.422*(hfbarCO2 + deltahbarCO2) + 0.578*(hfbarCO + deltahbarCO) + 0.289*(hfbarO2 + deltahbarO2)- (hfbarCO2r + deltahbarCO2r)
# Result
print '-> The heat transfer to the reactor, in kJ per kmol of CO2 entering is: '
print Qcvdot
Example 14.8 Page no-701
In [15]:
%matplotlib inline
In [16]:
# Given:-
# The ionization of cesium to form a mixture of Cs, Cs+, and e- is described by
# Cs ----> (1-z)Cs + zCs+ + Ze-
K = 15.63
z = 0.95
pref =1 # in atm
# Calculation
p = pref*K*((1-z**2)/z**2)
# Results
print '-> The pressure if the ionization of CS is 95 percent complete is: ',p,'atm'
x = []
y = []
from numpy import linspace
from pylab import plot, show
x = linspace(0,10,100)
for i in range(0,100):
y.append(i)
y[i]= 100*((1/(1+x[i]/K))**0.5)
plot(x,y)
xlabel("Pressure (atm)")
ylabel("Ionization")
show()
Example 14.9 Page no-703
In [17]:
# Given:-
# The overall reaction can be written as
# CO2 + .5O2 + .5N2 ----> aCO + bNO + (1-a)CO2 + .5(1+a-b)O2 + .5(1-b)N2
# At 3000 K, Table A-27 provides
log10K1 = -0.485 # equilibrium constant of the reaction CO2 <--> CO + .5O2
log10K2 = -0.913 # equilibrium constant of the reaction .5O2 + .5N2 <-->NO
# Solving equations K1 = (a/(1-a))*((1+a-b)/(4+a))^.5 and K2 = 2b/((1+a-b)*(1-b))^.5
a = 0.3745
b = 0.0675
# Calculations
K1 = 10**log10K1
K2 = 10**log10K2
# Result
print '-> The composition of the equilibrium mixture, in kmol per kmol of CO2 present initially, is then 0.3745CO, 0.0675NO, 0.6255CO2, 0.6535O2, 0.4663N2.'
Example 14.10 Page no-707
In [18]:
# Given:-
# With data from Table A-2 at 20C,
vf = 1.0018e-3 # in m^3/kg
psat = 0.0239 # in bar
p = 1.0 # in bar
T = 293.15 # in kelvin
Rbar = 8.314 # universal gas constant in SI units
M = 18.02 # molat mass of water in kg/kmol
e=2.715
# Calculations
pvbypsat = e**(vf*(p-psat)*10**5/((1000*Rbar/M)*T))
percent = (pvbypsat-1)*100
# Result
print '-> The departure, in percent, of the partial pressure of the water vapor from the saturation pressure of water at 20 is: ',round(percent,3)