Chapter 9 : Chemical reaction Equilibria¶
Example 9.1 Page No : 570¶
In [1]:
# Variables
n_o_CH3OH = 1. ; #[mol]
n_o_H2O = 3. #[mol]
S = 0.87 ;
# Calculations
n_CH3OH = 1 - S ;
n_H2O = 2 - S ;
n_CO2 = S ;
n_H2 = 3 * S ;
n_v = n_CH3OH + n_CO2 + n_H2O + n_H2 ;
y_H2 = n_H2 / n_v ;
# Results
print "No of moles of H2 produced for 1mol of CH3OH = %.3f mol" %( n_H2)
print "Mole fraction of H2 = %.2f" %( y_H2) ;
Example 9.2 Page No : 574¶
In [6]:
import math
# Variables
del_gf_0_CO2 = -394.36 ; #[kJ/mol],From Appendix A.3
del_gf_0_H2 = 0 ; #[kJ/mol],From Appendix A.3
del_gf_0_H2O = -228.57 ; #[kJ/mol],From Appendix A.3
del_gf_0_CH3OH = -161.96 ; #[kJ/mol],From Appendix A.3
n_CO2 = 1 ;
n_H2 = 3 ;
n_CH3OH = 1 ;
n_H2O = 1 ;
T = 298.15 ; #[K]
R = 8.314 ; #[J/molK]
# Calculations
del_g0_rxn = (n_CO2 * del_gf_0_CO2 + n_H2 * del_gf_0_H2 - n_H2O * del_gf_0_H2O - n_CH3OH * del_gf_0_CH3OH) * 10**3 ; # [J/mol]
K_298 = math.exp( - del_g0_rxn / (R * T)) ;
# Results
print "The equillibrium constant K298 = %.2f "%(K_298) ;
Example 9.3 page no : 575¶
In [1]:
import math
# Variables
K_298 = 4.69
# calculations
deltaH = (-393.51) + 3 * 0 - (-241.82) - (-200.66)
lnK333_469 = round((-deltaH*1000/8.314 )*(1./333 - 1./298),2)
lnK333 = lnK333_469 + math.log(K_298)
K333 = math.e**lnK333
# Results
print "Equilibrium constant = %.2f "%K333
# Note: answer is slightly different because of rounding off error.
Example 9.4 Page No : 577¶
In [5]:
import math
# Variables
del_gf_0_CH2O = -110.0 ; #[kJ/mol],From Appendix A.2 & A.3
del_gf_0_H2 = 0 ; #[kJ/mol],From Appendix A.2 & A.3
del_gf_0_CH4O = -162.0 ; #[kJ/mol],From Appendix A.2 & A.3
del_hf_0_CH2O = -116.0 ; #[kJ/mol],From Appendix A.2 & A.3
del_hf_0_H2 = 0 ; #[kJ/mol],From Appendix A.2 & A.3
del_hf_0_CH4O = -200.7 ; #[kJ/mol],From Appendix A.2 & A.3n_CH20 = 1 ;
n_H2 = 1. ;
n_CH4O = 1. ;
n_CH2O = 1. ;
T1 = 298. #[K]
T2 = 873. ; # [K]
R = 8.314 ; #[J/molK]
Del_A = 3.302 ;
Del_B = -4.776 * 10**-3 ;
Del_C = 1.57 * 10**-6 ;
Del_D = 0.083 * 10**5 ;
# Calculations and Results
del_g_rxn_298 = n_CH2O * del_gf_0_CH2O + n_H2 * del_gf_0_H2 - n_CH4O * del_gf_0_CH4O ;
K_298 = math.exp( - del_g_rxn_298 * 10**3 / (R * T1)) ;
print "a) K_298 = %.2e As the equilibrium constant is very small very little amount of\
formaldehyde will be formed ."%(K_298) ;
#Solution(b)
del_h_rxn_298 = (n_CH2O * del_hf_0_CH2O + n_H2 * del_hf_0_H2 - n_CH4O * del_hf_0_CH4O) * 10**3 ; #[J/mol]
K_873 = K_298 * math.exp((-del_h_rxn_298 * (1/T2 - 1/T1)) / R) ;
print "b) i) K_873 = %.2f "%(K_873) ;
#Solution(c)
x = ( -del_h_rxn_298 / R + Del_A * T1 + Del_B / 2 * T1**2 + Del_C /3 * T1**3 - Del_D / T1 ) *(1/T2 - 1/T1) + Del_A * math.log(T2 / T1) + Del_B / 2 * (T2 -T1) + Del_C / 6 * (T2**2 -T1**2) + Del_D / 2 * (1/(T2**2) -1/(T1**2)) ;
K_873 = K_298 * math.exp(x) ;
print " ii) K_873 = %.2f "%(K_873) ;
# Note : answer are slightly different because of rounding off error.
Example 9.6 Page No : 581¶
In [1]:
# Variables
import math
del_g0_f_C6H6 = -32.84 ; #[kJ/mol] , From Table E9.6
del_g0_f_C2H4 = 68.15 ; #[kJ/mol] , From Table E9.6
del_g0_f_H2 = 0 ; #[kJ/mol] , From Table E9.6
del_h0_f_C6H6 = -84.68 ; #[kJ/mol] , From Table E9.6
del_h0_f_C2H4 = 52.26 ; #[kJ/mol] , From Table E9.6
del_h0_f_H2 = 0 ; #[kJ/mol] , From Table E9.6
T1 = 298.2 ; #[K]
P = 1. #[bar]
R = 8.31 ;
T2 = 1273. # [K]
# Calculations
del_g0_f_rxn = del_g0_f_C2H4 + del_g0_f_H2 - del_g0_f_C6H6 ;
K_298 = math.exp ( - (del_g0_f_rxn * 10**3) / (R * T1)) ;
del_h0_f_rxn = (del_h0_f_C2H4 + del_h0_f_H2 - del_h0_f_C6H6) * 10**3 ;
K_1273 = K_298 * math.exp( - del_h0_f_rxn / R * (1/T2 - 1/T1)) ;
x = math.sqrt( K_1273 / ( K_1273 + P)) ;
# Results
print "n_C2H6 = %.2f mol n_C2H4 = %.2f mol n_H2 = %.2f mol"%(1-x,x,x) ;
Example 9.7 Page No : 583¶
In [10]:
from matplotlib.pyplot import *
from numpy import *
from scipy.integrate import *
import math
# Variables
del_h0_f_NH3 = -46.11 ; # [kJ/mol],From table E9.7
del_h0_f_N2 = 0 # [kJ/mol],From table E9.7
del_h0_f_H2 = 0 ; # [kJ/mol],From table E9.7
del_g0_f_NH3 = -16.45 ; # [kJ/mol],From table E9.7
del_g0_f_N2 = 0 ; # [kJ/mol],From table E9.7
del_g0_f_H2 = 0 ; # [kJ/mol],From table E9.7
n_NH3 = 2. ;
n_N2 = -1. ;
n_H2 = -3. ;
A_NH3 = 3.578
B_NH3 = 3.02 * 10**-3
D_NH3 = -0.186 * 10**5 ;
A_N2 = 3.280
B_N2 = 0.593 * 10**-3
D_N2 = 0.040 * 10**5 ;
A_H2 = 3.249
B_H2 = 0.422 * 10**-3
D_H2 = 0.083 * 10**5 ;
R = 8.314 ;
T = 298. ;
T2 = 773. ;
P = 1. ; #[bat]
# Calculations
Del_h0_rxn = (n_NH3 * del_h0_f_NH3 + n_N2 * del_h0_f_N2 + n_H2 * del_h0_f_H2) * 10**3 ;
Del_g0_rxn = (n_NH3 * del_g0_f_NH3 + n_N2 * del_g0_f_N2 + n_H2 * del_g0_f_H2) * 10**3 ;
del_A = n_NH3 * A_NH3 + n_N2 * A_N2 + n_H2 * A_H2 ;
del_B = n_NH3 * B_NH3 + n_N2 * B_N2 + n_H2 * B_H2 ;
del_D = n_NH3 * D_NH3 + n_N2 * D_N2 + n_H2 * D_H2 ;
K_298 = math.exp( - Del_g0_rxn / ( R * T)) ;
K_T = K_298 * math.exp( - Del_h0_rxn / R * (1 / T2 - 1 / T)) ;
A = K_T * P**2 *27 -16 ;
B = 64 - K_T * P**2 * 108 ;
C = -64 + K_T * P**2 * 162 ;
D = -108 * K_T * P**2 ;
E = 27 * K_T * P**2 ;
mycoeff =[E , D ,C , B ,A];
M = roots([A,B,C,D,E]);
for i in range(3):
ans = isreal(M[i]) ;
if ans == True:
y = M[i] / M[i+1] - 1 ;
ans = sign(y) ;
if ans == 1 :
x = M[i]
else:
x = M[i+1]
# Results
print "a)Extent of reaction = %.3f "%(x);
#(b)
X = (-Del_h0_rxn / R + del_A * T + del_B / 2 * T**2 - del_D / T) * (1/T2 - 1/T) + del_A * math.log(T2 / T) + del_B / 2 * (T2 - T) + del_D / 2 * (1/(T2**2) - 1/(T**2) );
K_T = K_298 * math.exp(X) ;
A = K_T * P**2 *27 -16 ;
B = 64 - K_T * P**2 * 108 ;
C = -64 + K_T * P**2 * 162 ;
D = -108 * K_T * P**2 ;
E = 27 * K_T * P**2 ;
mycoeff =[E , D ,C , B ,A];
M1 = roots([A,B,C,D,E]);
for i in range(3):
ans = isreal(M1[i])
if ans == True :
y = M1[i] / M1[i+1] - 1 ;
ans = sign(y) ;
if ans == True:
x1 = M1[i]
else:
x1 = M1[i+1]
print "b) Extent of reaction = %.3f"%(x1);
print "Under these conditions we do not expect to produce an appreciable amount of ammonia ."
Example 9.9 Page No : 585¶
In [8]:
from scipy.optimize import fsolve
import math
# Variables
K_T = 1.51 * 10**-5 ;
P = 300. ; #[bar]
T = 500 + 273.2 ; #[K]
R = 8.314 ;
# Calculations and Results
def f991(k):
return ((2 * k)**2 * (4 - 2 * k)**2 / ((1 - k) * (3 - 3*k)**3)) * P**-2 - K_T
z1 = fsolve(f991,0.3)[0]
print "a) Extent of reaction = %.2f "%(z1);
P_c = [111.3 * 101325 , 33.5 * 101325 , 12.8 * 101325] ;
T_c = [405.5 , 126.2 , 33.3] ;
a = zeros(3)
b = zeros(3)
V = zeros(3)
sai = zeros(3)
for i in range(3):
a[i] = 27./ 64 * (R * T_c[i])**2 / P_c[i] ;
b[i] = (R * T_c[i]) / (8 * P_c[i]) ;
def f992(v):
return (R * T) / (v - b[i]) - a[i] / (v**2) - P * 100000 ;
V[i] = fsolve(f992,.0002)
sai[i] = math.exp( - math.log((V[i] - b[i]) * P * 10**5/ ( R * T)) + b[i] / (V[i] - b[i]) - 2 * a[i] / (R * T * V[i])) ;
def f993(k):
return ((2 * k)**2 * sai[0]**2 * (4 - 2 * k)**2 * 3 / ((1 - k) * sai[1]* (3 - 3*k)**3 * sai[2]**3 ))* P**-2 - K_T
z2 = fsolve(f993,0.3)
x = (z1 - z2) / z1 * 100 ;
print "b) Extent of reaction = %.2f "%(z2);
print " A correction of about %d%% is observed from accounting for nonideal behaviour . "%(x)
Example 9.10 Page No : 586¶
In [3]:
# Variables
import math
del_g0_f_1 = 31.72 ; #[kJ/mol]
del_g0_f_2 = 26.89 ; #[kJ/mol]
R = 8.314 ;
T = 298 ; #[K]
# Calculations
del_g0_rxn = del_g0_f_2 - del_g0_f_1 ;
K = math.exp( - del_g0_rxn * 10**3 / (R * T) ) ;
x = K / (1 + K) ;
# Results
print "x = %.3f \nAt equilibrium %.1f %% of the liquid exists as cyclohexane."%(x ,x * 100) ;
Example 9.11 Page No : 587¶
In [10]:
# Variables
import math
del_g0_f_CaCO3 = -951.25 ;
del_g0_f_CaO = -531.09 ;
del_g0_f_CO2 = -395.81 ;
R = 8.314 ;
T = 1000. ; # [K]
# Calculations
del_g0_rxn = del_g0_f_CaO + del_g0_f_CO2 - del_g0_f_CaCO3 ;
K = math.exp(-del_g0_rxn * 10**3 / (R * T)) ;
p_CO2 = K ;
# Results
print "Equilibrium pressure = %.3f bar "%(p_CO2) ;
Example 9.12 Page No : 588¶
In [11]:
# Variables
import math
del_g0_f_B = 124.3 ; #[kJ/mol] , From Appendix A.3
del_g0_f_Ac = 209.2 ; #[kJ/mol] , From Appendix A.3
R = 8.314 ;
T = 298. ; # [K]
A = 9.2806 ;
B = 2788.51 ;
C = -52.36 ;
# Calculations
del_g0_rxn = del_g0_f_B - 3 * del_g0_f_Ac ;
K = math.exp( - del_g0_rxn * 10**3 / (R * T)) ;
P = 1 / K**(1./3) ;
X = A - B / (T + C) ;
P_b = math.exp(X) ;
# Results
print ("At equilibrium , the cylinder is almost completely filled with Benzene .")
print "System pressure = %.3f bar "%(P_b)
Example 9.13 Page No : 596¶
In [5]:
# Variables
import math
E_0_c = 0.153 ; #[V]
E_0_a = -0.521 ; # [v]
T = 298. ; #[K]
z = 1. ;
F =96485 ; #[C/mol e-]
R =8.314 ; #[J/mol K ]
# Calculations
E_0_rxn = E_0_c + E_0_a ;
del_g_0_rxn = - z * F * E_0_rxn ;
K = math.exp( - del_g_0_rxn / ( R * T )) ;
# Results
print "The equilibrium constant = %.3g "%(K)
print "The equilibrium constant is small . So the etching will not proceed\
spontaneously . However if we apply work through application of an electrical potential\
, we can etch the copper ."
# Note: answer is slightly different because of rounding off error.
Example 9.14 Page No : 596¶
In [12]:
# Variables
import math
E_0_c = 0.34 ; #[V]
E_0_a = -1.23 #[V]
T = 298. ; # [K]
pH = 1. ;
z = 2. ;
Cu2 = 0.07
F = 96485. #[C/mol e-]
R = 8.314 ;
# Calculations
E_0_rxn = E_0_c + E_0_a ;
E = E_0_rxn + 2.303 * R * T * 2 * pH / (z * F) + R * T * math.log(Cu2) / (z * F) ;
# Results
print "Del_E_0_rxn = %.2f "%(E_0_rxn ) ;
print "We have to apply potential greater than %.1f V"%(-E) ;
Example 9.17 Page No : 602¶
In [14]:
# Variables
m = 4. ;
T = 2. ;
Pai = 1. ;
S = 1. ;
# Calculations
R = m - T + 2 - Pai - S ;
# Results
print "We must specify %g independent equations ."%(R)
Example 9.19 Page No : 603¶
In [2]:
%matplotlib inline
from numpy import *
from scipy.optimize import fsolve
from matplotlib.pyplot import *
import math
# Variables
del_g_f_CH4 = -50.72 ;
del_g_f_H2 = 0 ;
del_g_f_H2O = -228.57 ;
del_g_f_CO = -137.17 ;
del_g_f_CO2 = -394.36 ;
del_h_f_CH4 = -74.81 ;
del_h_f_H2 = 0 ;
del_h_f_H2O = -241.82 ;
del_h_f_CO = -110.53 ;
del_h_f_CO2 = -393.51 ;
v1_CH4 = -1. ;
v1_H2 = 3. ;
v1_H2O = -1. ;
v1_CO = 1. ;
v1_CO2 = 0. ;
v2_CH4 = -1. ;
v2_H2 = 4. ;
v2_H2O = -2. ;
v2_CO = 0. ;
v2_CO2 = 1. ;
A_CH4 = 1.702 ;
B_CH4 = 9.08 * 10**-3 ;
C_CH4 = -2.16 * 10**-6 ;
D_CH4 = 0 ;
A_H2 = 3.249 ;
B_H2 = 4.22 * 10**-4 ;
C_H2 = 0 ;
D_H2 = 8.30 * 10**3 ;
A_H2O = 3.47 ;
B_H2O = 1.45 * 10**-3 ;
C_H2O = 0 ;
D_H2O = 1.21 * 10**4 ;
A_CO = 3.376 ;
B_CO = 5.57 * 10**-4 ;
C_CO = 0 ;
D_CO = -3.10 * 10**3 ;
A_CO2 = 5.457 ;
B_CO2 = 1.05 * 10**-3 ;
C_CO2 = 0 ;
D_CO2 = -1.16 * 10**5 ;
M = zeros((12,10))
M[:,0] = linspace(600,1150,12)
R = 8.314 ;
P = 1 ; #[bar]
T_ref = 298.15 ; #[K]
# Calculations
del_g_f_1 = (v1_CO * del_g_f_CO + v1_H2 *del_g_f_H2 + v1_CH4 * del_g_f_CH4 + v1_H2O * del_g_f_H2O) * 1000 ;
del_h_f_1 = (v1_CO * del_h_f_CO + v1_H2 *del_h_f_H2 + v1_CH4 * del_h_f_CH4 + v1_H2O * del_h_f_H2O) * 1000 ;
del_g_f_2 = (v2_CO2 * del_g_f_CO2 + v2_H2 *del_g_f_H2 + v2_CH4 * del_g_f_CH4 + v2_H2O * del_g_f_H2O) * 1000 ;
del_h_f_2 = (v2_CO2 * del_h_f_CO2 + v2_H2 *del_h_f_H2 + v2_CH4 * del_h_f_CH4 + v2_H2O * del_h_f_H2O) * 1000;
Del_A_1 = v1_CO * A_CO + v1_H2 * A_H2 + v1_CH4 * A_CH4 + v1_H2O * A_H2O ;
Del_B_1 = v1_CO * B_CO + v1_H2 * B_H2 + v1_CH4 * B_CH4 + v1_H2O * B_H2O ;
Del_C_1 = v1_CO * C_CO + v1_H2 * C_H2 + v1_CH4 * C_CH4 + v1_H2O * C_H2O ;
Del_D_1 = v1_CO * D_CO + v1_H2 * D_H2 + v1_CH4 * D_CH4 + v1_H2O * D_H2O ;
Del_A_2 = v2_CO2 * A_CO2 + v2_H2 * A_H2 + v2_CH4 * A_CH4 + v2_H2O * A_H2O ;
Del_B_2 = v2_CO2 * B_CO2 + v2_H2 * B_H2 + v2_CH4 * B_CH4 + v2_H2O * B_H2O ;
Del_C_2 = v2_CO2 * C_CO2 + v2_H2 * C_H2 + v2_CH4 * C_CH4 + v2_H2O * C_H2O ;
Del_D_2 = v2_CO2 * D_CO2 + v2_H2 * D_H2 + v2_CH4 * D_CH4 + v2_H2O * D_H2O ;
K_298_1 = math.exp( - del_g_f_1 / (R * T_ref)) ;
K_298_2 = math.exp( - del_g_f_2 / (R * T_ref)) ;
for i in range(12):
X = (-del_h_f_1 / R + Del_A_1 * T_ref + Del_B_1 / 2 * T_ref**2 + \
Del_C_1 /3* T_ref**3- Del_D_1 / T_ref) * (1./M[i,0] - 1./T_ref) + \
Del_A_1*math.log(M[i,0] / T_ref)+ Del_B_1 / 2 * (M[i,0] - T_ref) + \
Del_C_1 / 6 *(M[i,0]**2 - T_ref**2) + Del_D_1 / 2 * (1./(M[i,0]**2) - 1./(T_ref**2))
M[i,1] = K_298_1 * math.exp(X) ;
Y = (-del_h_f_2 / R + Del_A_2 * T_ref + Del_B_2 / 2 * T_ref**2 + Del_C_2/3* T_ref**3- Del_D_2 / T_ref) * \
(1/M[i,0] - 1/T_ref) + Del_A_2 * math.log(M[i,0] / T_ref)+ Del_B_2 / 2 * (M[i,0] - T_ref) + \
Del_C_2 / 6 *(M[i,0]**2 - T_ref**2) + Del_D_2 / 2* (1/(M[i,0]**2) - 1/(T_ref**2));
M[i,2] = K_298_2 * math.exp(Y) ;
def f918(R):
s1 = R[0] ;
s2 = R[1] ;
y = [0,0]
y[0] = (s1 * (3 * s1 + 4 * s2)**3) / ((5 + 2 * s1 + 2 * s2)**2 * (1 - s1 -s2) * (4 - s1 - 2 * s2)) * P**2 - M[i,1] ;
y[1] = (s2 * (3 * s1 + 4 * s2)**4) / ((5 + 2 * s1 + 2 * s2)**2 * (1 - s1 -s2) * (4 - s1 - 2 * s2)**2) * P**2 - M[i,2] ;
return y
z = fsolve(f918,[0.0001,0.0001])
M[i,3] = z[0] ;
M[i,4] = z[1] ;
M[i,5] = (1 - M[i,3] - M[i,4]) / (5 + 2 * M[i,3] + 2 * M[i,4]) ;
M[i,6] = (4 - M[i,3] - 2 * M[i,4]) / (5 + 2 * M[i,3] + 2 * M[i,4]) ;
M[i,7] = (3 * M[i,3] + 4 * M[i,4]) / (5 + 2 * M[i,3] + 2 * M[i,4]) ;
M[i,8] = M[i,3] / (5 + 2 * M[i,3] + 2 * M[i,4]) ;
M[i,9] = M[i,4] / (5 + 2 * M[i,3] + 2 * M[i,4]) ;
n1 = zeros([12,7])
for i in range(12):
for j in range(7):
n1[i,j] = M[i,j] ;
n2 = zeros([12,3])
for i in range(12):
for j in range(3):
n2[i,j] = M[i,j+7]
print " T K1 K2 S1 S2 y_CH4 y_H2"
for row in n1:
print " %5d %7.2e %7.2e %7.3f %7.3f %7.3f %7.3f "%(row[0],row[1],row[2],row[3],row[4],row[5],row[6])
#print (n1) ;
print (" y_H20 y_CO y_CO2 ") ;
for row in n2:
print " %7.3f %7.3f %7.3f "%(row[0],row[1],row[2])
#print (n2) ;
N = zeros([10,10])
for i in range(10):
for j in range(10):
N[i,j] = M[i,j]
'''
plot(N[3],N[0],"+") ;
plot(N[4], N[0],".") ;
plot(N[5] , N[0], "o-") ;
plot(N[6] , N[0], "s-");
plot(N[7] , N[0], "*-") ;
plot(N[8] , N[0], "x-") ;
plot(N[9] , N[0], ".-") ;
show()
'''
suptitle("Figure E9.18 Extent of reaxn vs temp")
xlabel("Temperature(K)")
ylabel("S") ;
#legend("S1","S2") ;
plot(N[:,0] , N[:,5], "o-") ;
plot(N[:,0] , N[:,6], "s-");
plot(N[:,0] , N[:,7], "^-") ;
plot(N[:,0] , N[:,8], "x-") ;
plot(N[:,0] , N[:,9], ".-") ;
# Note : some answers are different because of rounding off error.
Out[2]:
Example 9.21 Page No : 607¶
In [69]:
# Variables
import math
del_g_0_f_SiCl2 = - 216012. ;
del_g_0_f_SiCl4 = - 492536. ;
del_g_0_f_SiCl3H = -356537. ;
del_g_0_f_SiCl2H2 = -199368. ;
del_g_0_f_SiClH3 = -28482. ;
del_g_0_f_SiH4 = -176152. ;
del_g_0_f_HCl = -102644. ;
del_g_0_f_H2 = 0. ;
del_g_0_f_Si = 0. ;
R = 8.314 ;
T = 1300. ; #[K]
# Calculations
Del_g_rxn_1 = del_g_0_f_SiCl2 + 2 * del_g_0_f_HCl - del_g_0_f_SiCl4 - del_g_0_f_H2 ;
Del_g_rxn_2 = del_g_0_f_SiCl3H + del_g_0_f_HCl - del_g_0_f_SiCl4 - del_g_0_f_H2 ;
Del_g_rxn_3 = del_g_0_f_SiCl2H2 + del_g_0_f_HCl - del_g_0_f_SiCl3H - del_g_0_f_H2 ;
Del_g_rxn_4 = del_g_0_f_SiClH3 + del_g_0_f_HCl - del_g_0_f_SiCl2H2 - del_g_0_f_H2 ;
Del_g_rxn_5 = del_g_0_f_SiH4 + del_g_0_f_HCl - del_g_0_f_SiCl3H - del_g_0_f_H2 ;
Del_g_rxn_6 = del_g_0_f_Si + 4 * del_g_0_f_HCl - del_g_0_f_SiCl4 - 2 * del_g_0_f_H2 ;
M = zeros([6,4])
M[0,0] = math.exp( - Del_g_rxn_1 / (R * T)) ;
M[1,0] = math.exp( - Del_g_rxn_2 / (R * T)) ;
M[2,0] = math.exp( - Del_g_rxn_3 / (R * T)) ;
M[3,0] = math.exp( - Del_g_rxn_4 / (R * T)) ;
M[4,0] = math.exp( - Del_g_rxn_5 / (R * T)) ;
M[5,0] = math.exp( - Del_g_rxn_6 / (R * T)) ;
S = [0.0763,0.1979,0.0067,0.0001,0.0000,-0.0512] ;
K_cal = [.00137,0.0457,0.00644,0.00181,0.000752,0.000509] ;
for i in range(6):
M[i,1] = S[i] ;
M[i,2] = K_cal[i] ;
M[i,3] = M[i,0] - M[i,2] ;
# Results
print (" K_i S K_i_cal K_i - K_i_cal") ;
for row in M:
print " %5.2e %7.4f %7.2e %7.2e"%(row[0],row[1],row[2],row[3])
# Readers can refer figure E9.19 .
# Note : answers are different because of rounding off error.