# Chapter 13 : Free Energy and Chemical Reactions¶

## Example 13.1 Page No : 285¶

In [1]:
# Variables
import math
R = 4.576 			#cal deg**-1 mole**-1
T = 700 			#C
Kp = 0.71
p1 = 1.5 			#atm
p2 = 5   			#atm

# Calculations
dF = -R*(273+T)*(math.log(Kp)-math.log((p1*p2)/(10*p2)))*0.77

# Results
print  'Free energy  = %.f cal'%(dF-10)

Free energy  = -5340 cal


## Example 13.1.1 Page no : 287¶

In [2]:
import math

# variables and calculations

# for CO2
Tc = 304.      #K
Pc = 72.9       # atm
a = 873.       # K
b = 500.       # atm

theta1 = a/Tc
pi2 = b/Pc
YCO2 = 1.09

# for H2
Tc = 33.2      #K
Pc = 12.8       # atm

theta1 = a/(Tc+8)
pi2 = b/(Pc+8)
YH2 = 1.10

# for CO
Tc = 134.      #K
Pc = 34.6       # atm

theta1 = a/Tc
pi2 = b/Pc
YCO = 1.23

# for H20
Tc = 647.      #K
Pc = 218       # atm

theta1 = a/Tc
pi2 = b/Pc
YHO2 = 0.77

Jy = YCO2*YH2/(YCO*YHO2)
# results
print "Jy = %.2f"%Jy

Jy = 1.27


## Example 13.2 Page No : 293¶

In [4]:
# Variables
import math
k1 = 4600.
k2 = -8.64
k3 = 1.86*10**-3
k4 = -0.12*10**-6
k5 = 12.07
T = 600. 			#K

# Calculations
Kf = math.e**(k1*(1/T)+k2*math.log10(T)+k3*T+k4*T**2+k5)

# Results
print  'Kf  = %.3f '%(Kf)

Kf  = 0.041


## Example 13.3 Page No : 294¶

In [5]:
# Variables
import math
k = -8810 			#cal
k1 = -7.46 			#cal K**-1
k2 = 3.69*10**-3 			#cal K**-2
k3 = -0.47*10**-6 			#cak K**-3
T = 298 			#K

# Calculations
dH = k+k1*T+k2*T**2+k3*T**3

# Results
print  'Standard heat of reaction  = %.f cal'%(dH)

Standard heat of reaction  = -10718 cal


## Example 13.4 Page No : 296¶

In [1]:
# Variables
import math
k1 = -9130 			#cal
k2 = 7.46 			#cal K**-1
k3 = -3.69*10**-3 			#K**-2
k4 = 0.235*10**-6 			#K**-3
k5 = -12.07
T = 298 			#K
R = 1.987 			#cal deg**-1 mole**-1

# Calculations
dF = k1+k2*T*math.log(T)+k3*T**2+k4*T**3+k5*R*T

# Results
print  'Free energy  = %.f cal'%(dF)

# rounding off error is there. please check.

Free energy  = -3933 cal


## Example 13.5 Page No : 297¶

In [7]:
# Variables
T = 25       			#C
dF1 = 61.44 			#kcal
dF = 54.65 			#kcal
R = 4.576 			#cal deg**-1 mole**-1

# Calculations
Kf = 10**(-(dF1-dF)*10**3/(R*(273.2+T)))

# Results
print  'Kf at this temperature  = %.2e '%(Kf)

Kf at this temperature  = 1.06e-05


## Example 13.6 Page No : 300¶

In [10]:
# Variables
import math
R = 4.576 			#cal mole**-1 K**-1
T = 25. 			#C
p1 = 122. 			#mm
F1 = -5.88 			#kcal
F2 = -33 			#kcal

# Calculations
dF = R*(273.2+T)*math.log10(p1/760)
F = F2+F1+(dF/1000)

# Results
print  'Standard free energy change  = %.f kcal'%(F)

Standard free energy change  = -40 kcal


## Example 13.7 Page No : 303¶

In [11]:
# Variables
r = 3.38*10**-4 			#volt deg**-1
F = 23070.       			#cal volt**-1 deg**-1
Sagcl = 23 	        		#E.U.mole**-1
Shg = 18.5 	        		#E.U.mole**-1
Sag = 10.2 		        	#E.U.mole**-1

# Calculations
dS = F*r
shgcl = 2*-(dS-Sagcl-Shg+Sag)

# Results
print  'dS   = %.1f E.U.cal deg**-1'%(dS)
print  ' molar entropy   = %.f E.U.mole**-1'%(shgcl)

dS   = 7.8 E.U.cal deg**-1
molar entropy   = 47 E.U.mole**-1


## Example 13.8 Page No : 304¶

In [13]:
# Variables
s1 = 44.5 			#cal deg**-1 mole**-1
s2 = 49 			#cal deg**-1 mole**-1
s3 = 51.06 			#cal deg**-1 mole**-1
s4 = 16.75 			#cal deg**-1 mole**-1
h1 = -17.9 			#kcal mole**-1
h2 = 0 			#kcal mole**-1
h3 = -94 			#kcal mole**-1
h4 = -68.3 			#kcal mole**-1
T = 25 			#C
n = 2

# Calculations
dS = s3+2*s4-s1-n*s2
dH = h3+n*h4-h1-n*h2
dF = -0.001*(273.2+T)*dS+dH

# Results
print  'Entropy Change  = %.1f E.U'%(dS)
print  ' Enthalpy Change  = %.1f E.U'%(dH)
print  ' Standard free energy  = %.1f kcal'%(dF)

Entropy Change  = -57.9 E.U
Enthalpy Change  = -212.7 E.U
Standard free energy  = -195.4 kcal


## Example 13.9 Page No : 304¶

In [7]:
# Variables
import math
a = -15.84
b = 22.84*10**-3
c = -80.97*10**-7
T = 25. 			#C
H1 = -48.1 			#kcal
H2 = -26.4
dS = 53.09
T1 = 327. 			#C
r1 = 0.58
r2 = 1.1
r3 = 1.13

# Calculations
dH = round((H1-H2)*1000-a*(273.2+T)-0.5*b*(273.2+T)**2-0.33*c*(273.2+T)**3,-1)
dF = round((H1-H2)*1000+(273.2+T)*dS,-1)
I = (dF-dH+a*(273.2+T)*math.log(273.2+T)+0.5*b*(273.2+T)**2+0.166*c*(273.2+T)**3)/(273.2+T)
dF1 = (dH-a*(273+T1)*math.log(273+T1)-0.5*b*(273+T1)**2-0.166*c*(273+T1)**3)+I*(273+T1)
Kf = 10**(-dF1/(4.576*(273+T1)))
Jr = r1/(r2**2*r3)
Kp = Kf/Jr

# Results
print  'heat of formation  = %d cal'%(dH)
print  ' Entropy  = %.f cal'%(dF)
print  ' Inertia  = %.2f gm cm**2'%(I)
print  ' Entropy  = %.f cal'%(dF1)
print  ' Kf  = %.1e '%(Kf)
print  ' Kp  = %.1e '%(Kp)

# rounding off error is there. please check.

heat of formation  = -17920 cal
Entropy  = -5870 cal
Inertia  = -46.56 gm cm**2
Entropy  = 11121 cal
Kf  = 8.9e-05
Kp  = 2.1e-04


## Example 13.10 Page No : 309¶

In [15]:
# Variables
F1 = 24.423			#cal deg**-1
F2 = 21.031 			#cal deg**-1
F3 = 37.172 			#cal deg**-1
H1 = 2.024 			#kcal
H2 = 1.035 			#kcal
H3 = 2.365 			#kcal
H = -57.8 			#kcal
T = 25.  			#C

# Calculations
dF = F3-F1-F2
dH = H3-H1-H2
Hf = H-dH
F = Hf-((273.2+T)*dF*10**-3)

# Results
print  'Standard free energy  = %.2f kcal'%(F)

Standard free energy  = -54.64 kcal


## Example 13.11 Page No : 311¶

In [17]:
# Variables
import math
T = 1000 			    #C
j = 1.5
Q = 35840    			#cal
I = 743*10**-40 		#g cm**2
w = 214 	    		#cm**-2
Kf = 0.184

# Results
print  'Equilibrium constant  = %.3f '%(Kf)

# Note :NO SOLUTION IS GIVEN TO SOLVE Kf INCOMPLETE SOLUTION IN THE TEXTBOOK

Equilibrium constant  = 0.184


## Example 13.12 Page No : 313¶

In [13]:
# Variables
import math
dH = 83 			#cal
R = 1.98 			#cal mole K**-1
T = 25 			#C
M1 = 128 			#gms
M2 = 4 			#gms
M3 = 2 			#gms
M4 = 129 			#gms
I1 = 4.31 			#g cm**2
I2 = 0.920 			#g cm**2
I3 = 0.459 			#g cm**2
I4 = 8.55 			# g cm**2

# Calculations
K = 1+10**((-dH/(2.303*R*(298)))+1.5*math.log(M1**2*M2/(M3*M4**2))+math.log(I1**2*I2/(I3*I4**2)))

# Results
print  'Equilibrium constant  = %.1f '%(K)

# note : rounding off error.

Equilibrium constant  = 1.2

In [ ]: