Chapter 19 : EMF and the thermodynamics of ions

Example 19.1 Page No : 466

In [1]:
# Variables
h = 23070. 			#cal volt**-1 g equiv**-1
n = 2.   			#electrons
e = 1.005 			#volts
T = 25. 			#C
e1 = 1.015 			#volts

# Calculations
dH = (-n*h*(e-((273.2+T)*(e-e1)/T)))/1000

# Results
print  'Heat change in the cell reaction  = %.2f kcal'%(dH)

Heat change in the cell reaction  = -51.87 kcal

Example 19.2 Page No : 482

In [2]:
# Variables
E = -0.344   			#volt
E1 = -0.401 			#volt
R = 0.05914 			#volt
n = 4.
T = 25. 	    		#C
H = -7300. 		    	#cal

# Calculations
po2 = 10**(-n*(E-E1)/R)
dH = -0.5*n*H+0.5*n*(273+T)

# Results
print  'Pressure of Oxygen  = %.1e atm'%(po2)
print  ' Change in Enthalpy  = %.f cal'%(dH+4)

Pressure of Oxygen  = 1.4e-04 atm
 Change in Enthalpy  = 15200 cal

Example 19.3 Page No : 489

In [4]:
# Variables
H = -60.15 			#kcal
e = 2.924 			#volt
v = 23070. 			#cc
T = 25. 			#C
Sm = 15.2 			#E.U.mole**-1
Sg = 31.2 			#E.U.mole**-1

# Calculations
dS = (H*1000-(-e*v))/(273.2+T)
Sk = (dS+Sm)-0.5*Sg

# Results
print  'Standard entropy of pottasium ion   = %.1f E.U.g ion**-1'%(Sk)

Standard entropy of pottasium ion   = 24.1 E.U.g ion**-1

Example 19.4 Page No : 490

In [5]:
# Variables
dS = -4.61 			#E.Ugm ion**-1
SH = 31.21 			#E.U gm ion**-1
Sm = 9.95 			#E.U gm ion**-1

# Calculations
Szn = dS-SH+Sm

# Results
print  'Standard entropy of zinc ion   = %.1f E.U.g ion**-1'%(Szn)

Standard entropy of zinc ion   = -25.9 E.U.g ion**-1

Example 19.5 Page No : 491

In [7]:
# Variables
import math 
n = 2.
T = 25. 			#C
R = 4.576 
is_ = 9.57*10**-6
n1 = 4.
f = 0.509 			#volts
dH = 5970. 			#cal
SBa = 2.3 			#E.U. gm ion**-1
Sba = 31.5 			#E.U. gm ion6-1

# Calculations
r = 10**(-n1*f*math.sqrt(n1*is_))
dF = -n*R*(273.2+T)*math.log10(is_*r)
dS = (dH-dF)/(273.2+T)
Sso = Sba-SBa+dS

# Results
print  'Standard entropy of sulfate ion   = %.1f E.U.g ion**-1'%(Sso)

Standard entropy of sulfate ion   = 3.2 E.U.g ion**-1

Example 19.6 Page No : 496

In [8]:
# Variables
f1 = 20.66 			#kcal
h1 = 21.6 			#kcal
e1 = 50.34 			#kcal
f2 = 0   			#kcal
f3 = -56.70 			#kcal
f4 = -26.25 			#kcal
h2 = 0 	    		#kcal
h3 = -68.32 			#kcal
h4 = -49.5 			#kcal
e2 = 49.00 			#kcal
e3 = 16.75 			#kcal
e4 = 35 			#kcal
n1 = 2
n2 = 1.5
n3 = 1
T = 25 			#C

# Calculations
dF = n1*f4-(n1*f1+f3)
dH = n1*h4-(n1*h1+h3)
dS = n1*e4-(n1*e1+e3+n2*e2)
dS1 = (dH-dF)*1000/(273.2+T)

# Results
print  'free energy   = %.2f kcal'%(dF)
print  ' heat of formation   = %.1f kcal'%(dH)
print  ' Entropy  = %.1f E.U'%(dS)
print  ' Entropy umath.sing heat of formation and free energy   = %.1f E.U'%(dS1)

free energy   = -37.12 kcal
 heat of formation   = -73.9 kcal
 Entropy  = -120.9 E.U
 Entropy umath.sing heat of formation and free energy   = -123.3 E.U
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