# Chapter 5 : Thermochemistry¶

## Example 5.1 Page No : 71¶

In [1]:
# Variables
Q1 = -1227 	    		#kcal
R = 2*10**-3 			#kcal
T = 25       			#C
dn = -2

# Calculations
Qp = Q1+R*(273+T)*dn

# Results
print  'Heat of reaction  = %.1f kcal'%(Qp)

Heat of reaction  = -1228.2 kcal


## Example 5.2 Page No : 72¶

In [2]:
# Variables
H1 = -337.3 			#kcal
H2 = -68.3   			#kcal
H3 = -372.8 			#kcal

# Calculations
Ht = H1+H2-H3

# Results
print  'Heat change of reaction  = %.1f kcal'%(Ht)

Heat change of reaction  = -32.8 kcal


## Example 5.3 Page No : 74¶

In [3]:
# Variables
dH = -1228.2 			#kcal
n1 = 10
n2 = 4
dH1 = -94.05 			#kcal
dH2 = -68.32 			#kcal

# Calculations
x = n1*dH1+n2*dH2-dH

# Results
print  'Heat of formation  = %.1f kcal'%(x)

Heat of formation  = 14.4 kcal


## Example 5.4 Page No : 75¶

In [4]:
# Variables
H1 = -29.6 			#kcal
H2 = -530.6 			#kcal
H3 = -94 			#kcal
H4 = -68.3 			#kcal

# Calculations
dH1 = -(H1+H2-3*H3-4*H4)
dH2 = -dH1+3*H3+3*H4

# Results
print  'Heat of combustion  = %.f kcal'%(dH1)
print  ' Standard heat of formation  = %.1f kcal'%(dH2)

Heat of combustion  = 5 kcal
Standard heat of formation  = -491.9 kcal


## Example 5.5 Page No : 79¶

In [5]:
# Variables
T1 = 25.     			#C
T2 = 100.    			#C
dH1 = -57.8 			#kcal
Cp1 = 8.03  			#cal deg**-1
Cp2 = 6.92  			#cal deg**-1
Cp3 = 7.04  			#cal deg**-1

# Results
Cp = Cp1-(Cp2+0.5*Cp3)
dH2 = Cp*10**-3*(T2-T1)+dH1

# Results
print  'Standard heat of formation  = %.2f kcal mole**-1'%(dH2)

Standard heat of formation  = -57.98 kcal mole**-1


## Example 5.6 Page No : 80¶

In [6]:
# Variables
a = -2.776
b = 0.947*10**-3
c = 0.295*10**-6
T1 = 373     			#K
T2 = 298 	    		#K
dH1 = -57.8 			#kcal

# Calculations
dH = a*(T1-T2)+0.5*b*(T1**2-T2**2)+0.33*c*(T1**3-T2**3)
dH2 = dH1+(dH/1000)

# Results
print  'Heat obtained  = %.f cal '%(dH)
print  ' Smath.tanadard heat of formation  = %.2f kcal mole**-1'%(dH2)

Heat obtained  = -182 cal
Smath.tanadard heat of formation  = -57.98 kcal mole**-1


## Example 5.7 Page No : 81¶

In [1]:
# Variables
a1 = 6.189
a2 = 3.225
a3 = 10.421
b1 = 7.787*10**-3
b2 = 0.707*10**-3
b3 = -0.3*10**-3
c1 = -0.728*10**-6
c2 = -0.04014*10**-6
c3 = 0.7212*10**-6
dH = -9.13 			#kcal

# Calculations
a = -(a2+a3-a1)*10**-3
b = -0.5*(b2+b3-b1)*10**-3
c = -0.33*(c2+c3-c1)*10**-3

# Results
print  'a  = %.2e kcal mole**-1'%(a)
print  ' b  = %.2e kcal mole**-1'%(b)
print  ' c  = %.2e kcal mole**-1'%(c)
print  ' dH  = %.2f kcal mole**-1'%(dH)

# note : rounding off error.

a  = -7.46e-03 kcal mole**-1
b  = 3.69e-06 kcal mole**-1
c  = -4.65e-10 kcal mole**-1
dH  = -9.13 kcal mole**-1


## Example 5.8 Page No : 81¶

In [8]:
# Variables
dH = 31.39       			#kcal
k1 = 3.397*10**-3 			#kcal K**-1
k2 = -1.68*10**-6 			#kcal K**-2
k3 = -0.022*10**-9 			#kcal K**-3
k4 = 1.17*10**2 			#kcal K
T = 25 			#C
#CALCULTIONS
H = dH-(k1*(273+T)+k2*(273+T)**2+k3*(273+T)**3+k4*(273+T)**-1)

# Results
print  'Change in enthalpy = %.2f kcal'%(H)

Change in enthalpy = 30.13 kcal


## Example 5.9 Page No : 85¶

In [10]:
from numpy import *
# Variables
dH = 214470     			#kcal mole**-1
a = 72.43    	    		#calmole**-1deg**-1
b = 13.08*10**-3 			#kcalmole**-1
c = -1.172*10**-6 			#kcalmole**-1

# Calculations
vec =roots([c,b,a,-dH])
T = vec[2]

# Results
print  'Temperature  = %.f C'%(T+15)

# note : rounding off error because of roots()

Temperature  = 2253 C


## Example 5.10 Page No : 88¶

In [8]:
# Variables
c1 = 9.3 			#cal deg**-1
c2 = 6.3 			#cal deg**-1
n = 2.
dH = -57500. 			#cal
V = 3. 		    	#cc
v1 = 3.5 			#cc
T1 = 25. 			#C
p1 = 1. 			    #atm

# Calculations
T2 = (-dH/(c1+n*c2))+298
p2 = p1*V*T2/(v1*(273+T1))

# Results
print  'Temperature final  = %.f K'%(round(T2,-1))
print  ' pressure final  = %.1f atm'%(p2)

Temperature final  = 2920 K
pressure final  = 8.4 atm


## Example 5.11 Page No : 92¶

In [19]:
# Variables
Hc = 234.4 			#kcal
Hdc = 300 			#kcal
Hch = 436.5 			#kcal
Hco = 152 			#kcal
Hsco = 70 			#kcal
Hoh = 110.2 			#kcal
Hoo = 885 			#kcal
Hb = 38 			#kcal
Hc = 28 			#kcal
Ha = 206 			#kcal
H1co = 2128 			#kcal
H1oh = 661 			#kcal
H1c = 231 			#kcal

# Calculations
dH = Hc+Hdc+Hch+Hco+Hsco+Hoh+Hoo+Ha+Hb+Hc-H1co-H1oh-H1c

# Results
print  'Heat of combustion  = %.f kcal'%(dH)

Heat of combustion  = -766 kcal

In [ ]: