University of Babylon /College Of Engineering
Electrochemical Engineering Dept.
Second Stage /Thermodynamics
Effect of temperature on standard heat of reaction
Estimation of heat of reaction when reactants and products are not at standard state (298.15
○
K) but on T
○
K
N
2
+ 3 H
2
→ 2NH
3
ν N
2
= - 1 ; ν H
2
= - 3 ; and ,ν NH
3
= 2
     i
 i
    i
  f i
4 HCl
( g )

O
2 ( g )

2 H
2
O
( g )

2 Cl
( g )
  
2
  fH
2
O

4
  fHCl

( 2 )(

241 , 818 )

( 4 )(

92 , 302 )
 
114 , 408 J
For standard reactions, products and reactants are always at the standard -state pressure of 1 bar. Standard-state enthalpies are therefore functions of temperature only, and by Eq. d
 i
 
C

Pi dT i identifies a particular product or reactant , multiplying by ν i where , ν


 d i d
 i
 
(
 i
 i

) i i ,
positive ( + ) for products and , negative ( - ) for reactants

  d i
C

Pi i
  i dT
 i
    i i
C

Pi dT aA+bB→cC+dD





T d
2 9 8 .
1 5
   
T
298 .
15

C
P
 dT
This is the fundamental equation relating heats of reaction to temperature .
Integrations gives ;
 
T
  
298 .
15
 
C

Pmh
( T

298 .
15 ) ( 1)

University of Babylon /College Of Engineering
Electrochemical Engineering Dept.
Second Stage /Thermodynamics
If the temperature dependence of heat capacity of each product and reactant then

C

Pmh
is given by the analog of equation :
Cp mean
R
 
A

(

B ) T am


3
C
( 4 T 2 am

T
1
T
2
)


D
T
1
T
2
Where :
∆A≡∑ν i
A i with analogous definitions for
∆B,∆C
, and
∆D
.
Ex -1 : Calculate the standard heat of the methanol –synthesis reaction at 800
○
C
CO
( g )

2 H
2 ( g )

CH
3
OH
By use table 4-4 we can calculate heat of reaction at 25
( g
○
)
C , as below
 
298
 
200 , 660

(

110 , 525 )
 
90 , 135 J
From table 4-1 , take constants value for heat capacity equation i
ν i
A B×10
3
C×10
6
D×10
-5
CH
3
OH 1 2.211 12.216 -3.450 0.000
CO -1 3.376 0.557 0.000 -0.031
H
2
-2 3.249 0.422 0.000 0.083
∆A= (1)(2.211)+(-1)(3.376)+(-2)(3.249) = - 7.663
∆ B= 10.815×10
-3
∆C= -3.450× 10 -6
∆D = -0.135 × 10 -5
By use T1= 298
○
K, T2 = 1073
○
C and R = 8.314 J mol

C

Pmh
 
17 .
330 J

K

1
-1 ○
K
-1
 
298
 
90 , 135

( 17 .
330 )( 1073

298 )
 
103566 J
J- factor method
The alternative for the equation (1) is general integration to give

T


Cp

J

Eliminating
 
C
P
 dT

C

P
from equation (2) by used the below equation
 
A

(

B ) T
 
CT 2 

D
R T
And this lead to get the below
 
T

J

R [

AT


2
B
T 2


C
3
T 3


D
]
T

University of Babylon /College Of Engineering
Electrochemical Engineering Dept.
Second Stage /Thermodynamics
Ex -2 : For the reaction; N
2
+ 3 H
2
→ 2NH
3
 
298
 
92220 J i
N
2
H
2
NH
3
ν i
-1
-3
2
A
3.28
3.249
3.578
B×10
3
0.593
0.422
3.020
C×10
6
0
0
0
D×10
-5
0.04
0.083
-0.186
∆A=-5.871
∆ B= 4.181×10 -3
∆C= 0.0
∆D = -0.711×10 5

92220

J

J

8 .
314 [

5 .
871


878151 .
267
298

4 .
181

10

3
2
( 298 )
2 

0 .
711

10 5
298
]
  
T
 
878151 .
267

8 .
314 [

5 .
871 T

4 .
181

10

3
2
T
2 

0 .
711

10 5
]
T
Ex -3 : Repeat Ex-1 by use J- factor method
CO
( g )

2 H
2 ( g )

CH
3
OH
( g )
 
T

J

R [

AT


B
T 2
2


C
T 3
3


D
]
T
 
298
 
90 , 135 J

90 , 135

J

8 .
314 [(

7 .
663 )( 298 )

10 .
815

10

3
2
( 298

0 .
135

10
  
298
1073
 
5

J
75259



75259
63710

1073

44 .
962

10

3 
298
2
) 2 
3 .
450
3

10

9 .
561

10

6

6

( 298
298
3
) 3

1 .
122

10
5
298
  
1073
 
103566 J