University of Babylon /College Of Engineering
Electrochemical Engineering Dept.
Second Stage /Thermodynamics
Adiabatic reaction
Industrial reactions are rarely carried out under standard state conditions, because :
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Reaction may not be present in stoichiometric properties.
Reaction may not go to compilation.
Final temperature differ from initial temperature .
Inert species may be present.
Other reactions may occur simultaneously.
This reaction mean that there is no heat absorbed or ejected , if the reaction
exothermic the products temperature raised while for endothermic the products
temperature decreased , a good example for this kind of reaction is fuel burned
with oxygen.
Adiabatic flame temperature , the highest temperature reached for products during
oxidation reaction for the fuel, in reality these kind of reaction described that is an
adiabatic because they are occurred too fast and in this case there is no time for
heat exchange between system and surroundings
Ex-4 :What is the maximum temperature that can be reached by combustion of
CH4 with 20% air excess ? both methane and air enter the burner at 25 ○C
CH4 + 2O2 → CO2 + 2H2O ( g )
 298  393509  (2)(241818)  (74520)  802625J
Assumptions
 Assume complete adiabatic combustion ( Q = 0)
 Kinetic and potential energy change are negligible
For overall energy balance result ∆H = 0
From chemical equation reaction for 1 mole methane
2 Mole of O2 required ,mole of O2 excess = (0.2 ) ( 2 ) = 0.4 ,and mole of N2
entering = ( 2.4 ) ( 79 / 21) = 9.03
Gas leaving the burner contain 1 mole CO2 ,2 moles H2O ,0.4 mole O2 and 9.03 N2
Since enthalpy change must be same regardless of path
University of Babylon /College Of Engineering
Electrochemical Engineering Dept.
Second Stage /Thermodynamics
 298   P    0
 298  802625J
 P   (ni CPmh ,i )(T2  298.15)
 Because mean heat capacity depend on T2 ,and it is sought
 Assume T2 ( because of T1 = 298 , thus the value of assumed T2 is ≥ 298)
 ni Di ]
 Use equation  (ni CPmh ,i )  R[ ni Ai  ( ni Bi )Tam 
T1T2
 298
 298.15
 Calculate T2 by use T2 
n
C
 I pmh,i
 Repeat the above steps until calculated T2 converges with assumed T2
 n A  1 5.457  2  3.470  0.4  3.639  9.03  3.280  43.489
i
i
n B
i
i
n D
i
i
 9.504  103
 0.645  105
  ni C Pmh ,i
0.645  105
 8.314(43.489  9.504  10 Tam 
T1T2
By continue try and error , T2 = 2066 ○K
3