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Skip to 3.4.2 Degree of Freedom Table for Energy Balances

3.4: Examples of Energy Balances

3.4.1 Sabatier Reactor Example

Problem: Determine the heat that must be added to a reactor in which CO2 and H2 react to produce methane and water by:

CO2 + 4H2 -> CH4 + 2H2O

The inlet stream contains 1 mol CO2, 3 mols H2 and 0.5 mols N2 and is at 500K and 1 atm. The exit stream contains 0.5 mols of H2 and is at 600K and 1 atm.

We will assume that the specific enthalpy of each compound depends only on the compound and the temperature since the pressure is fixed and all streams are gaseous. We will designate the specific enthalpy of compound A at temperature T by A(T). The reaction rate must be 2.5/4 or 0.625 so the flow rates and specific enthalpies of the two streams look like:

                   Inlet                  Exit
  Compound   Flow     Enthalpy       Flow    Enthalpy
 1  CO2      1.000    1(500)       0.375   1(600)
 2  H2       3.000    2(500)       0.500   2(600)
 3  N2       0.500    3(500)       0.500   3(600)
 4  CH4      0.000    4(500)       0.625   4(600)
 5  H2O      0.000    5(500)       1.250   5(600)

Two different methods will be used to find Q. Both are based on:


The Heat of Reaction Procedure

The heat of reaction with all reactants and products as ideal gases depends only on the temperature and is:

Hr(T) = 4(T) + 25(T) - 1(T) - 42(T)

Then we may find Q as:

The problem then is to determine the heat of reaction at 600K. To do so we need to use the heats of formation for each of the reactants and products in the reaction defined by:

where the formula for A is:
(E1)e1(E2)e2 .........
and Ej is the jth element

Normally the pressure, P, and the state, s, in the specific enthalpy of A are taken to be 1 atm and the state that the compound would normally be found in at 25°C (298K) and 1 atm. This is then designated as the "standard" heat of formation of A. The heats of formation in the Appendix in Reklaitis are for all compounds as ideal gases regardless of their normal states at 25°C and 1 atm. This makes it particularly simple to determine the heat of reaction for all gas reactants and products by writing for our specific reaction:

Hr(T) =


- 4(298)

+ 4(298)

+ 2(5(T)

- 5(298)

+ 5(298))

- (1(T)

- 1(298)

+ 1(298))

- 4(2(T)

- 2(298)

+ 2(298))

Each of the differences in enthalpy for the same compound at T and at 298K may be evaluated by integrating the specific heat for that compound. The specific enthalpies at 298K may then be given in terms of the heats of formation for the compounds and the enthalpies of the elements. Thus using:

we find:

+ Hf,4(T)

- C(298)

+ H2(298)

+ 2(Hf,5(T)

- H2(298)

+ O2(298))

- (Hf,1(T)

- C(298)

+ O2(298))

- 4H2(298)

Note that the enthalpies of the elements at 298K cancel out in this relation since the equation is balanced. Thus we are left with:

This same procedure extends to the determination of the heat of any reaction, but we must be careful to account for phase changes if either:

a) the heats of formation are "standard" or
b) the compounds in the reaction are not gaseous.

Use of Enthalpies Relative to Elements

An alternate procedure that gives the same results starts with the same original expression for Q and determines the specific enthalpies directly:

In any steady state system the molar flow rates of the elements into and out of the system must be equal. Thus the specific enthalpies of the elements can not appear in the final result for Q. An equivalent way to calculate Q is to determine the specific enthalpies of all compounds with the enthalpies of the elements taken as 0. This leaves us with:


This expression works for any gas in the system, but must be modified for a liquid. In all the integrals used to determine sensible heat effects up to now the specific heat always referred to the gas. Now we need to distinguish between the gas and liquid heat capacities and to use the latent heat of vaporization. Since the latter is normally known at the boiling point for the compound we determine the specific enthalpy for a liquid by:


For either liquids or gases the procedure for determining Q is the same:

a) Calculate the enthalpy for all species relative to the elements by either 2) or 3) depending on the state of the stream that it is in.
b) Determine Q from 1) using these specific enthalpies.

We can see how the two procedures work with the programs discussed in the last section of these notes. First we will actually find the heat of reaction at 600K and then add to it the amount of heat required to heat the inlet gases to 600K. In any case we need the names of our compounds so we use start301:

Input the name of your new file: sab

Here are your compounds' formulae and names:
No. Formula  Name
  1 CO2      carbon dioxide 
  2 H2       hydrogen       
  3 N2       nitrogen       
  4 CH4      methane        
  5 H2O      water          

Here are your reactions:
  1)  CO2  + 4H2    --> CH4  + 2H2O  

Enter the number of streams: 4

Now let's follow the heat of reaction approach.

The Heat of Reaction Approach

In this problem all compounds are vapors, so we need the heat capacity polynomials for the vapors:

cpv =
   19.0200    0.0796   -0.0001    0.0000   -0.0000
   17.6300    0.0670   -0.0001    0.0000   -0.0000
   29.4100   -0.0030    0.0000    0.0000   -0.0000
   38.3800   -0.0737    0.0003   -0.0000    0.0000
   34.0400   -0.0097    0.0000   -0.0000    0.0000

Our only reaction:

CO2 + 4H2 -> CH4 + 2H2O

has already been defined.

>> stoic
stoic =
    -1    -4     0     1     2

The heat of formation for all compounds is stored in:

>> StdhkJ'   <-- Note StdhkJ is stored as a column vector
ans =
 -393.5000         0         0  -74.8500 -241.8300

The heat of reaction at 298.15K is then:

>> dhr298=stoic*StdhkJ
dhr298 =

The difference in heat capacity coefficients for a stoichiometric amount of product compared to that of the reactant is found by:

>> cpt=stoic*cpv
cpt =
   16.9200   -0.4406    0.0010   -0.0000    0.0000
>> format long      
>> cpt 
cpt =
  Columns 1 through 4 
  16.92000000000001 -0.44058000000000 0.00095618000000 -0.00000076533000
  Column 5 

We may use this to find the heat of the reaction at 600K:

>> format short 
>> dhr600=dhr298+.001*enth1(298.15,600,cpt) 
dhr600 =

The heat capacity of the feed (per mol of CO2) has the coefficients:

>> format long 
>> cpin=[1 3 .5 0 0]*cpv
cpin =
  Columns 1 through 4 
  86.61499999999999 0.27911500000000 -0.00046517500000 0.00000035741550
  Column 5 

We may use this to calculate the heat that must be added per mol of CO2 fed:

>> q=(.625*dhr600)+.001*enth1(500,600,cpin) 
q =

The Enthalpy Relative to the Elements Approach

Use start301 as with the first approach, then set the feed stream with setne:

>> setne('v',1,500,[1 3 .5],1:3)

Then react this stream to form a product at 600K with fraction 0.625 of the CO2 converted:

>> reacte('v',600,1,2,.625) 
ans =

Our answer has the same value we found by the heat of reaction approach. We can also display our results:

>> showe(1,2,10,3) 
          Inlet     |  Outlet  
  Stream         1  |         2
  Tmp K     500.00  |    600.00
  State     vapor   |    vapor 
  Entlpy    -364.5  |    -461.6
Compound    Stream Flows      
CO2          1.000  |     0.375
H2           3.000  |     0.500
N2           0.500  |     0.500
CH4          0.000  |     0.625
H2O          0.000  |     1.250
Total        4.500  |     3.250


Equilibrium Reactions with kequil and reactee

The equilibrium constant, K, for a equilibrium reaction is a function of the standard Gibb's free energy change for the reaction, G°, and temperature T:

The standard state Gibb's free energies of formation for reactants and products are normally given at a temperature of 298.15 K and 1 atm. The Gibb's free energies of formation for compounds at 25°C and 1 atm is in the Appendix of the notes. G°(298.15K) is calculated in the same manner that H°(298.15K) is calculated from enthalpies of heats of formation. The equilibrium constant at 298.15K for a reaction can be easily calculated from the data for Gf°.

In order to calculate K at another temperature, an expression for G° as a function of temperature is needed. G is related to the change in enthalpy H, and change in the entropy, S, by:

If the derivative of G°/RT is taken with respect to T at constant entropy, and the expression for K is used, then the following relationship is obtained:

Consequently, the enthalpy as a function of temperature must be known, which we have shown previously:

When the above equation is integrated for H°, the references to the elements are eliminated. If the Sabatier reaction was the equilibrium reaction, the integrated relationship would be:


Remember that each Cp is a function of temperature. Substitution of this relationship into the expression for the derivative of ln K, then another integration will yield a relationship of the form:

where J is the integration constant from the relationship for Hr(T) and I is the integration constant for the last integration. The integration constants are determined by evaluation of each of the last two relationships at 298.15K. J can be determined uniquely form the expression for Hr(T) and I from the expression for ln K. Since G° = - RT ln K, I can be evaluated at 298.15K using Gibb's free energies at the standard state at 298.15K. Once the integration constants are known, then the expression for ln K can be evaluated for any temperature.

The function kequil will calculate equilibrium constants for gas phase reactions at any temperature as long as the reactions are set in stoic. Help on kequil gives:

>> help kequil

  kequil	- sets the equilibrium reactions for gas phase reactions
  function keq=kequil(t2,j)
  Argument list
  Argument    Gives
    t2      temperature in units of Tdeg
     j      indices of the reactions that will be at equilibrium
  Function return the variable: keq,
  for reactions specified and will put a zero for reactions in
  which equil. const. is _not_ solved for. If j is omitted,
  equil. constant for all reactions in stoic will be determined.
  Example : keq1=kequil(550,[1 3])

Here we will use kequil to solve example 15.5 in Smith and Van Ness (page 510) : Calculate the equilibrium constant for the vapor-phase hydration of ethylene at 145 and 320 C. The reaction is C2H4(g) + H2O(g) ---> C2H5OH(g).
After start301:

Input the name of your new file: eths
Here are your compounds' formulae and names:
No. Formula  Name
  1 C2H4     ethylene 
  2 H2O      water    
  3 C2H5OH   ethanol  

Here are your reactions:
  1)  C2H4    + H2O      --> C2H5OH  

Enter the number of streams: 2
>> kequil(418.1)
ans =
   0.1154		<--- value in Smith and VanNess, K = .1426
>> kequil(593.15)
ans =
   0.0022		<--- value in Smith and VanNess, K = .00291


The equilibrium constant K is related to the mol fractions at equilibrium and if a gas, pressure. For gases, this is:

For liquids, this is:

The mol fractions can be related to reaction rates and initial molar flows. The function reactee is a reactor module for a single equilibrium reaction. Help on reactee gives:

>> help reactee

  reactee: energy-balance equilibrium reactor module
  function er = reactee(s,tmp,in,out,rsx,jv,P)
  Argument List:
  Argument    Gives
     s      the state of the exit stream:  'v'
    tmp     the temperature of the exit stream in Tdeg.
    in      the index(ices) of the inlet streams.
    out     the index of the exit stream.
    rsx     guess for reaction rates
     jv     indices of reactions that will be at equilibrium
      P     pressure of the exit stream
   reactee returns the error  between the keq(thermo)
   and the product of partial pressures from kvalp.m
   Ex. cs = ssec2([0 .95],'reactee(''v'',1393,1,2,x,1,1)')
      would give the equilibrium extent of reaction in vapor
      stream 2 with temperature 1393Tdeg and 1(unit pressure)
      for reaction 1 and inlet stream 1

Now we will solve example 15.8 in Smith and Van Ness (page 521) using the modules: In a laboratory investigation, acetylene is catalytically hydrogenated to ethylene at 1120 C and 1 bar. If the feed is an equimolar ratio of acetylene and hydrogen, what is the composition? The reaction is C2H2(g) + H2(g) --> C2H4(g). After using start301.

Input the name of your new file: acet 
Here are your compounds' formulae and names:
No. Formula  Name
  1 C2H2     acetylene 
  2 H2       hydrogen  
  3 C2H4     ethylene  

Here are your reactions:
  1)  C2H2  + H2     --> C2H4  

Enter the number of streams: 2
>> setne('v',1,1393,[1 1 0]) 
>> kequil(1393)
ans =
   1.0725	<-- value in Smith and VanNess is ~1
		  for this reaction at 1393 K
>> cs = ssec2([0 .95],'reactee(''v'',1393,1,2,x,1,1)'
cs =
   0.3054	<---- value in Smith and VanNess is 0.2929
>> showe(1,2,10,3)
           Inlet       Outlet
   Stream         1           2
  Tmp K     1393.00     1393.00
  State      vapor       vapor
  Entlpy      327.7       274.4
Compound    Stream Flows           Equil. comp.    Reference value
acetylene     1.000       0.695       0.4097           0.4142
hydrogen      1.000       0.695       0.4097           0.4142
ethylene      0.000       0.305       0.1806           0.1716
Total         2.000       1.695

>> kvalp(1,2,1)
ans =
   1.0708		<--- very close to 1

3.4.2 Degree of Freedom Table for Energy Balances

The degree of freedom table for energy balances is very similar to the one for material balances. Only a few new variables such as dW/dT, dQ/dT, and the stream temperatures have to be considered. In addition, one also has to consider the addition of the energy balance equation. Finally, the new degree of freedom table will just combine material and energy balances.

Example 7.25 in Reklaitis

A process gas stream at 400°C is to be considered rapidly cooled to 200°C by direct quenching with cold liquid benzene at 200°C. The hot stream consists of 40% C6H6, 30% C6H5CH3, 10% CH4, and 20% H2. The gas feed rate is 1000 kgmol/hr.


Material Balance Sub problem

Combined Balance Problem






Stream Vars.


4 each in Streams 2&3
1 in Stream 1


4 each in Streams 2&3
1 in Stream 1



1 in each Stream



Always two




Balances and Specifications

Material Balance Sub problem

Combined Balance Problem






Material Bal.


4 Species


4 Species

Energy Bal.


See below



4 Comps are given in
a Stream of 4 Species


4 Comps in Stream
of 4 Species



All 3 Temps.
are given



Both = 0



given 1000 kgmol/hr


1000 kgmol/hr




Material Balance Sub problem

Combined Balance Problem

Degree of Freedom



Energy Balance

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Last modified August 6, 1998.