1. Energy Balance of Reactive Systems
Chapter 9

2. Types of reaction
Exothermic reaction: the product molecules have lower internal energies than the reactants at the same T and P.. ΔH = NEGATIVE, reaction produces heat
Endothermic reaction: the product molecules have higher internal energies than the reactants. ΔH = POSITIVE, reaction consumes heat

3. Heat of reaction
ΔH depends on how the stoichiometric equation is written
CH4 (g) + 2O2(g)  CO2(g) + 2H2O(l) ΔHr1 (25OC)= kJ/mol  for 1 CH4
2CH4 (g) + 4O2(g)  2CO2(g) + 4H2O(l) ΔHr2 (25OC)= kJ/mol  for 2 CH4

4. Heat of reaction (cont’d)
ΔH depends on the states of aggregation (gas, liquid, or solid)
CH4 (g) + 2O2(g)  CO2(g) + 2H2O(l)
ΔHr1 (25OC)= kJ/mol
CH4 (g) + 2O2(g)  CO2(g) + 2H2O(g)
ΔHr2 (25OC)= kJ/mol

5. Standard heat of reaction (ΔHro)
 heat of reaction when both reactants and products are at reference conditions (usually 25 C and 1 atm)
C4H10 (g) + 13/2O2(g)  4CO2(g) + 5H2O(l)
ΔHr1 (25OC)= kJ/mol
For 2400 mol/s CO2 produced,
ΔHr2=(2400/4 mol/s)*(-2878 kJ/mol)=-1.73X106kJ/s

6. Reaction in a REACTOR (constant V)
ΔUr(T)=Uproducts – Ureactants

7. Hess’s Law Look at this reaction… C(s) + ½ O2(g)  CO(g) : ΔHr = ?
(1) C(s) + O2(g)  CO2(g) : ΔHr1 = kJ/mol
(2) CO(g) + ½ O2(g)  CO2(g): ΔHr2 = kJ/mol
C(s) + ½ O2(g)  CO(g) ΔHr = ?
ΔH = ΔHr1
ΔH = -ΔHr2
CO2
Then...ΔHr = ΔHr1 + (- ΔHr2) = ( ( )) = kJ/mol

8. Working Session
Example 9.2-1

9. Heat of formation (ΔHf)
 Enthalpy change associated with the formation of 1 mole of a compound from its elemental constituents (in nature) at a reference T and P
N2 (g) + 2 H2 (g) + 3/2 O2 (g)  NH4NO3 (c) ΔHro = kJ/mol
6 C(s) + 3 H2 (g)  C6H6 (l) ΔHro = kJ/mol
Note: standard heat of formation of an elemental species is ZERO.

10. Working Session
Example 9.3-1

11. Heat of combustion (ΔHc)
 the heat of the combustion of a substance with oxygen to yield specific products
C2H5OH (l) + 3 O2 (g)  2 CO2 (g) + 3 H2O (l) :
ΔHc (25oC, 1 atm) = kJ/mol
See Table B.1 Felder-Rousseau

12. Working Session
Example 9.4-1

13. Energy Balances (general procedures)
Heat of reaction method:
Reactants
Tin
Products
Tout
T=25 oC
ΔHro ΔH
ΔH1
ΔH2

14. Energy Balances (general procedures)
Heat of formation method ΔH
Reactants
Tin
Products
Tout
ΔH1
ΔH2
Elements
25 oC

15. Some notes…. (extent of reaction)
For a single reaction, extent of reaction (ξ) can be calculated:

16. Some notes… (inlet-outlet enthalpy table)
Components
nin
Hin
nout
Hout A B C

17. Some notes…
Latent heat  heat transferred without change of T. It could be as heat of vaporization (or condensation). There is a phase change
Sensible heat  heat transferred due to the T difference. There is a change of T, but no phase change
Since Cp = f(T), then don’t forget to integrate it. See Table B.4

18. STEP-by-STEP procedure of calculation
Material balance calculation of reactor
Choose reference states (usually 25 C, 1 atm)
Calculate extent of reaction
Prepare inlet-outlet enthalpy table
Calculate unknown component enthalpy
Calculate ΔH for the reactor

19. Working Session
Example 9.5-1

20. Working Session
Examples
9.5-2
9.5-3
9.5-4

21. Thermochemistry of solutions
the enthalpy change associated with the formation of a solution from the solute elements and the solvent
Standard heat of formation of solution
Heat of solution at 25 C

22. Working Session
Example 9.5-5