1. Chapter 5
Energy Balances
with reaction

2. Energy out = Energy in + generation – consumption – accumulation
CONSERVATION OF ENERGY
A general equation can be written for the conservation of energy:
Energy out = Energy in + generation – consumption – accumulation
This is a statement of the first law of thermodynamics.
An energy balance can be written for any process step.
Chemical reaction will evolve energy (exothermic) or consume energy (endothermic).
For steady-state processes the accumulation of both mass and energy will be zero.
Energy can exist in many forms and this, to some extent, makes an energy balance more complex than a material balance.

5. ΔH°rxn = Σ ΔH°f (products) – Σ ΔH°f (reactants)

6. EXAMPLE: determination of a heat of formation from heat transfer measurements
Suppose that you want to find the standard of formation of CO from experimental data. Can you prepare pure CO from reaction of C and O2 and measure the heat transfer. This wou1d be far too difficult. It would easier experimentally to find first the heat of reaction at standard conditions for the two reactions shown below for the flow process as shown in Figure E25.1.

7. for which the net heat of reaction per gram mole of CO is the heat of formation of CO:

8. Page: 631

9. Page: 631

10. Solution

11. The Heat (Enthalpy) of Reaction ΔH°rxn
You can obtain the heat of reaction from experiments, of course, but it is easier to first see if you can calculate the standard heat of reaction from the known tabulated values of the heats of formation as follows: Consider a steady-state flow process with no work involved, such as the one shown in Figure 25.2 in which benzene (C6H6) reacts with the stoichiometric amount of H2 to produce cyclohexane (C6H12) in the standard state:
Figure Reaction of benzene to form cyclohexane
C6H6(g) + 3 H2(g) → C6H12(g)

12. The energy balance for the process reduces to Q = ΔH, where ΔH is by definition ΔH°rxn for the specified chemical reaction equation. Because we adopt for the heat of reaction the same reference conditions (0 enthalpy for the elements at 25°C and 1 atm) as used in defining the heats of formation, the values of the specific enthalpies associated with each species involved in the reaction are just the values of the respective heats of formation. For the process shown in Figure 25.2 the data are as follows:
The ΔH°rxn for the reaction as written, and actually has the units of energy per moles reacting for the specified chemical equation. In general for complete reaction:

13. EXAMPLE: calculation of Heat of Reaction from heat of formation
Calculate the heat of reaction of 4 gmol NH3 at standard conditions for the following reaction:
Per gram mole of NH3,
NH3

14. ore خامة
cinder نفاية المعادن

16. 4 FeS2 + 11 O2 ======> 2 Fe2O3 + 8 SO2

18. Out In

22. Since the amount of N2 and gangue are the same at the inlet and out let and there are no changes in T and P conditions at the inlet and outlet, then

24. From excess air information:
An iron pyrite containing 85% FeS2 and 15% gangue (inert dirt, rock) is roasted with an amount of air equal to 200% excess air according to the reaction
MW of Fe = 55.85
Fe2O3 = 159.7
FeS2 = 120
4FeS2 + 11O Fe2O3 + 8SO2 4 11
0.7083
Required O2
n (moles) of FeS2 = 0.85 × 100 kg/ 120
Mol FeS2 = (85/120) = kg mol
Required O2 = ×11/4 = kg mol
Excess O2 = ×200% = kg mol
Total O2 in = = kg mol
Total N2 in = ×0.79/0.21 = kg mol

25. MW of Fe = 55.85 Fe2O3 = 159.7 FeS2 = 120 ……………….. [1] ……….... [2]
……………….. [1]
……….... [2]
…….…….….. [3]
Also,
……….... [4]
Solving the last 4 equations simultaneously,

28. N2 (g) + C2H2 (g) 2HCN (g) N2 (g) + C2H2 (g) 2HCN (g)
Calculation of the Heat of Reaction at a Temperature Different from the Standard Conditions
ΔH rxn500 °C
N2 (g) + C2H2 (g) HCN (g)
N2 (g) + C2H2 (g) HCN (g)
500 °C
500 °C
ΔH1
ΔH2
ΔHrxn25°C
25 °C
25 °C
ΔH rxn500 °C = ΔH1(25 – 500) + ΔH rxn25 °C + ΔH2(500 – 25)
ΔH rxnT °C = ΔH1(25 – T) + ΔH rxn25 °C + ΔH2(500 – 25)

29. Calculation of Heat of Reaction at different temperatures

30. CO2(g) + 4H2(g) → 2H2O(g) + CH4(g)
EXAMPLE: Calculation of the Heat of Reaction at a Temperature Different from the Standard Conditions
Public concern about the increase in the carbon dioxide in the atmosphere has led to numerous proposals to sequester or eliminate the carbon dioxide. An inventor believes he has developed a new catalyst that can make the gas phase reaction:
CO2(g) + 4H2(g) → 2H2O(g) + CH4(g)
proceed with 100% conversion of the CO2. The source of the hydrogen would be from the electrolysis of water using electricity generated from solar cells. Assume stoichiometric amounts of the reactants enter the reactor. Determine the heat of reaction if the gases enter and leave at 1 atm and 500oC.
SOLUTION Basis: 1 g mol CO2(g) at 500 °C and 1 atm

31. CO2(g) + 4H2(g) → 2H2O(g) + CH4(g)

32. Calculation of Heat of Reaction at different temperatures

33. Application of Energy Balances with Reactions Processes
In this section we primarily illustrate the solution of continuous, steady-state processes for which the general energy balance reduces to two choices:
With the effects of chemical reaction merged with the sensible heats
With the effects of chemical reaction lumped in the heat of reaction
merged اندمجت
مع آثار تفاعل كيميائي جمعها في حرارة التفاعل

34. Application of Energy Balances with Reactions Processes
The steady-state with Q = 0 reduces to just ΔH = 0. If you use tables such as in Appendix D than heat capacity equations to calculate the "sensible heats" of the various streams entering and leaving the reactor, the calculations will involve trial and error. To find the exit temperature for which ΔH = 0, if tables are used as the source of the Δ values, the simplest procedure is to:

35. EXAMPLE: Calculation of an Adiabatic Reaction (Flame) Temperature
Calculate the theoretical flame temperature for CO gas burned at constant pressure with 100% excess air, when the reactants enter at 100°C and 1 atm.
Solution
The system is shown in Figure E26.2. We will use data from Appendix. The process is a steady-state flow system. Ignore any equilibrium effects.

38. T ΔH
2000
36740
TFT
1750
– 16657
Theoretical Flame Temperature

44. ANEROBIC اللاهوائية
mannitol = compound is found in many plants and is used in various foods and medical products.

46. Green chemistry refers to the adoption of chemicals in commercial processes that reduce concern with respect to the environment. An example is the elimination of methyl isocyanate, a very toxic gas, in the production of carbaryl (1-napthalenyl methyl carbamate).
In 1984 in Bhopal, India, the accidental release of methyl isocyanate in a residential area led to the death of thousands of people and the injury of many thousands more. The Bhopal process can be represented by the reaction equations (a) and (b):

55. Heat of Combustion

56. Heat of Combustion

57. The heating value is the amount of heat released during the combustion of a specified amount of it.

58. EXAMPLE Heating Value of coal

59. Q = C H S (O)

62. Component
lbmol
T (°F)
ΔHf
(Btu/lbmol)
ΔH (Btu/lbmol) CO 50 –
945.83 O2
1000 N2
CO2
800 –

69. 3 glucose O BM CA CO2

70. 3 glucose O BM CA CO2

71. 3 glucose O BM CA CO2

72. 3 glucose O BM CA CO2

73. 3 glucose + 7.8 O2 5.35 BM + 2.22 CA + 4.50 CO2 3 2.22 X 52.05 X =
Mass = n × Mwt=

74. Mass = n × Mwt=

75. 3 glucose O BM CA CO2

76. 3 glucose O BM CA CO2

77. 3 glucose + 7.8 O2 5.35 BM + 2.22 CA + 4.50 CO2 3 7.8 5.35 2.22 X Z Y
52.05

78. 3 glucose O BM CA CO2 3
7.8
4.5
117.32
O2 accum
CO2 accum

79. work done = power × time