Energy Balance of Reactive Systems Chapter 9

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

Heat of reaction ΔH depends on how the stoichiometric equation is written CH4 (g) + 2O2(g)  CO2(g) + 2H2O(l) ΔHr1 (25OC)= -890.3 kJ/mol  for 1 CH4 2CH4 (g) + 4O2(g)  2CO2(g) + 4H2O(l) ΔHr2 (25OC)= -1780.6 kJ/mol  for 2 CH4

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)= -890.3 kJ/mol CH4 (g) + 2O2(g)  CO2(g) + 2H2O(g) ΔHr2 (25OC)= -802.3 kJ/mol

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)= -2878 kJ/mol For 2400 mol/s CO2 produced, ΔHr2=(2400/4 mol/s)*(-2878 kJ/mol)=-1.73X106kJ/s

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

Hess’s Law Look at this reaction… C(s) + ½ O2(g)  CO(g) : ΔHr = ? (1) C(s) + O2(g)  CO2(g) : ΔHr1 = -393.51 kJ/mol (2) CO(g) + ½ O2(g)  CO2(g): ΔHr2 = -282.99 kJ/mol C(s) + ½ O2(g)  CO(g) ΔHr = ? ΔH = ΔHr1 ΔH = -ΔHr2 CO2 Then...ΔHr = ΔHr1 + (- ΔHr2) = (-393.51 + (-282.99)) = -110.52 kJ/mol

Working Session Example 9.2-1

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 = -365.14 kJ/mol 6 C(s) + 3 H2 (g)  C6H6 (l) ΔHro = 48.66 kJ/mol Note: standard heat of formation of an elemental species is ZERO.

Working Session Example 9.3-1

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) = -1366.9 kJ/mol See Table B.1 Felder-Rousseau

Working Session Example 9.4-1

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

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

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

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

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

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

Working Session Example 9.5-1

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

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

Working Session Example 9.5-5