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Lecture 3: Hess’ Law

Lecture 3: Hess’ Law. Reading: Zumdahl 9.5 Outline Definition of Hess’ Law Using Hess’ Law (examples). First Law: Energy of the Universe is Constant E = q + w q = heat. Transferred between two bodies w = work. Force acting over a distance (F x d).

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Lecture 3: Hess’ Law

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  1. Lecture 3: Hess’ Law • Reading: Zumdahl 9.5 • Outline • Definition of Hess’ Law • Using Hess’ Law (examples)

  2. First Law: Energy of the Universe is Constant E = q + w q = heat. Transferred between two bodies w = work. Force acting over a distance (F x d) First Law of Thermodynamics

  3. Definition of Enthalpy • Thermodynamic Definition of Enthalpy (H): H = E + PV E = energy of the system P = pressure of the system V = volume of the system

  4. Changes in Enthalpy • Consider the following expression for a chemical process: DH = Hproducts - Hreactants If DH >0, then qp >0. The reaction is endothermic If DH <0, then qp <0. The reaction is exothermic

  5. Ideal Monatomic Gas Cv = 3/2R Cp = Cv + R = 5/2 R Polyatomic Gas Cv > 3/2R Cp > 5/2 R Heat Capacity, energy and enthalpy • DE = q + w • w = -PextDV (for now) • DE = nCvDT = qV • DH = nCpDT = qP

  6. Thermodynamic State Functions • Thermodynamic State Functions: Thermodynamic properties that are dependent on the state of the system only. (Example: DE and DH) • Other variables will be dependent on pathway (Example: q and w). These are NOT state functions. The pathway from one state to the other must be defined.

  7. Hess’ Law Defined • From lecture 3: Enthalpy is a state function. As such, DH for going from some initial state to some final state is pathway independent. • Hess’ Law: DH for a process involving the transformation of reactants into products is not dependent on pathway. Therefore, we can pick any pathway to calculate DH for a reaction.

  8. Hess’ Law: An Example

  9. Using Hess’ Law • When calculating DH for a chemical reaction as a single step, we can use combinations of reactions as “pathways” to determine DH for our “single step” reaction.

  10. Example (cont.) • Our reaction of interest is: N2(g) + 2O2(g) 2NO2(g) DH = 68 kJ • This reaction can also be carried out in two steps: N2 (g) + O2 (g) 2NO(g) DH = 180 kJ 2NO (g) + O2 (g) 2NO2(g) DH = -112 kJ

  11. Example (cont.) • If we take the previous two reactions and add them, we get the original reaction of interest: N2 (g) + O2 (g) 2NO(g) DH = 180 kJ 2NO (g) + O2 (g) 2NO2(g) DH = -112 kJ N2 (g) + 2O2 (g) 2NO2(g) DH = 68 kJ

  12. Changes in Enthalpy • Consider the following expression for a chemical process: DH = Hproducts - Hreactants If DH >0, then qp >0. The reaction is endothermic If DH <0, then qp <0. The reaction is exothermic

  13. Example (cont.) • Note the important things about this example, the sum of DH for the two reaction steps is equal to the DH for the reaction of interest. • We can combine reactions of known DH to determine the DH for the “combined” reaction.

  14. Hess’ Law: Details • Once can always reverse the direction of a reaction when making a combined reaction. When you do this, the sign of DH changes. N2(g) + 2O2(g) 2NO2(g) DH = 68 kJ 2NO2(g) N2(g) + 2O2(g) DH = -68 kJ

  15. Details (cont.) • The magnitude of DH is directly proportional to the quantities involved (it is an “extensive” quantity). • As such, if the coefficients of a reaction are multiplied by a constant, the value of DH is also multiplied by the same integer. N2(g) + 2O2(g) 2NO2(g) DH = 68 kJ 2N2(g) + 4O2(g) 4NO2(g) DH = 136 kJ

  16. Using Hess’ Law • When trying to combine reactions to form a reaction of interest, one usually works backwards from the reaction of interest. • Example: What is DH for the following reaction? 3C (gr) + 4H2 (g) C3H8 (g)

  17. Example (cont.) 3C (gr) + 4H2 (g) C3H8 (g) DH = ? • You’re given the following reactions: C (gr) + O2 (g) CO2 (g) DH = -394 kJ C3H8 (g) + 5O2 (g) 3CO2 (g) + 4H2O (l) DH = -2220 kJ H2 (g) + 1/2O2 (g) H2O (l) DH = -286 kJ

  18. Example (cont.) • Step 1. Only reaction 1 has C (gr). Therefore, we will multiply by 3 to get the correct amount of C (gr) with respect to our final equation. Initial: C (gr) + O2 (g) CO2 (g) DH = -394 kJ Final: 3C (gr) + 3O2 (g) 3CO2 (g) DH = -1182 kJ

  19. Example (cont.) • Step 2. To get C3H8 on the product side of the reaction, we need to reverse reaction 2. Initial: C3H8 (g) + 5O2 (g) 3CO2 (g) + 4H2O (l) DH = -2220 kJ Final: 3CO2 (g) + 4H2O (l) C3H8 (g) + 5O2 (g) DH = +2220 kJ

  20. Example (cont.) • Step 3: Add two “new” reactions together to see what is left: 3C (gr) + 3O2 (g) 3CO2 (g) DH = -1182 kJ 3CO2 (g) + 4H2O (l) C3H8 (g) + 5O2 (g) DH = +2220 kJ 2 3C (gr) + 4H2O (l) C3H8 (g) + 2O2 DH = +1038 kJ

  21. Example (cont.) • Step 4: Compare previous reaction to final reaction, and determine how to reach final reaction: 3C (gr) + 4H2O (l) C3H8 (g) + 2O2 DH = +1038 kJ H2 (g) + 1/2O2 (g) H2O (l) DH = -286 kJ 3C (gr) + 4H2 (g) C3H8 (g) Need to multiply second reaction by 4

  22. Example (cont.) • Step 4: Compare previous reaction to final reaction, and determine how to reach final reaction: 3C (gr) + 4H2O (l) C3H8 (g) + 2O2 DH = +1038 kJ 4H2 (g) + 2O2 (g) 4H2O (l) DH = -1144 kJ 3C (gr) + 4H2 (g) C3H8 (g)

  23. Example (cont.) • Step 4 (cont.): 3C (gr) + 4H2O (l) C3H8 (g) + 2O2 DH = +1038 kJ 4H2 (g) + 2O2 (g) 4H2O (l) DH = -1144 kJ 3C (gr) + 4H2 (g) C3H8 (g) DH = -106 kJ

  24. Changes in Enthalpy • Consider the following expression for a chemical process: DH = Hproducts - Hreactants If DH >0, then qp >0. The reaction is endothermic If DH <0, then qp <0. The reaction is exothermic

  25. Another Example • Calculate DH for the following reaction: H2(g) + Cl2(g) 2HCl(g) Given the following: NH3 (g) + HCl (g) NH4Cl(s) DH = -176 kJ N2 (g) + 3H2 (g) 2NH3 (g) DH = -92 kJ N2 (g) + 4H2 (g) + Cl2 (g) 2NH4Cl(s) DH = -629 kJ

  26. Another Example (cont.) • Step 1: Only the first reaction contains the product of interest (HCl). Therefore, reverse the reaction and multiply by 2 to get stoichiometry correct. NH3 (g) + HCl (g) NH4Cl(s) DH = -176 kJ 2NH4Cl(s) 2NH3 (g) + 2HCl (g) DH = 352 kJ

  27. Another Example (cont.) • Step 2. Need Cl2 as a reactant, therefore, add reaction 3 to result from step 1 and see what is left. 2NH4Cl(s) 2NH3 (g) + 2HCl (g) DH = 352 kJ N2 (g) + 4H2 (g) + Cl2 (g) 2NH4Cl(s) DH = -629 kJ N2 (g) + 4H2 (g) + Cl2 (g) 2NH3(g) + 2HCl(g) DH = -277 kJ

  28. Another Example (cont.) • Step 3. Use remaining known reaction in combination with the result from Step 2 to get final reaction. N2 (g) + 4H2 (g) + Cl2 (g) 2NH3(g) + 2HCl(g) DH = -277 kJ ( N2 (g) + 3H2(g) 2NH3(g) DH = -92 kJ) H2(g) + Cl2(g) 2HCl(g) DH = ? Need to take middle reaction and reverse it

  29. Another Example (cont.) • Step 3. Use remaining known reaction in combination with the result from Step 2 to get final reaction. N2 (g) + 4H2 (g) + Cl2 (g) 2NH3(g) + 2HCl(g) DH = -277 kJ 2NH3(g) 3H2 (g) + N2 (g) DH = +92 kJ 1 H2(g) + Cl2(g) 2HCl(g) DH = -185 kJ

  30. Changes in Enthalpy • Consider the following expression for a chemical process: DH = Hproducts - Hreactants If DH >0, then qp >0. The reaction is endothermic If DH <0, then qp <0. The reaction is exothermic

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