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Lecture 4: Standard Enthalpies

Lecture 4: Standard Enthalpies. Reading: Zumdahl 9.6 Outline What is a standard enthalpy? Defining standard states. Using standard enthalpies to determine D H° rxn. Hess’ Law.

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Lecture 4: Standard Enthalpies

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  1. Lecture 4: Standard Enthalpies • Reading: Zumdahl 9.6 • Outline • What is a standard enthalpy? • Defining standard states. • Using standard enthalpies to determine DH°rxn

  2. Hess’ Law • Enthalpy is a state function. As such, DH for going from some initial state to some final state is pathway independent. 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.

  3. Definition of DH°f • We saw in the previous lecture the utility of having a series of reactions with known enthalpies. • What if one could deal with combinations • of compounds directly rather than dealing • with whole reactions to determine DHrxn?

  4. Definition of DH°f (cont.) • Standard Enthalpy of Formation, DH°f: “The change in enthalpy that accompanies the formation of 1 mole of a compound from its elements with all substances at standard state.” • We envision taking elements at their standard state, and combining them to make compounds, also at standard state.

  5. What is Standard State? • Standard State: A precisely defined reference state. It is a common reference point that one can use to compare thermodynamic properties. • Definitions of Standard State: • For a gas: P = 1 atm (or 100,000 Pascal; 1 atm=101325) • For solutions: 1 M (mol/l). • For liquids and solids: pure liquid or solid • For elements: The form in which the element exists under conditions of 1 atm and 298 K.

  6. Definitions (cont.) • Standard elemental states (cont.): • Hydrogen: H2 (g) (not atomic H) • Oxygen: O2 (g) • Carbon: C (gr) … graphite as opposed to diamond • We will denote the standard state using the subscript “°”. • Example: DH°rxn

  7. Importance of Elements • We will use the elemental forms as a primary reference when examining compounds. • Pictorially, for chemical reactions we envision taking the reactants to the standard elemental form, then reforming the products.

  8. Example: Combusion of MethaneCH4(g) + 2O2(g) CO2(g) + 2H2O(l)

  9. Elements and DH°f • For elements in their standard state: DH°f = 0 • With respect to the graph, we are envisioning chemical reactions as proceeding through elemental forms. In this way we are comparing DH°f for reactants and products to a common reference (zero)

  10. Tabulated DH°f • In Zumdahl, tables of DH°f are provided in Appendix 4. • The tabulated values represent the DH°f for forming the compound from elements at standard conditions. C (gr) + O2 (g) CO2 (g) DHf° = -394 kJ/mol

  11. Using DH°f to determine DH°rxn • Since we have connected the DH°f to a common reference point, we can now determine DH°rxn by looking at the difference between the DH°f for products versus reactants. • Mathematically: • DH°rxn = SDH°f (products) - SDH° f (reactants)

  12. How to Calculate DH°rxn • When a reaction is reversed, DH changes sign. • If you multiply a reaction by an integer, • DH is also multiplied (it is an extensive variable) • Elements in their standard states are not included in calculations since DH°f = 0.

  13. Example • Determine the DH°rxn for the combustion of methanol. CH3OH (l) + 3/2O2(g) CO2(g) + 2H2O(l) • DH°f in Appendix 4: CH3OH(l) -238.6 kJ/mol CO2(g) -393.5 kJ/mol H2O(l) -286 kJ/mol

  14. Example (cont.) CH3OH (l) + 3/2O2(g)CO2(g) + 2H2O(l) DH°rxn=SDH°f (products)-SDH° f (reactants) = DH°f(CO2(g)) + 2DH°f(H2O(l)) - DH°f(CH3OH(l)) = -393.5 kJ - (2 mol)(286 kJ/mol) - (-238.6 kJ) = -728.7 kJ Exothermic!

  15. Another Example • Using the following reaction: 2ClF3(g) + 2NH3(g) N2(g) + 6HF(g) + Cl2(g) DH°rxn = -1196 kJ Determine the DH°f for ClF3(g) DH°f NH3(g) -46 kJ/mol HF(g) -271 kJ/mol

  16. Another Example (cont.) • Given the reaction of interest: 2ClF3(g) + 2NH3(g) N2(g) + 6HF(g) + Cl2(g) DH°rxn = -1196 kJ DH°rxn= SDH°f (products) - SDH° f (reactants) DH°rxn= 6DH°f(HF(g)) -2DH°f (NH3(g)) -2DH° f (ClF3(g))

  17. Another Example (cont.) DH°rxn= 6DH°f(HF(g)) -2DH°f (NH3(g)) -2DH°f (ClF3(g)) -1196 kJ = (6mol)(-271 kJ/mol) -(2mol)(-46 kJ/mol) -(2mol)DH°f (ClF3(g)) -1196 kJ = -1534 kJ -(2mol)DH°f (ClF3(g)) 338 kJ=-(2mol)DH°f (ClF3(g)) -169 kJ/mol=DH°f (ClF3(g))

  18. Bonus Example • Problem 9.81 DHvap for water at the normal boiling point (373.2 K) is 40.66 kJ. Given the following heat capacity data, what is DHvap at 340.2 K for 1 mol? • CP H20(l) = 75 J/mol.K • CP H20(g) = 36 J/mol.K

  19. Bonus Example (cont.) (373.2 K) H2O(l) H2O(g) DH = 40.66 kJ/mol nCP(H2O(g))DT nCP(H2O(l))DT H2O(l) H2O(g) DH = ? (340.2 K)

  20. Bonus Example (cont.) (373.2 K) H2O(l) H2O(g) DH = 40.66 kJ/mol nCP(H2O(l))DT nCP(H2O(g))DT H2O(l) H2O(g) (340.2 K) DHrxn (340.2K) = DH(product) - DH(react) = DHwater(g) (373.2K)+ nCP(H2O(g))DT - DHwater(l) (373.2K)- nCP(H2O(l))DT

  21. Bonus Example (cont.) DHrxn (340.2K) =DHwater(g) (373.2K)+ nCP(H2O(g))DT - DHwater(l) (373.2K)- nCP(H2O(l))DT DHrxn (340.2K) = 40.66 kJ/mol + 36 J/mol.K(-33 K) - 75 J/mol.K(-33 K) DHrxn (340.2K) = 40.66 kJ/mol + 1.29 kJ/mol = 41.95 kJ/mol

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