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Chemistry 231 Thermodynamics in Reacting Systems.

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Presentation on theme: "Chemistry 231 Thermodynamics in Reacting Systems."— Presentation transcript:

1 Chemistry 231 Thermodynamics in Reacting Systems

2 Enthalpy Changes for Reactions The shorthand form for a chemical reaction  J = chemical formula for substance J J = stoichiometric coefficient for J

3 Reaction Enthalpy Changes The enthalpy change for a chemical reaction H m [J] = molar enthalpies of substance J n J = number of moles of J in the reaction

4 The Enthalpy Change Reaction beginning and ending with equilibrium or metastable states Note – Initial and final states have the same temperature and pressure!

5 Reaction Enthalpies (cont’d) We note that 1 mole of a reaction occurs if

6 A Standard State Reaction A reaction that begins and ends with all substances in their standard states The degree sign, either  or  P = 1.00 bar [aqueous species] = 1.00 mol/ kg T = temperature of interest (in data tables - 25  C or 298 K).

7 Standard Reaction Enthalpies We note that for 1 mole of a reaction under standard conditions

8 The Formation Reaction A "chemical thermodynamic reference point." For CO and CO 2 C (s) + O 2 (g)  CO 2 (g) C (s) + ½ O 2 (g)  CO (g)

9 The Formation Reaction The formation reaction 1 mole of a compound constituent elements stable state of aggregation at that temperature. Formation of 1.00 mole of Na 2 SO 3 (s) 2 Na(s) + S(s) + 3/2 O 2 (g)  Na 2 SO 3 (s) ‘Formation enthalpy of Na 2 SO 3 (s)’,  f H°[Na 2 SO 3 (s)]

10 The Significance of the Formation Enthalpy  f H° is a measurable quantity! Compare CO (g) with CO 2 (g) C (s) + 1/2 O 2 (g)  CO (g)  f H° [CO(g)] = -110.5 kJ/mole C (s) + O 2 (g)  CO 2 (g)  f H° [CO 2 (g)] = - 393.5 kJ/mole

11 Formation Enthalpies Formation enthalpies - thermodynamic reference point! H o m [J] =  f H  [J] H m  [elements] = 0 kJ / mole. Use the tabulated values of the formation enthalpies

12 The General Equation The enthalpy change for a given reaction is calculated from the formation enthalpies as Notes Reverse a reaction Multiply a reaction by an integer

13 The Calorimeter A calorimeter - device containing water and/or another substance with a known heat capacity Calorimeters – either truly or approximately adiabatic systems

14 Two major types of calorimeters. The constant volume (bomb) calorimeter.  U = q v. The constant pressure calorimeter.  H = q p.

15 The Constant Volume (Bomb) Calorimeter

16 The Constant Pressure Calorimeter

17 Relating H and U The enthalpy and the internal energy both represent quantities of heat.  U = q v.  H = q p. Relate the two state functions using the following relationship  U =  H -  PV

18 Other Important Enthalpy Changes Enthalpy of solution Enthalpy of dilution Enthalpy of fusion Enthalpy of vapourisation

19 The Solution Enthalpy  sol H - heat absorbed or released when a quantity of solute is dissolved in fixed amount of solvent  sol H = H m (sol’n) – H m (component) H(component) = H m (solid) + H m (solvent) Two definitions Standard Limiting

20 The Dilution Enthalpy For the process, HCl (aq, 6 M)  HCl (aq, 1 M). The Enthalpy of dilution of the acid.  dil H = H m (sol’n 2) – H m (sol’n,1)

21 Reaction Enthalpy Changes With Temperature Differentiate the reaction enthalpy with temperature

22 The Result  r C  p - the heat capacity change for the reaction

23 Internal Energy Changes in Chemical Reactions Examine a chemical reaction. C (s) + O 2 (g)  CO 2 (g)  U = U[CO 2 (g)] – U[C(s)] – U[O 2 (g)] Note -  r H  = -393.5 kJ/mole

24 Enthalpies and Hess’s Law Use tabulated values of formation enthalpies to obtain  r H°. May also estimate reaction enthalpies using an indirect method.

25 Hess’s Law Hess’s Law – the enthalpy change for a given reaction is the same whether the reaction occurs in a single step or in many steps.

26 The Entropy Change in a Chemical Reaction Burning ethane! C 2 H 6 (g) + 7/2 O 2 (g)  2 CO 2 (g) + 3 H 2 O (l) The entropy change is calculated in a similar fashion to that of the enthalpies

27 Some Generalizations For any gaseous reaction (or a reaction involving gases).  n g > 0,  r S  > 0 J/(K mole).  n g < 0,  r S  < 0 J/(K mole).  n g = 0,  r S   0 J/(K mole). For reactions involving only solids and liquids – depends on the entropy values of the substances.

28 The Gibbs Energy Change for a Chemical Reaction The standard Gibbs energy change for a chemical reaction is obtained as follows

29 The Gibbs Energy Change For the methane combustion reaction 1 CH 4 (g) + 2 O 2 (g)  1 CO 2 (g) + 2 H 2 O(l)  r G  =  n p  f G  (products) -  n r  f G  (reactants) = 2  f G  [H 2 O(l)] + 1  f G  [CO 2 (g)] - (7/2  f G  [O 2 (g)] + 1  f G  [CH 4 (g)] )

30 The Formation Gibbs Energies  f G  (elements) = 0 kJ / mole. Tabulated values at SATP used to obtain the Gibbs energy changes for chemical reactions.

31 A Caveat!!!  r G° refers to standard conditions only! For non-standard conditions -  r G  r G < 0 - reaction moves in the forward direction  r G > 0 - reaction moves in the reverse direction  r G = 0 - reaction is at equilibrium

32 Bond Energies Examine the following reactions H 2 (g)  H (g) + H (g)  U° = 433.9 kJ Cl 2 (g)  Cl (g) + Cl (g)  U° = 239.5 kJ Bond dissociation energies. Enthalpy changes are designated D (H- H) and D (Cl-Cl).

33 For Polyatomic Molecules CO 2 (g)  C (g) + 2 O (g)  U  = 740 kJ  H of this reaction D(C=O) What about dissociating methane into C + 4 H’s? CH 4 (g)  C(g) + 4 H(g)  U° = 1640 kJ 4 C-H bonds in CH 4  D (C-H)  410 kJ/mol

34 Make or Break!! Note: all chemical reactions involve the breaking and reforming of chemical bonds Bonds break - we add energy. Bonds form - energy is released.  r U°   D(bonds broken) -  D(bonds formed)

35 A Word of Caution These are close but not quite exact. Why? The bond energies we use are averaged bond energies ! This is a good approximation for reactions involving diatomic species. Can only use the above procedure for GAS PHASE REACTIONS ONLY!!!


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