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Thermochemistry Internal Energy Kinetic energy Potential energy.

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Presentation on theme: "Thermochemistry Internal Energy Kinetic energy Potential energy."— Presentation transcript:

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2 Thermochemistry Internal Energy Kinetic energy Potential energy

3 Thermochemistry Internal Energy Kinetic energy Potential energy

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5 Chemical Energy Changes System and Surroundings Exothermic Reaction

6 Endothermic Reaction

7 Thermochemistry Thermochemisty is the study of the relationship between heat and chemical reactions. 1. Kinetic energy is energy possessed by matter because it is in motion –Thermal energy-- random motion of the particles in any sample above 0 K –Heat -- causes a change in the thermal energy of a sample. Flows from hot to cold

8 Heat

9 Potential Energy 2. Potential energy is energy possessed by matter because of its position or condition. –A brick on top of a building has potential energy that is converted to kinetic energy when it is dropped on your head –Chemical energy is energy possessed by atoms as a result of forces which hold the atoms together (Boxes!)

10 Where is the Energy? Definitions we will use: –System: Reaction (bonds) –Surrounding: solvent, reaction vessel, air, etc. An everyday example: burning wood –Initially, much energy stored as potential in C-H bonds, little kinetic energy in the air –Finally, lower potential energy in the C=O bonds, higher kinetic energy in the air

11 Total Energy Total Energy = kinetic + potential Law of Conservation of Energy - The total energy of universe is constant Internal Energy - E - the sum of all the kinetic and potential energies of all the atoms and molecules in a sample.

12 Change in Energy of System Change in internal energy of system = heat + work Convention: point of view of system

13 Change in Internal Energy  E = q + w Work = Force x distance What happens to your internal energy when you push a boulder? What happens to your internal energy when you push a boulder on a rough surface?

14 Chemical Work ׀W׀ = ׀ F x  h ׀ P = F/A ׀W׀ = ׀P x A x h׀ ׀W׀ = ׀PV׀ Sign Convention: W = -P  V

15 Test Your Understanding For the following three reactions: –Are they performed under constant pressure or not? –What is the sign of work in each case?

16 State Function –Internal Energy –Pressure –Volume Path Dependent –Work –heat Property depends only on present state

17 Enthalpy Most reactions are done in open containers, so P is constant Need a term for constant pressure where only work is PV At constant pressure, q p ΔE = q p – PΔV q p = ΔE + PV ΔH = ΔE + PΔV (definition) ΔH = q p

18 Enthalpy IF pressure is constant and only work is PV Change in enthalpy is equal to flow of energy in form of heat –Measurable by Temperature –Change in enthalpy is “heat of reaction” If no net change in moles of gas, enthalpy ~ energy

19 Enthalpy Signs on ΔH –+ heat is taken in by system –- heat is given off by system Endothermic ΔH = + Exothermic ΔH = -

20 Exothermic 2 Al (s) + Fe 2 O 3 (s)  Al 2 O 3 (s) + 2 Fe (s) + energy Which bonds have more potential energy? Which bonds are stronger?

21 Endothermic Ba(OH) 2. 8H 2 O (s) + 2 NH 4 SCN (s) + energy  Ba(SCN) 2 (aq) + 2 NH 3 (g) + 10 H 2 O (l)

22 Reaction Enthalpy CH 4 (g) + 2 O 2 (g)  CO 2 (g) + 2 H 2 O (l) + ENERGY It is useful to know how much energy is released CH 4 (g) + 2 O 2 (g)  CO 2 (g) + 2 H 2 O (l) ΔH = kJ This is a stoichiometric amount 890 kJ released = 1 mol CH 4 = 2 mol O 2 = 1 mol CO 2 = 2 mol H 2 O

23 Reaction Enthalpy Depends on coefficients, direction, and phases –2 CH 4 (g) + 4 O 2 (g)  2 CO 2 (g) + 4 H 2 O (l) ΔH = kJ –CO 2 (g) + 2 H 2 O (l)  CH 4 (g) + 2 O 2 (g) ΔH = kJ –CH 4 (g) + 2 O 2 (g)  CO 2 (g) + 2 H 2 O (g) ΔH = -802 kJ

24 Reaction Enthalpy

25 Standard Reaction Enthalpy Standard State - a compound in its pure state at 1 atm pressure, all solutions are 1 M Temperature can vary but usually K Standard Reaction Enthalpy ( ΔH r o )- reaction enthalpy when all products and reactants are in the standard state CH 4 (g) + 2 O 2 (g)  CO 2 (g) + 2 H 2 O (l) ΔH o = kJ

26 Enthalpy and Stiochiometry

27 Application: Enthalpies of Combustion Standard Enthalpy of Combustion ( ΔH c o) -- change in enthalpy for the combustion of one mole substance at standard conditions Combustion is combination with O 2 to give CO 2 and water Specific Enthalpy -- the enthalpy of combustion per gram enthalpy density -- enthalpy of combustion per liter

28 Enthalpies of Combustion

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31 How Do We Determine Standard Reaction Enthalpies? Tabular Data: Hess’s Law –Combining appropriate reactions –Heat of Formation –Bond enthalpies Experimental –Calorimetry: constant pressure –Calorimetry: constant volume

32 Hess’s Law Just a restatement of the first law

33 How Do We Determine Standard Reaction Enthalpies? Tabular Data: Hess’s Law –Combining appropriate reactions –Heat of Formation –Bond enthalpies Experimental –Calorimetry: constant pressure –Calorimetry: constant volume

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35 Using Hess’s Law Find ΔH o for C (s) + ½ O 2 (g)  CO (g) C (s) + O 2 (g)  CO 2 (g) ΔH o = kJ 2 CO (g) + O 2 (g)  2 CO 2 (g) ΔH o = kJ

36 How Do We Determine Standard Reaction Enthalpies? Tabular Data: Hess’s Law –Combining appropriate reactions –Heat of Formation –Bond enthalpies Experimental –Calorimetry: constant pressure –Calorimetry: constant volume

37 Standard Enthalpies of Formation The standard enthalpy of formation is the enthalpy change when one mole of a substance in its standard state is formed from the elements in their standard states. H o f Write an equation for the standard heat of formation of carbon dioxide –C(s) + O 2 (g)  CO 2 (g) H o f (CO 2 ) Write an equation for  H o f of CH 3 OH Write an equation for  H o f of N 2 (g) and explain why its value is zero.

38 Standard Enthalpies of Formation  H o f can be compiled in table form

39 Application of Heat of Formation Hess’s Law Without the limitations of combining limited number of reactions Common starting point: elements

40 Calculating Heat of Reaction ΔH o r = ΣH o f (products) - ΣH o f (reactants) CH 4 (g) + 2 O 2 (g)  CO 2 (g) + 2 H 2 O (l)

41 How Do We Determine Standard Reaction Enthalpies? Tabular Data: Hess’s Law –Combining appropriate reactions –Heat of Formation –Bond enthalpies Experimental –Calorimetry: constant pressure –Calorimetry: constant volume

42 Using Bond Enthalpies Most versatile Least exact Must be able to draw Lewis Dot structures Textbook/Bond-Enthalpies-718.html

43 Bond Enthalpies Bond Dissociation Energies –Positive values Hess’s Law –In principle, “free atoms” formed  H rxn =  BDE (broken) –  BDE (formed)

44 Bond Enthalpies Calculate the heat of reaction for the combustion of formamide (CH 3 NO). 1.Equation 2.Lewis Dot 3.Calculate

45 How Do We Determine Standard Reaction Enthalpies? Tabular Data: Hess’s Law –Combining appropriate reactions –Heat of Formation –Bond enthalpies Experimental –Calorimetry: constant pressure –Calorimetry: constant volume

46 Heat Capacity Heat Capacity (C) - the heat required to raise the temperature of an object by 1 K C = q/ΔT extensive property

47 Specific Heat Capacity Specific heat capacity (C s ) - the heat required to raise 1 g of a substance by 1 K Specific heat C s = C/m intensive property q = m C s ΔT

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49 Each Substance “Stores” Heat Differently Putting the same amount of heat into two different substances will raise their temperature differently based on the specific heat of each substance q=mC s  T 100 J 1 g copper 1 g water 25 o C Increased 263 o C Increased 24 o C

50 Calorimetry Calorimeter - an insulated container fitted with a thermometer Open to atmosphere, so P is constant q p = m C s ΔT q p = ΔH

51 Calorimetry Problem In a coffee cup calorimeter, 50.0 mL of M silver nitrate and 50.0 mL of M HCl are mixed. The following reaction occurs: Ag + (aq) + Cl - (aq)  AgCl (s). If the two solutions are initially at o C, and if the final temperature is o C, calculate the change in enthalpy for the reaction. (What assumptions need to be made?)

52 Bomb Calorimeter Volume is constant q = ΔE q system = -q surroundings q rxn = -(q bomb + q water ) q water = m water C s ΔT q bomb = m bomb C s ΔT or C  T

53 Thermodynamics of Ideal Gas Heat capacity of monoatomic gas From KMT, KE = 3/2RT KE = translational energy Heat required to raise temp 1 degree is 3/2R At constant volume, no work is done –Molar heat capacity C v = 3/2R = J/mol K Is this constant for all gases?

54 Consider polyatomic gases For polyatomic molecules, _____ energy has to be put into the same amount of gas to raise it by 1 degree Where does this energy go? (Not translational!) Trend??

55 Monoatomic Gas at Constant Pressure Energy input does two things: increase translational energy (T) and expand gas (w) w = P  V = nR  T Molar heat capacity –C p = 3/2R + R = 5/2R

56 Polyatomic at Constant Pressure Explain physical basis of this equation: C p =C v + R This is observed! When would you see a deviation from C p -C v = R?

57 Summary

58 Conceptual Understanding of Gas Cycle

59 Thermochemistry of Physical Change Vaporization - endothermic process Vapor has higher H that a liquid at the same temperature Enthalpy of vaporization ΔH vap -- ΔH vap = H vapor - H liquid

60 Freezing, Melting and Sublimation Enthalpy of fusion ΔH fus (melting) ΔH fus = H liquid - H solid Enthalpy of freezing = - ΔH fus Enthalpy of sublimation, ΔH sub ΔH sub = H vapor - H solid

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63 Heating/Cooling Curve

64 Enthalpy of Solution Enthalpy (Heat) of Solution – ΔH soln – heat change associated with the dissolution of a known amount of solute in a known amount of solvent. ΔH soln = H soln – H components

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66 Lattice Energy STEP 1 Lattice Energy (U) – the energy required to separate 1 mole of a solid ionic compound into gaseous ions. NaCl (s)  Na + (g) + Cl - (g) U = 788 kJ/mol

67 Heat of Hydration Heat of hydration – ΔH hydr – Enthalpy change associated with the hydration process. Na + (g) + Cl - (g)  Na + (aq) + Cl - (aq) ΔH hydr = -784 kJ

68 ΔH soln = U +ΔH hydr = 788 kJ/mol + (-784 kJ) = 4 kJ/mol


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