Unit 5 - Energetics. Exo vs Endo, Enthalpy, Hess s Law, Born-Haber, Entropy, Spontaneity (Gibbs Free Energy)

Transcription

1 Unit 5 - Energetics Exo vs Endo, Enthalpy, Hess s Law, Born-Haber, Entropy, Spontaneity (Gibbs Free Energy)

2 Heating some water... You re job is to figure out how we can find the heat change for one mole of butane in the canister...as accurately as possible you have 20 minutes to brainstorm (need materials list too!) and write a procedure...this is not a formal lab (because I know some of you started freaking out!)

3 Homework!! Write a paragraph discussing the outcome of your experiment. Mention REALISTIC random AND systematic errors that (actually) occurred. Please discuss realistic suggestions for improvement.

4 Thoughts.. What type of change is this? What equations might we need to use? What can we do to minimize error? How many times should we do it (or different setups)?

5 Energy and Heat Energy - a measure of the ability to do work Heat - a form of energy which is transferred as a result of a temperature difference System vs. Surroundings water + (NH 4 )NO 3 - describe

6 Enthalpy H - overall energy change in a system when H is negative... when H is positive...

7 Hess s Law if H 1 = -277kJ mol -1 H 2 =-1367kJ mol -1 H 3 = mol -1 what is the relationship between the 3?

8 H 1 = H 2 + H 3 Hess s Law

9 Example... Calculate the H value of the following rxn. H 2 O + C CO + H 2 Given Steps: H 2 + 1/2O 2 H 2 O H = kJ 2CO 2C + O 2 H = 221.0kJ

10 One more... Calculate the H value of the following rxn. C 3 H 8 + 5O 2 3CO 2 + 4H 2 O Given: 3C + 4H 2 C 3 H 8 H = kj 2H 2 + O 2 2H 2 O H = kj C + O 2 CO 2 H = kj

11 One more... Using the equations below: C (s) + O 2(g) CO 2(g) H = -390 kj Mn (s) + O 2(g) MnO 2(s) H = -520 kj What is H (in kj) for the following reaction? MnO 2(s) + C (s) Mn (s) + CO 2(g)

12 Breaking bonds... What do you know about the energetics of breaking a bond? Endothermic! Bond Enthalpy - the energy it takes to break one mole of bonds in a gaseous atom why only gaseous atoms? need to eliminate the energy needed to overcome IMFs Ex. Cl 2(g) 2Cl (g) H = 242.0kJ mol -1

13 Remember... in terms of bond strength: A. single < double < triple B. bond enthalpies will increase accordingly

14 Bond Formation bond making involves bringing together atoms that are attracted by an electrostatic force and releases energy... Exothermic!

15 Energy changes in reactions -using bond ethalpies Given an equation (find the overall H): CH 4 + 2O 2 CO 2 + 2H 2 O Step 1: Write down the equation showing all the bonds + 2 O=O O=C=O + 2 How many of each bond?

16 Example Continued How many of each bond? 4 C-H bonds, 2 O=O bonds, 2 C=O bonds, 4 H-O bonds IB Data Booklet - Table 11 Bond Enthalpies broken bonds - endothermic - positive formed bonds - exothermic - negative

17 Calculations C-H +412 x 4 bonds = kj O=O +496 x 2 bonds = +992 kj C=O -743 x 2 bonds = kj H-O -463 x 4 bonds = kj H = = -698 kj mol -1

18 Another Practice C 2 H 4 + H 2 C 2 H 6 Bond Enthalpy kj mol -1 C-C +347 C=C +612 C-H +413 H-H +436 Find H for the reaction

19 Standard Enthalpy changes of reaction enthalpy of a reaction depends on the physical state of the reactants and products Standard enthalpy changes 298K and 1.00 x 10 5 Pa Standard Enthalpy Change (ΔH ) - when 1 mole of a substance is formed from its elements in their standard states (what they 298K and 100kPa [1.00 x 10 5 Pa] )

20 Standard enthalpy of formation a substance can have a ΔH f of zero if it exists in its standard state ex. Li(s), H 2 (g), S(s), Hg(l) an element will have a ΔH f of something other than zero if its not in its standard state ex. Ca(l), Br 2 (s), F(g) compounds ΔH f 0

21 Example The Enthalpy of formation (ΔH f ) of chlorobenzene is given on table 12 in your data booklet. Give the thermochemical equation which represents the ΔH f of chlorobenzene. Steps: 1. What elements created this compound? 2. Balance the equation

22 Example Continued C + H 2(g) + Cl 2(g) C 6 H 5 Cl (l) **Remember: only 1 mole of chlorobenzene can be made!!** 6C + 5/2H 2 + 1/2Cl 2 C 6 H 5 Cl ΔH o = 11 kj mol 1-

23 ΔH reaction reactants products H f (reactants) H f (products) Elements

24 Standard heat of formation (any rxn) ΔH o reaction = H f (products) - H f (reactants)

25 Example Calculate the enthalpy change for the reaction: C 3 H 8(g) + 5O 2(g) 3CO 2(g) +4H 2 O (g) given the following information: ΔH f / kj mol -1 C 3 H 8(g) -105 CO 2(g) -394 H 2 O (g) -286

26 Standard Enthalpy of Combustion the enthalpy change that occurs when one mole of a substance burns completely under standard conditions Can be used to more directly compare the output of different fuels

27 Example Enthalpy of combustion - Table 13 in Data booklet Give the thermochemical equation which represents the standard enthalpy of combustion of hexane what happens during combustion? reactants/products? C 6 H 14(l) + O 2(g) CO 2(g) + H 2 O (l)

28 Continued C 6 H 14(l) + O 2(g) CO 2(g) + H 2 O (l) C 6 H 14(l) + O 2(g) 6CO 2(g) + 7H 2 O (l) C 6 H 14(l) + 9½O 2(g) 6CO 2(g) + 7H 2 O (l)

29 Example Give an equation for the combustion of glucose (products - reactants) Use table 12 in the IB data booklet to calculate the enthalpy of combustion (the enthalpy of formation for glucose is kj mol -1 )

30 Born - Haber Cycles

31 Some Vocabulary Review: First Ionization energy - ΔH i - the energy needed to lose one electron in one mole of atoms (forming a positive ion) New: Electron affinity: (ΔH e ) the enthalpy change when one mole of atoms attracts one mole of electrons (forming a negative ion). Both on TABLE 8 of IB Data Booklet!!!

32 Some math Ionization energy of sodium Na (g) Na + (g) + e - (g) ΔH i = 496 kj mol -1 Electron affinity of chlorine Cl (g) + e - (g) Cl - (g) ΔH e = -349 kj mol -1 Add the equations to find the total energy

33 Math Continued Na (g) + Cl (g) Na + (g) + Cl - (g) ΔH = +147 kj mol -1 What do you realize about this process? Endothermic!! Even though they are forming noble gas electron configurations!!

34 Something else must happen oppositely charged gaseous ions come together to form an ionic lattice structure Na + (g) + Cl - (g) NaCl (s) ΔH = -790 kj mol -1 exothermic! because there is a strong attraction between the ions

35 Lattice enthalpy shows the enthalpy change in terms of the reverse, endothermic process - one mole of solid crystal breaking into gaseous ions NaCl (s) Na + (g) + Cl - (g) ΔH lat = +790 kj mol -1 How do we get from solid to an ionized gas? What processes might need to occur?

36 Born-Haber Cycle Step ΔH /kj mol -1 Sodium is atomized to form 1 mole of gaseous sodium (Na (s) Na (g) ) One mole of Cl atoms are formed as ½ mole of Cl-Cl bonds break energy of atomization ΔH atom = +107 ½E(Cl-Cl) = ½(+243) (Table 11) One electron is removed from the outer shell of a sodium atom Na(g) Na + (g) + e - ΔH i (Na) = +496 One electron is added to the outer shell of a Chlorine atom Cl (g) + e - Cl - (g) ΔH e = -349 Gaseous ions come together to form one mole of NaCl (s) Na + + Cl - NaCl (s) ΔH lat =?

37 figure on page ΔH i = +496 Na + (g) + 1e - + Cl (g) ΔH e = -349 Na + (g) + Cl - (g) Na (g) + Cl (g) ½E(Cl-Cl) = ½(243) Na (g) + ½Cl 2(g) ΔH atom = +107 Na (s) + ½Cl 2(g) ΔH lat =?? ΔH f = -411 NaCl (s)

38 Your job now... Split into groups of 2-3 (MAX!) work through the example on page 241 #45 Find out how a lattice enthalpy can be calculated from the ionic model Do practice problem #21 Compare the difference between theoretical and experimental lattice energies (discuss ionic character)

39

40

41 Entropy degree of disorder in a system solid < liquid < aqueous < gas energy and matter tend to disperse and the universe becomes more disordered as time moves forward, matter and energy become more disordered, and the total entropy of the universe increases second law of thermodynamics

42 How does the entropy change? Br 2(l) Br 2(g) 2Cu (s) + O 2(g) 2CuO (s) Ag + (aq) + Br - (aq) AgBr (s) H 2(g) + Cl 2(g) 2HCl (g) CH 4(g) + O 2(g) CO 2(g) + H 2 O (l) Cu 2+ (aq) + Zn (s) Cu (s) + Zn 2+ (aq)

43 Absolute Entropy (S ) can be calculated (@ Standard condition) depends on temperature, pressure a perfectly ordered absolute zero has an absolute entropy of zero ΔS reaction = S products - S reactants

44 Example... Calculate the entropy change for the hydrogenation of ethene using the entropy values given in Table 12 of your data booklet ΔS Hydrogen = 131 J K -1 mol -1 C 2 H 4(g) + H 2(g) C 2 H 6(g)

45 Spontaneity

46 Thoughts? Cu 2+ (aq) + Zn (s) Cu (s) + Zn 2+ (aq) ΔH sys = -217 kj mol -1 ΔS sys approximately equal to zero

47 Some relationships... entropy directly proportional to enthalpy ΔS surroundings -ΔH system entropy inversely proportional to absolute temperature ΔS surroundings 1/T

48 Explaining the units of entropy ΔS surroundings = -ΔH system /T (must be in Kelvin) J mol -1 K -1

49 Using ΔG sys to predict feasibility of change ΔG sys = ΔH sys - TΔS sys < 0 Temperature can be thought of as the variable that adjusts the significance of ΔS sys At low temperature: ΔG sys ΔH sys (as TΔS 0) all exothermic reactions can occur at low T At high temperature: ΔG sys make ΔH sys negligible) -TΔS sys (the T is high enough to all reactions which have a positive ΔS sys can be feasible at high temperatures even if they are endothermic

50 Gibbs Free Energy H<0 H>0 S>0 Spontaneous at all T ( G<0) Spontaneous at high T (when T S is large) S<0 Spontaneous at low T (when T S is small) Nonspontaneous at all T ( G>0)

51 Page 257 in your books Complete worked example for the boiling of water.

52 Everyone - do this now! Copy the chart on page 258 into your notes Not! for memorization - should be able to work through the data to decide spontaneity

53 Calculating ΔG Two Routes: At temp other than 298K ΔG sys = ΔH sys - TΔS sys At standard conditions ΔG rxn = ΔG f (products) - ΔG f (reactants) (same way ΔH f can be calculated)

54 Example: Calculate ΔG rxn for the following reaction: 2Al (s) + Fe 2 O 3(s) 2Fe (s) + Al 2 O 3(s) COMPOUND ΔG f kj mol -1 Fe 2 O 3(s) -742 Al 2 O 3(s) -1582

55 One more example.. Calculate the ΔG rxn for the thermal decomposition of calcium 298K from the following data Compound ΔH f kj mol -1 ΔS f J K -1 mol -1 CaCO 3(s) CaO (s) CO 2(g) Hint: calculate ΔH, ΔS then ΔG

56 Enthalpies of Solution NH 4 Cl (s) + H 2 O > NH 4 + (aq) + Cl - (aq) H 0 sol = kj mol -1 Calculated by measuring temperature change in solution. Enthalpy of solution refers to the ideal situation of infinite dilution.

57 Enthalpies of Solution...continued NaCl and NH 4 Cl dissolve readily due to strong attractions to the polar solvent water. Ions separate from the lattice are said to be hydrated. The strength of the interaction is given by hydration enthalpy.

58 Hydration Enthalpy Enthalpy changes when one mole of GASEOUS ions dissolve. Difficult to measure directly so... We use H 0 hyd = H + H + (g) > H + (aq) H 0 hyd = kj mol -1 Combine this value with hydration enthalpy of different compounds for individual ions.

59 Hydration Enthalpy Enthalpy change that occurs when one mole of gaseous ions is dissolved to form an infinitely dilute solution of one mole of aqueous ions. M n+ (g) > M n+ (aq) X m- (g) > X m- (aq) H 0 hyd(m n+ ) H 0 hyd(x m- ) Exothermic process so enthalpy changes are negative.

60 Hydration Enthalpy Values become less exothermic as ionic radius increases. As charge increases, ionic radius decreases for Period 3 metals

61 Hydration Enthalpies hydration enthalpy (kj.mol) I - Br - Cl - Rb + K + Group 1 Metals Halogens Na+ Li F - 1/R (10*-12 m)

62 Hydration & Lattice Enthalpies Solid sublimed to gaseous ions and then plunged into water. H 0 sol(nacl) = H 0 lattice(nacl) + H 0 hyd(na + ) + H 0 hyd(cl - ) H 0 sol(nacl) = kj mol -1 H 0 sol(nacl) = +7 kj mol -1 Compare this value to that found in Table 19 of your data booklet.

63 Ozone Depletion #2 Oxygen present in two forms O 2 and O 3 What can you tell me about bond order and bond strength between the two allotropes as it relates to ultraviolet radiation? Example: The bond energy in ozone is 363 kj mol -1. Calculate the wavelength of UV radiation needed to break the bond.

64 Example Continued Remember the following relationships: E photon = hv (Planck s Equation) The wavelength λ is related to the frequency: v = c/λ where c is the speed of light. Substituting for v in Plank s equation: E photon = hc/λ One mole of photons is needed to break one mole of bonds. L x E photon = 363 kj = J E photon = /6.02x10 23 Substitue and solve: 330 nm

65 Chapman Cycle Level of ozone in the atmosphere (less than 10 ppm) stays constant if the rate of formation is balanced by rate of removal (steady state).