slide1 n.
Download
Skip this Video
Loading SlideShow in 5 Seconds..
Thermodynamics PowerPoint Presentation
Download Presentation
Thermodynamics

Loading in 2 Seconds...

play fullscreen
1 / 61

Thermodynamics - PowerPoint PPT Presentation


  • 41 Views
  • Uploaded on

Thermodynamics. Temperature K, ° C Measure of average KE of motion of particles. Heat kJ, kcal (Cal) 1kcal=4.184 kJ Measure of total energy transferred from an object of high E to low E. Temperature Vs. Heat. Note: A change in T is accompanied by a transfer of energy.

loader
I am the owner, or an agent authorized to act on behalf of the owner, of the copyrighted work described.
capcha
Download Presentation

Thermodynamics


An Image/Link below is provided (as is) to download presentation

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.


- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -
    Presentation Transcript
    1. Thermodynamics

    2. Temperature K, °C Measure of average KE of motion of particles Heat kJ, kcal (Cal) 1kcal=4.184 kJ Measure of total energy transferred from an object of high E to low E Temperature Vs. Heat Note: A change in T is accompanied by a transfer of energy

    3. Specific Heat • c or cp • specific for each substance = amount of energy required to raise the temperature of 1 g of a substance by 1°C

    4. Thinking about specific heat List the following substances in terms of specific heat, from lowest to highest: water, gold, ethanol, granite

    5. Thinking about specific heat List the following substances in terms of specific heat, from lowest to highest: Answer: gold, granite, ethanol, water 0.129 0.803 2.44 4.184 J/g·oC

    6. Relationship among temperature, heat and mass Change inheat = specific heat x mass x change in T q = cp x m x Δt Units: heat absorbed or released, J cp inJ/g·°C m in g ΔT in °C or K

    7. Example Problems • How much heat energy is released to your body when a cup of hot tea containing 200. g of water is cooled from 65.0oC to body temperature, 37.0oC? (cpH2O = 4.18 J/g·oC) A: 23.4 kJ

    8. Example Problems • A 4.50-g nugget of pure gold absorbed 276 J of heat. What was the final temperature of the gold if the initial temperature was 25.0oC? (cp of gold = 0.129 J/g·oC) Hint: begin by solving for DT A: 500. oC

    9. Example Problems • A 155 g sample of an unknown substance was heated from 25.0oC to 40.0oC. In the process, the substance absorbed 5696 J of energy. What is the specific heat of the substance? A: 2.45 J/g·oC

    10. Using a Calorimeter (WS) Put dried food sample in “bomb” Burn it Use ∆T of water to calculate ∆Hcomb of the food.  The amount of energy in food is measured in Cal, rather than kJ. 1 Calorie = 1 kcal = 4.184 kJ

    11. Calorimetry Step 1 Calculate q from calorimeter data. q = cpH2O x mH2O x DtH2O Step 2 Relate q to the quantity of substance in the calorimeter.

    12. Calorimetry Problem What is the heat of combustion, DHc , of sucrose, C12H22O11, if 1.500 g of sugar raises the temperature of water (3.00 kg) in a bomb calorimeter from 18.00oC to 19.97oC? (cpH2O = 4.18 J/g·oC) A: 16.5 kJ/g sucrose, 5650 kJ/mol sucrose

    13. Calorimetry Practice Problem You burned 1.30 g of peanuts below an aluminum can which contained 200.0 mL of ice water (DH2O = 1.00 g/mL). The temperature of the water increased from 6.7oC to 28.5oC. Assume all of the heat energy produced by the peanuts went into the water in the can. (cpH2O = 4.18 J/goC) What was the energy content of peanuts, in kJ/g (i.e. DHcomb of peanuts? A: 14.0 kJ/g

    14. Calorimetry Problem, p. 881, #6(Honors) • A 75.0 g sample of a metal is placed in boiling water until its temperature is 100.0oC. • A calorimeter contains 100.0 g of water at a temperature of 24.4oC. • The metal sample is removed from the boiling water and immediately placed in water in the calorimeter. • The final temperature of the metal and water in the calorimeter is 34.9oC. • Assuming that the calorimeter provides perfect insulation, what is the specific heat of the metal? • A: 0.900 J/g·oC

    15. ΔH = change in enthalpy (i.e. change in the amount of heat) The heat change which occurs during processes can be described as ΔHx, x=process name ΔHcomb= … combustion ΔHfus= … fusion ΔHvap= … vaporization ΔHrxn= … reaction ΔHf°= … formation ΔHsol= … solution Heats of … Units: All in kJ/mol except heats of reaction and solution, which are just in kJ

    16. Thermodynamics Word Problems Using Heats of _____ Data Two of the following three factors will be given: • amount of substance of interest (usually in grams), • heat of _____ for the substance (kJ/g or kJ/mol) • total kJ

    17. Thermodynamics Word Problems Using DHxData e.g. Calculate the heat required to melt 25.7 g of solid methanol at its melting point. DHfus of methanol = 3.22 kJ/mol (from Table 16-6 on p. 502) 1. Convert mass of methanol to moles. • Multiply moles of methanol by DHfus. A: 2.58 kJ

    18. Thermodynamics Word Problems Using DHxData • How much heat is needed to vaporize 343 g of acetic acid (CH3COOH)? DHvap = 38.6 kJ/mol (A = 220. kJ)

    19. Calculating E change using ΔHxLab Data Determining ΔHx • Measure total # kJ (calorimetry) using q = cp x m x ΔT • Measuremass of substance • Divide # kJ by mass of substance  kJ/g 4. Convert to kJ/mol using massmole conversion

    20. Combustion of a Candle Lab What is the DHcomb of paraffin wax? • Measure total # kJ using qH2O = cpH2O x mH2O x ΔTH2O (i.e. q of water in can) 2. Measuremass of substance: mwax burned = mcandle+card before and after expt.

    21. Combustion of a Candle Lab What is the DHcomb of paraffin wax? • Divide # kJ by mass of substance: #1 ¸ #2  kJ/g (note units include both kJ and g) • Convert to kJ/mol using massmole conversion __ kJ (MM of paraffin in g)  ____ kJ/mol paraffin g 1 mol paraffin paraffin = C26H54

    22. Temperature Changes of Water (WS)

    23. Using Heat of Fusion Use Table 16-6 on p. 502 for DHvapo and DHfuso for different substances. Calculate the energy required to melt 8.5 g of ice at 0oC. The molar heat of fusion for ice is 6.02 kJ/mol. A: 2.8 kJ

    24. Using Heat of Vaporization Calculate the total energy change (in kJ) required to a) heat 25 g of liquid water from 25oC to 100.oC, then b) change it to steam at 100.oC. cpliquid water = 4.18 J/goC, DHvapH2O = 40.6 kJ/mol. A: a) 7.8 kJ b) 56 kJ Total = 64 kJ

    25. Thermochemical Stoichiometry • The amount of energy (in kJ) can be incorporated into mole ratios. • C6H12O6(aq) + 6 O2(g) 6 CO2(g) + 6 H2O(l) + 2870 kJ • ΔH = -2870 kJ/mol glucose • Mole Ratios Examples 1 mol C6H12O66 mol CO2 6 mol O2 6 mol O2 2870 kJ6 mol H2O 1 mol C6H12O6 2870 kJ

    26. Thermochemical StoichiometryProblems C6H12O6(aq) + 6 O2(g) 6 CO2(g) + 6 H2O(l) + 2870 kJ 1. How much energy (in kJ) will be released when 675 g of glucose is burned? A: 10,800 kJ

    27. Thermochemical StoichiometryProblems C6H12O6(aq) + 6 O2(g) 6 CO2(g) + 6 H2O(l) + 2870 kJ 2. If 398 kJ is released when a certain amount of glucose is burned, how many grams of oxygen are consumed? A: 26.6 g

    28. Thermochemical StoichiometryProblems C6H12O6(aq) + 6 O2(g) 6 CO2(g) + 6 H2O(l) + 2870 kJ • If 5782 kJ is released when a certain amount of glucose is burned, how many liters of carbon dioxide are released, assuming the reaction takes place at STP? A: 271 L

    29. Enthalpy • enthalpien= to warm (Greek) • Enthalpy (H) = total E of a substance • Enthalpy change (ΔH) = comparison between H of products and H of reactants in a reaction (ΔH=Hproducts - Hreactants)

    30. Enthalpy • ΔHrxn is + if endothermic (energy of products > energy of reactants) • ΔHrxn is - if exothermic (energy of reactants > energy of products) 2 H2 + O2 2 H2O + 483.6 kJ DH = - 483.6 kJ

    31. Hess’s Law • The total enthalpy change for a chemical or physical change is the same whether it takes place in one or several steps. • “state function”

    32. Hess’s Law • Version 1: Add equations together • Version 2: Use heats of formation (DHfo) of products and reactants

    33. Hess’s Law Version 1 Example: What is the change in enthalpy for converting graphite to diamond? Given: C(graphite) + O2(g) CO2(g)DH = -393.5 kJ/mol C C(diamond) + O2(g) CO2(g)DH = -395.4 kJ/mol C Step 1: Write rxn: C(graphite)  C(diamond) Step 2: Add up rxns to get total rxn, canceling similar terms. C(graphite) + O2(g) CO2(g)DH = -393.5 kJ/mol C CO2(g) C(diamond) + O2(g)DH = +395.4 kJ/mol C ______________________________________________ C(graphite) C(diamond)DH = +1.9 kJ/mol C Note: Focus is on the DH. Adding equations is the way you get there.

    34. Hess’s Law Version 1 Practice Calculate the heat of formation of pentane (C5H12). Given: C(s) + O2(g) CO2(g)DH = -393.51 kJ/mol H2(g) + ½ O2(g) H2O(l)DH = -285.83 kJ/mol C5H12(l) + 8 O2(g) 5 CO2(g) + 6 H2O(l) DH = -3536.1 kJ/mol Final equation: 5 C(s) + 6 H2(g) C5H12(l)DHfo = ? (A: -146.4 kJ)

    35. Hess’s Law (Version 2) The enthalpy change of a rxn is equal to the heats of formation of the products minus the heats of formation of the reactants (and taking stoichiometric relationships into account). DH = SDHfo (products) – SDHfo(reactants)

    36. Hess’s Law (Version 2) Example: Calculate ΔH for this reaction CH4(g) + 2O29g) CO2(g) + 2H2O(l) Given: DHfo CO2(g)= -393.5 kJ/mol DHfo H2O(l) = -285.8 kJ/mol DHfo CH4(g) = -74.86 kJ/mol DHfo O2(g) = 0 (Note: DHfo of free elements = 0)

    37. Hess’s Law (Version 2) Step 1: Calculate the total DHfo for the products and the reactants. DHproducts = (-393.5 kJ)(1 mol CO2) + (-285.8 kJ)(2 mol H2O) 1 mol 1 mol = -965.1 kJ DHreactants = (-74.86 kJ)(1 mol CH4) 1 mol = -74.86 kJ

    38. Hess’s Law (Version 2) Step 2: Plug these values into the following equation. DHrxn = DHproducts – DHreactants = -965.1 kJ – (-74.86 kJ) = -890.2 kJ

    39. Hess’s Law (Version 2) Practice Use standard enthalpies of formation from Appendix C, Table C-13 on p. 921 to calculate DHorxn for the following reaction: 4 NH3(g) + 7 O2(g) 4 NO2(g) + 6 H2O(l) A: -1397.9 kJ

    40. Hess’s Law (Version 2) Practice Use standard enthalpies of formation from Appendix C, Table C-13 to calculate DHorxn for the thermite reaction: 2 Al(s) + Fe2O3(s) 2 Fe(s) + Al2O3(s) A: - 851.5 kJ

    41. Entropy: Sen- (Greek, = in) trope (Greek, = a turning) Entropy: degree of disorder or chaos In nature, things tend toward disorder It is usually much easier to go from low S to high S All chemical/physical changes involve changes in entropy. This natural tendency to disorder can be overcome by enthalpy.

    42. Change in Entropy: ∆S Entropy change (∆S) = comparison in entropy of the products and reactants in a chemical equation ∆S = Sproducts – Sreactants +∆S increased entropy: S products > S reactants -∆S decreased entropy: S products < S reactants Units of S = J/mol•K Always use K in S calculations

    43. Factors that Increase Entropy • state change from solid to liquid to gas • increased temperature (more collisions, more disorder) • more particles of product than reactant (decomposition, dissolving)

    44. Calculating ∆S from So ΔS = S°Products- S°Reactants Use the information in the table below to calculate ∆S for the following reaction: 2 SO2(g) + O2(g)  2 SO3(g) [2 mol SO3(257 J/mol K)] – [2 mol SO2 (248 J/mol K) + 1 mol O2(205 J/mol K)] = 514 J/K– 701 J/K = -187 J/K Would this be favored?

    45. Spontaneity Spontaneous = happens Spontaneity depends on: ∆S and ∆H of reaction Example: Combustion of octane is spontaneous 2 C8H18(l) + 25 O2(g) 16 CO2(g) + 18 H2O(g) + kJ Exothermic Reaction (-∆H) Entropy increased (+∆S) due to • More particle energy (l)  (g) • More product particles

    46. Spontaneous Reactions Spontaneity depends on • Entropy of reaction (DS) • Enthalpy of reaction (DH) • Temperature (T in K)  Gibbs Free Energy calculation

    47. Example • Burn octane: 2 C8H18(l) + 25 O2(g)  16 CO2(g) + 18 H2O(g) + kJ • increased entropy ­due to (l)  (g) favored and more product particles • exothermic - favored

    48. Gibbs Free Energy ΔG = ΔH - (TΔS) ΔG = - kJ; yes! spontaneous ΔG = + kJ; no! nonspontaneous

    49. Gibbs Free Energy: ∆G = ∆H - T∆S Reaction is spontaneous if the sign of ∆G is negative. Is this reaction spontaneous @ 25oC? 2 SO2(g) + O2(g)  2 SO3(g) To take both the change in enthalpy AND entropy into account, calculate ∆G: ∆G = ∆H - T∆S = -198 kJ – 298 K (-0.187 kJ/K) watch the units… = -198 kJ – (-55.7 kJ) = -142 kJ So, is this reaction spontaneous at 298 K? YES.

    50. Gibbs Free Energy: ∆G = ∆H - T∆S Is this reaction spontaneous @ 2000.oK? 2 SO2(g) + O2(g)  2 SO3(g) ∆G = ∆H - T∆S = -198.2 kJ – 2000. K (-0.187 kJ/K) = -198.2 kJ – (-374 kJ) = 176 kJ So, is this reaction spontaneous at 2000. K? NO. The reaction is spontaneous at low temps (enthalpy wins) but not spontaneous at very high temps (entropy wins) (decomposition occurs….)