1 / 58

THERMOCHEMISTRY

THERMOCHEMISTRY. CHAPTER 6. 6.1- The Nature of Energy . Ch. 6 Unit Essential Question: What does the thermodynamics branch of chemistry study and how is it involved in chemical reactions ? Lesson Essential Question: What aspects must be considered when studying the area of thermodynamics?

tayten
Download Presentation

THERMOCHEMISTRY

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. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. THERMOCHEMISTRY CHAPTER 6

  2. 6.1- The Nature of Energy • Ch. 6 Unit Essential Question: • What does the thermodynamics branch of chemistry study and how is it involved in chemical reactions? • Lesson Essential Question: • What aspects must be considered when studying the area of thermodynamics? • Temperature vs. heat • Temperature: measure of average KE of particles. • Heat: E transfer between two objects at different temperatures. • Flows from higher energy (hotter object) to lower energy (cooler object).

  3. 6.1- The Nature of Energy • State function: depends on the current state of the substance, not how the current state of the substance was reached. • Ex. 1: The net distance from A to B is shown, and does not change (does not depend on the path taken)- state function. But, the total distance travelled from A to B may change depending upon the path taken- path function. A B

  4. State Functions According to this diagram, does it matter how sodium chloride is formed with respect to the net amount of heat energy involved (ΔH)? Ex. 2 No matter the path that’s taken to form a compound, the net amount of heat energy involved will always be the same for that compound- state function! http://chemwiki.ucdavis.edu/Physical_Chemistry/Thermodynamics/State_Functions

  5. Thermodynamics • Thermodynamics: study of heat E and its transformations. • Examines internal energy changes of systems- sum of PE and KE. • Use simplest system to examine internal energy: ideal gas. • No attraction between particles = no PE. • Thus internal E is all KE, which we said in the last chapter is directly related to T. • We can say that the internal E of a gas increases with T. • Recall: E = 3/2RT • Bonds in reactants and products have PE. • Stronger bonds = more PE.

  6. Thermodynamics • Recall system and surroundings. • System = object(s) being studied- for us reactions • Surroundings = everything outside of the object(s) being studied- for us reaction container • Signs always reflect what happens to the system! • Heat involved is given in terms of the system (reaction). • Positive (+q) = system absorbs heat (takes in E). • Negative (-q) = system releases heat (lets E out). • Same notation applies for work. • +w = surroundings do work on system (system ‘takes in’ work). • -w = system does work on surroundings (system ‘puts out’ work).

  7. Thermodynamics • First Law of Thermodynamics: The E of the universe is constant (law of conservation of E). • ΔEuniverse = ΔEsystem + ΔEsurroundings = 0 • KE + PE of all particles in a system = internal E of a system. • Internal E can be changed by doing work, heat flow, or both: ΔE = q + w (take signs into account!)

  8. Thermodynamics & Work • Work specification: V of system (gas!) can expand during a reaction- the system is doing work on the surroundings. • w can be replaced with –PΔV • Expansion: ΔV is positive, w is negative: gas (system) does work. • Compression: ΔV is negative, w is positive: gas (system) has work done on it. • ΔE = q – PΔV • To convert units: -PΔV will give you Latm, but q will be in J so you need both units to be in J. • 101.3J = 1Latm  use as a conversion factor!

  9. AP Practice Question The average ________ is the same for any ideal gas at a given temperature. • Free energy • Lattice energy • Kinetic energy • Activation energy

  10. AP Practice Question- Review What is the energy released when the gaseous ions combine to form an ionic solid? • Free energy • Lattice energy • Kinetic energy • Activation energy

  11. AP Practice Question When ammonium chloride dissolves in water, the temperature drops. Which of the following conclusions may be related to this? • Ammonium chloride is more soluble in hot water. • Ammonium chloride produces an ideal solution. • The heat of solution for ammonium chloride is exothermic. • Ammonium chloride has a low lattice energy. *Ideal solution = follows Raoult’s Law (assumptions are made for liquids like there were for gases)

  12. 6.1- The Nature of Energy • HW: pg. 267 #6,10, 21, 24

  13. Lesson Essential Questions • What is enthalpy and how does it relate to chemical reactions? • How is calorimetry performed and what does it study?

  14. 6.2- Enthalpy & Calorimetry • Enthalpy is represented by H: heat involved in a reaction. • ΔH = q at constant pressure; the change in enthalpy of a system is equal to the E flow as heat. • For chemical reactions: ΔH = Hproducts – Hreactants • Enthalpy of reaction shows the change in enthalpy from reactants to products. • This is the E flow as heat, what you feel as hot or cold! • Enthalpy and stoichiometry: ΔH for a reaction can be used to develop conversion factors. • The given ∆H is often the enthalpy of reaction for one mole of a substance involved in the reaction. But it could be for another amount (such as per gram)!

  15. 6.2- Enthalpy & Calorimetry • Ex: When 1mol of CH4 burns at constant P, 890kJ of E is released as heat. What is ΔH if 5.8g of CH4 is burned at constant P? • *If not told whether P is constant or not in a problem, assume P is constant.* • Determine necessary conversion factor(s): • 890kJ/molCH4 • Convert to moles and solve; remember signs!!: • 5.8gCH4 x (1molCH4/16.05gCH4) x (890kJ/molCH4) = -320kJ • Negative because you’re told E is released.

  16. Calorimetry • Lab technique to measure heat involved in a chemical or physical change. • Heat capacities: • Cp = q/ΔT (heat capacity) • No specified amount. • c = q/mΔT (specific heat capacity) • For 1 gram of a substance. • C = q/nΔT (molar heat capacity) • For 1 mole of a substance. • Two types of calorimeters: coffee cup and bomb.

  17. Calorimetry at Constant Pressure • Coffee cup calorimetry. • At constant P: ΔH = c x m x ΔT = q • This is the formula for specific heat; now you’re solving for q since ΔH = q at constant P. • May need concentration, D, etc. to find mass in grams. • Remember- AP questions like to combine various topics/knowledge/calculations!

  18. Calorimetry at Constant Pressure • Example: Solid BaSO4 forms when 1.00L of 1.00M Ba(NO3)2 is mixed with 1.00L of 1.00M Na2SO4, and the temperature increases from 25.0°C to 28.1°C. If the amount of heat absorbed by the calorimeter is negligible, the specific heat of the mixed solutions is 4.18J/g°C, and the density of the mixed solutions is 1.0g/mL, what is the enthalpy change per mole of BaSO4 formed? Answer: 26,000J/mol BaSO4 **Note stoichiometric relationships that can be made from reactions: 4Fe (s) + 3O2 (g)  2Fe2O3 (s) ΔH = -1652kJ (1) 4mol Fe/1652kJ (3) 2molFe2O3/1652kJ (2) 3mol O2/1652kJ

  19. Calorimetry at Constant Volume • Bomb calorimeter is used. • Reactants placed in steel container (constant V) and ignited. • Energy change determined by measuring T increase of water and other calorimeter parts. • ΔH = ΔT  ccalorimeter + cwatermwaterΔT • Need to account for all heat gained- by the water and the calorimeter!

  20. AP Practice Question A 1.5886g sample of glucose (C6H12O6) is ignited in a bomb calorimeter. The temperature increased by 3.682°C. The heat capacity of the calorimeter was 3.562kJ/°C and the calorimeter had 1.000kg of water inside. Find the molar heat of the reaction: C6H12O6 (s) + 6O2 (g)  6CO2 (g) + 6H2O (l) Answer: -3,234kJ/mol

  21. 6.2- Enthalpy & Calorimetry • HW: pg. 267 #: 32, 34, 36, 42(a), 52, 54

  22. Lesson Essential Question • How is Hess’ Law used to determine the change in enthalpy of a chemical reaction?

  23. 6.3- Hess’ Law • Hess’ Law: for a chemical equation that can be written as the sum of two or more steps, the enthalpy change for the overall equation equals the sum of the enthalpy changes for the individual steps. • Add together ΔH’s of each reaction involved. • This works because H is a state function! • Ex: making nitrogen dioxide. • N2 + 2O2 2NO2ΔH1 = 68kJ VS. • N2 + O2 2NO ΔH2 = 180kJ 2NO + O2 2NO2ΔH3 = -112kJ N2 + 2O2 2NO2ΔH = 180kJ – 112kJ = 68kJ Add reactions together to obtain same reaction above; add ΔH together to get net ΔH. +

  24. Using Hess’ Law • Two important characteristics about ΔH: • 1. If a reaction is reversed, ΔH’s sign is reversed. • Ex: C(s) + O2(g)  CO2(g) ΔH = -394kJ CO2(g)  C(s) + O2(g) ΔH = 394kJ • 2. Size of ΔH is directly proportional to amounts of reactants and products. If a reaction is multiplied by some integer, ΔH is also multiplied by the same integer. • Ex: [C(s) + O2(g)  CO2(g)]x2 ΔH = (-394kJ)x2 2C(s) + 2O2(g)  2CO2(g) ΔH = -788kJ • Watch for ∆H given as kJ/mol of a substance- if the balanced reaction has more than 1mol of the substance, multiply H by the appropriate coefficient! • Ex: Combustion of C4H4 has ∆H = -585.2kJ/mol CO2. C4H4(g) + 5O2(g)  4CO2(g) + 2H2O(g) ∆H = -2341kJ

  25. AP Practice Question Given the following information: C (s) + O2 (g)  CO2 (g) ΔH = -393.5kJ H2 (g) + 1/2O2 (g)  H2O ΔH = -285.8kJ C2H2 (g) + 5/2 O2  2CO2 (g) + H2O (l) ΔH = -1299.8kJ Find the enthalpy change for: 2C (s) + H2 (g)  C2H2 (g) • 454.0kJ • -227.0kJ • 0.0kJ • 227.0kJ

  26. 6.3- Hess’ Law • HW: pg. 268 #: 57, 60, 62

  27. AP Practice Question Determine ΔH for the reaction if CH3OH (l) were formed instead of CH3OH (g). The ΔH of vaporization for CH3OH is 37 kJ/mol. CO (g) + 2H2 (g)  CH3OH (g) ΔH = -91kJ • -128kJ • -54kJ • 128kJ • 54kJ

  28. Lesson Essential Question • How can standard enthalpies of formation be used to calculate enthalpy changes in reactions?

  29. 6.4- Standard Enthalpies of Formation • Standard enthalpy of formation: enthalpy change when one mole of a substance is formed from its elements when all substances are in their standard states. • Represented by ∆H°f. Degree sign indicates standard conditions, f indicates ‘formation’. • Standard states for compounds: • Gases- at a pressure of 1atm. • Solutions- concentration of 1M. • Pure substances (elements & compounds)- most stable form (how they’re typically found) at 1atm & 25°C. • Ex: oxygen- O2 (g), mercury- Hg (l), sodium chloride- NaCl (s)

  30. Standard Enthalpies of Formation • Always given per mole of substance in its standard state; formed from elements in standard states: • Ex: C(s) + 2H2(g) + ½O2(g)  CH3OH(l) • The enthalpy of formation of methanol is -239kJ/mol. • Enthalpy of formation for an element in its standard state is set to be zero. • The element is found this way; no energy is involved to put it into this form! • Besides Hess’ Law, Enthalpy changes for reactions can also be calculated by subtracting the enthalpies of formation of reactants from the enthalpies of formation of products: ∆H°rxn = ∑n∆H°products – ∑n∆H°reactants • Note coefficients/moles (n) are taken into account! ∆H° = -239kJ/mol f f f

  31. AP Practice Question The decomposition of CaCO3 is shown in the equation below. Using the data below, which of the following values is closest to the ∆Hrxn of the decomposition of CaCO3? CaCO3 (s)  CaO (s) + CO2 (g) ∆H°f (kJ/mol): -1,207.1 -635.5 -393.5 • -2,240 kJ/mol • -180 kJ/mol • 180 kJ/mol • 1,207 kJ/mol • 2,240 kJ/mol

  32. Bond Energies & Enthalpy • Heats of reaction can also be calculated using bond energies. • Similar formula is used as seen with standard enthalpy of formations: ∆H°rxn = ∑nbonds broken– ∑nbonds formed • Bonds broken are the reactants, and bonds formed are the products. • Take into account ALL bonds broken and formed- in other words multiply each bond type by how many there are. Drawing out molecules can help! Watch out for single, double, and triple bonds! • Ex: CH4 (g) + 2O2 (g)  CO2 (g) + 2H2O (g) ∆H°rxn = [4(C-H) + 2(O-O)] – [2(C=O) + 4(H-O)]

  33. AP Practice Question Use the values of average bond energies to find the enthalpy change for the following reaction: 2H2 (g) + O2 (g)  2H2O (g) Average bond energies (in kJ/mol): H-H: 436, O=O: 499, and H-O: 464. • 0kJ • 485kJ • -485kJ • 464kJ Note: the actual enthalpy of reaction for this reaction is -572kJ. Why such a big difference?

  34. Bond Energies & Enthalpy Cont. • Note that the same bond in different molecules impacts bond energy too. Thus ∆H°rxn is approximate here. • Ex: O-H bond in H2O does not have the exact same E as the O-H bond in CH3OH. • Note also that using bond energies to calculate enthalpies of reaction are only sufficient if all species are in the gaseous state! • Why might this be? • Only in the gaseous state can we assume interactions between particles are negligible (ideal gases), and so only bond energies need to be considered for energy changes. • In other states the interactions between particles are not negligible and cannot be ignored since these too contribute to energy changes.

  35. 6.4- Standard Enthalpies of Formation • HW: pg. 269 #: 65, 67

  36. Three Laws of Thermodynamics • Three laws of thermodynamics: • 1. Total E of the universe is constant (law of conservation!): ΔEuniverise = ΔEsystem + ΔEsurroundings = 0 • 2. All processes that occur spontaneously move in the direction of an increase in entropy of the universe (system + surroundings): ΔSuniverse = ΔSsystem + ΔSsurroundings > 0 • 3. The entropy of a pure, perfect crystal at 0K is zero. • Note: the third law has more to it, but this is all you need to take away. There is also a zeroth law that we will not cover. • We’ve already investigated #1, let’s look at #2!

  37. Entropy Changes • List all of the examples you can think of that would result in an increase in entropy. • Entropy increases when the number of molecules increases from reactants to products. • Entropy increases when T increases. • Entropy increases when a gas is formed from a liquid or a solid. • Entropy increases when a liquid is formed from a solid.

  38. AP Practice Question Choose the reaction expected to have the greatest increase in entropy. • H2O (g)  H2O (l) • 2KClO3 (s)  2KCl (s) + 3O2 (g) • Ca (s) + H2 (g)  CaH2 (s) • N2 (g) + 3H2 (g)  2NH3 (g)

  39. AP Practice Question Which of the following combinations is true when sodium chloride melts? • ΔH > 0 and ΔS > 0 • ΔH = 0 and ΔS > 0 • ΔH > 0 and ΔS < 0 • ΔH < 0 and ΔS < 0

  40. AP Practice Question Which of the following reactions would be accompanied by the greatest decrease in entropy? • N2 (g) + 3H2 (g)  2NH3 (g) • C (s) + O2 (g)  CO2 (g) • 2H2 (g) + O2 (g)  2H2O (g) • 2Na (s) + Cl2 (g)  2NaCl (s) Note: according to the practice book the reaction that produces the most gas will have the greatest increase in S and the one losing the most gas will have the greatest decrease in S.

  41. Standard Molar Entropies • Standard molar entropies: entropy associated with 1mol of a substance in its standard state. • Represented by S°. The change in entropy of a reaction can be calculated like ΔH° was: • ∆S°= ∑nS° products – ∑nS° reactants • Standard states are the same as previously discussed. • BUT entropies of elements in standard states are not zero! • Everything will have entropy values!

  42. AP Practice Question Calculate ΔS° for the following given the following S° values. a. H2 (g) + ½ O2 (g)  H2O (g) *Do you expect a positive or negative ΔS° value? Why? S° (J/molK): H2: 131.0 O2: 205.0 H2O: 188.7 ΔS° = -44.8J/molK b. CaCO3 (s) + H2SO4 (l)  CaSO4 (s) + H2O (g) + CO2 (g) *Do you expect a positive or negative ΔS° value? Why? S° (J/molK): CaCO3: 92.9 H2SO4: 157 CaSO4: 107 H2O: 188.7 CO2 213.6 ΔS° = 259J/molK

  43. Gibbs Free Energy • Are ΔH and ΔS the only thermodynamic quantities that determine if a reaction will occur? What if one is favorable and the other is not? • Recall favorable signs for each: -ΔH and +ΔS. • What do you recall about ΔG? • It combines ΔH and ΔS, and determines whether or not a reaction will occur or not. • Favorable sign: -ΔG. • In terms of spontaneity: -ΔG = spontaneous; +ΔG = nonspontaneous (energy needs to be added for the reaction to occur); ΔG = 0 the reaction is at equilibrium.

  44. AP Practice Question A sample of Ga metal is sealed inside a well-insulated, rigid container. The T inside the container is at the melting point of the Ga metal. What can be said about the E and the S of the system after equilibrium has been established? Assume the insulation prevents any E change with the surroundings. • ΔH > 0 and ΔS > 0 • ΔH = 0 and ΔS > 0 • ΔH > 0 and ΔS < 0 • ΔH < 0 and ΔS < 0

  45. Recall ΔG Formula • ΔG = ΔH -TΔS • ΔG°rxn = ΔH°rxn -TΔS°rxn • Remember- you can use these formulas to calculate values of ΔG, but you can also use it with signs for ΔH and ΔS to see if a reaction will occur and the effect T has on the reaction. • Ex: A salt is dissolved in a beaker of water and the temperature drops by 4.3°C. What are the signs of ΔH, ΔS, and ΔG for this reaction? • +ΔH, +ΔS, -ΔG

  46. AP Practice Question A certain reaction is nonspontaneous under standard conditions, but becomes spontaneous at higher temperatures. What conclusions may be drawn under standard conditions? • ΔH < 0, ΔS > 0, ΔG > 0 • ΔH > 0, ΔS < 0, ΔG > 0 • ΔH > 0, ΔS > 0, ΔG > 0 • ΔH < 0, ΔS < 0, ΔG > 0

  47. Gibbs Free Energies of Formation • Just like with ΔH and S, ΔG also has energies of formation: ΔG°f • ΔG°f for elements in their standard states are zero, just like they were for ΔH°f. • Same equation is used for ΔG°rxn as was used for ΔH°rxn: ∆G°rxn = ∑n∆G°f products – ∑n∆G°f reactants • This is another way to calculate ΔG, in addition to using the formula previously discussed.

  48. AP Practice Question Calculate ΔG° for the following reaction. 2NH4Cl (s) + CaO (s)  CaCl2 (s) + H2O (l) + 2NH3 (g) ΔG°f (kJ/mol) : -203.9 -604.2 -750.2 -237.2 -16.6 Answer: -8.6kJ/mol

  49. Ch. 17 Homework • Pg. 783 # 8, 20, 24, 25, 27(a&b), 28(a), 33, 36, 37(a), 43, 45(a), 52 • Note: problems in red are technically work for this information, but due to the large quantity these will be your class work for Friday when I’m out.

  50. Thermodynamics & Equilibrium • What if substances in a reaction are not under standard conditions? • Pressures and concentrations are no longer equal to one. • The new conditions must be taken into account: ΔG = ΔG° + RTlnQ • Q is the reaction quotient; used to indicate where a reaction is in terms of reaching equilibrium (we will do more with this later): Q = [products]x/[reactants]y • Note: x and y superscripts come from coefficients! • Ex: 2NO (g) + O2 (g)  2NO2 (g) • Q = [NO2]2/([NO]2[O2])

More Related