* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
Download Ppt19(PS8)_Thermo_Hess
Survey
Document related concepts
Equipartition theorem wikipedia , lookup
Second law of thermodynamics wikipedia , lookup
Heat equation wikipedia , lookup
Heat transfer wikipedia , lookup
Thermal conduction wikipedia , lookup
First law of thermodynamics wikipedia , lookup
Internal energy wikipedia , lookup
Conservation of energy wikipedia , lookup
Adiabatic process wikipedia , lookup
Thermodynamic system wikipedia , lookup
Heat transfer physics wikipedia , lookup
Gibbs free energy wikipedia , lookup
History of thermodynamics wikipedia , lookup
Transcript
Ppt19, Thermochemistry I. Basic Ideas and Definitions KE(review), heat, temperature, potential energy, thermal energy, enthalpy, etc. II. Calorimetry—Obtaining energy changes by measuring T changes. III. Thermochemical Equations: Stoichiometry with Energy! IV. Hess’s Law (and related Ideas) Using energy changes of known reactions to calculate energy changes of related ones V. Standard Enthalpies of Formation Using tabulated values to calculate energy changes 1 Quick “Quiz” (PS8a, Q1) True or false (correct if false): (i) When a chemical bond is broken, energy is released. Answer: FALSE. It “takes” (absorbs) energy to break any chemical bond! No exceptions!! Breaking a bond is like “pulling apart two magnets” or “lifting a book”!! • PE of system increases; energy (either in the form of heat or work) comes from outside of system (surroundings) Ppt19 2 Quick “Quiz” (PS8a, Q1) True or false (correct if false): (ii) When a chemical reaction takes place, energy is released. Answer: FALSE. Some chemical reactions absorb energy (called “endothermic”), and some reactions release energy (“exothermic”) • If only bond breaking occurs: endo Cl2 2 Cl • If only bond making occurs: exo Cl + Cl Cl-Cl • If some bonds are broken and some (other ones) are made, it could be endo or exo (it depends). Ppt19 3 Thermodynamics is the study of energy changes • Thermo = “heat” (a type of energy) • Dynamic = “motion” “changes” • Two basic KINDS of energy: Kinetic (KE): energy of motion (of a particle) Recall: For a sample of particles: • KEavg(per particle) TKelvin KEavg (x # particles) is called “thermal energy” • T is NOT an energy, but it is proportional to one kind of energy (thermal energy or avg. KE of particles) Ppt19 4 NOTE: T is not the same as “heat”! • Heat is a type of energy; T is not an “energy” • Things that are “hot” have a high T They have a high average KE per particle (relative concept) They do not “have” a lot of “heat” in them • Heat is energy that transfers from a hotter sample to a colder one • Confusing because we “sense” heat flow but brain interprets it as “temperature”: An object feels hot if heat transfers into our skin! An object feels cold if heat energy transfers out! Ppt19 5 Follow up: Difference between T and heat • You go to your car on a hot summer day after the car has been sitting out for several hours. Which is hotter, the metal belt buckle or the cloth seat? • They are the same temperature!! • The belt buckle feels hotter to you because it conducts heat well, so the amount of heat that transfers into your skin each ms is much greater than the amount of heat that transfers in from the cloth! Ppt19 6 Potential Energy (PE): The Second Type of Energy • PE is energy of position Results from forces (e.g., book in gravitational field) • Chemical potential energy results from forces between atoms or molecules It takes energy to pull bonded atoms apart It takes energy to pull molecules in a liquid apart (to turn into a gas) It takes energy to pull an electron away from a nucleus • When physical or chemical changes take place, positions of atoms or molecules change relative to one another PE changes! 7 Dx means “change in x” • Dx = xf – xi “final minus initial” • If T goes from 35 to 45 ºC, then: DT = 45 – 35 = +10. ºC • Did T increase or decrease? It increased • A positive delta means an increase in the variable • If T goes from 45 to 35 ºC, then: DT = 35 – 45 = -10. ºC • Did T increase or decrease? It decreased • A negative delta means a decrease in the variable Ppt19 8 Go to PS8a first Ppt19 9 1st Law: Energy is neither created nor destroyed. Euniverse is a constant • Energy can change forms E.g., from KE to PE or vice versa • Energy can transfer from one “place” to another Define a SYSTEM and a SURROUNDINGS • Universe = system + surroundings (See Board) DEuniv= 0 DEsys + DEsurr = 0 DEsys = - DEsurr (minus “opposite of”, not “negative!) • the amount of E that leaves the system equals the amount of E that enters the surroundings [and vice versa] Ppt19 10 What is Enthalpy? (Better question: what is the change in enthalpy?) • Internal Energy (Esys) = sum of all KE & PE of particles • Enthalpy (H) is a property of a system whose change is similar to the change in E for typical chemical processes The formal definition of H is not important. Its change is. • At constant P, the difference between DE and DH is the amount of work (w) that is done on (or by) the system: DH = DE – w (P const.) For typical chemical processes, w << DE, and DH DE Since DE = q + w, q = DE – w …. = DH! (IF P is constant) More about “work” later Ppt19 11 The change in Hsys (DHsys) equals the amount of heat (flow) if a process occurs at constant P • When a process occurs in the system at constant pressure, the change in H equals the amount of heat transferred: DHsys = qP (subscript “p” means “at const. P”) If DHsys > 0, heat flows INTO the system (to make H increase) “ENDOTHERMIC” If DHsys < 0, heat flows OUT of the system (to make H decrease) “EXOTHERMIC” PS8a, Q5! Also Q6 on PS8b • Note: if P is NOT constant during some process, the amount of heat (flow) is NOT equal to the change in H (DHsys)! “H” is “enthalpy”, NOT heat! The change in H just happens to equal q when a process is carried out at constant P. Ppt19 12 1st Law Applied, and convention for q • qsys = -qsurr Key equation (&concept) • qsys < 0 means heat flowed OUT of the system (and into the surroundings) & qsurr > 0 • qsys > 0 means heat flowed INTO the system (and out of the surroundings) & qsurr < 0 • E.g.: If 10 J flows from sys to surr: qsys = -10 J and qsurr = +10 J • E.g.: If 20 J flows into sys from surr: qsys = +20 J and qsurr = -20 J Ppt19 13 DHsys often represents a conversion of PE into KE or KE into PE (PE in sys; KE into or from surroundings) • For a chemical or physical process at constant P (and T), DHsys ↔ DPEsys DHsys = qsys AND qsys = - qsurr DHsys = - qsurr Key equation • Thus, if DHsys < 0 (exothermic), chemical PE in the system ends up getting converted into KE in the surroundings, and the energy transfer occurs as heat (warming up the surroundings)! Ppt19 14 Figure 6.2 (Zumdahl): Exothermic Process DHsys = - qsurr Ppt19 15 Endothermic Processes Generally convert KE of surroundings into PE in system DHsys = - qsurr • If endothermic (DH > 0), the rearrangement in system requires energy to occur, and that energy flows in from the surroundings (qsurr < 0) [imagine the REVERSE of the process on prior slide] • The “-” sign means “opposite of”, not “negative”!!! Concepts in Q’s 1-4, & 6 on PS8b have now been addressed. Try them! Ppt19 16 DHsys is not determined by the surroundings—it is “assessed” by it! • NOTE: The fact that the value of DH equals - qsurr should not be interpreted to mean that the value is determined by the surroundings—it is not!! • The value of DHsys is determined by the rearrangement (changes in position of atoms / molecules / ions) in the system. i.e., it is determined by the PROCESS in the system The surroundings is just a “reporter” of sorts Ppt19 17 II. Calorimetry • Used to obtain changes in enthalpy (DHsys’s) by measuring (changes in) temperature of the “surroundings” (DTsurr’s). • Use: DHsys = - qsurr A property of the surroundings; can be determined if Csurr is known via: qsurr = Cs, surr x msurr x DTsurr The “surroundings” is usually “reduced” to a calorimeter, a liquid, a solid, etc. (Assume no heat is lost to the “rest” of the surroundings) Ppt19 18 Reminder (PS 8a, Q3) “It takes energy to raise a substance’s T” How much energy? • Cs is the specific heat (capacity) of a substance amount of heat energy needed to raise 1 g of a substance by 1 C A large(r) Cs means “hard(er) to change its T” (Other abbreviations: s, S.H., c) • If the only thing that happens to a substance (A) is that it changes T, then: qA = Cs, A x mA x DTA Ppt19 Now try Q5 on PS8b! 19 Ppt19 20 Ppt19 21 Helpful Question—Is a chemical or physical process taking place, or “just” heat flow? • If there is NO process in the system (or surroundings): Heat flow is a result of different initial T’s in sys & surr T of both sys and surr changes because of heat flow q is related to DT by: • q = C x m x DT in both the sys and surr Thus, qsys = -qsurr reduces to: Csys x msys x DTsys = -(Csurr x msurr x DTsurr) (no chemical or physical change in system) Ppt19 22 Example—Calorimetry (case with heat transfer only—no phys or chm change) #1 on Handout Sheet: If a 40.1 g piece of iron at 652 °C is dropped into a sample of 328 g of water at 32.4 °C, what will be the final temperature after thermal equilibrium is established? Assume that no heat is lost during the process. CFe = 0.45 J/(gC) Q8, Q9, and Q10 on PS8b use similar ideas/skills. Try them now! Ppt19 23 Helpful Question—continued • If there IS a process in the system (but not in the surroundings) and Tsys is kept constant (common): q flow is ultimately caused by the process (see next slide) Because PE change in the system (DHsys) converted to KE qsurr is related to DT Thus, qsys = -qsurr reduces to: DHsys = -(Csurr x msurr x DTsurr) (IS a chemical or physical change in system) NOTE: qsys is NOT equal to Csys x msys x DTsys here!! The process dictates DHsys—surroundings responds to energy change in system Ppt19 24 III. Short but important interlude— Meaning of Thermochemical Equations • Before we work with the calorimetry eqn on the prior slide, recall ideas from PS8a (Q2 & next slide): 1) The amount of DH associated with a process depends on the amount of the process that occurs 2) The DH for a chemical equation is not the same as the DHsys associated with an actual chemical reaction. • Just like the coefficient in a chemical equation is not the same as the amount of moles of a substance that actually “reacts” or “forms” during an actual chemical reaction! 3) “Stoichiometry with energy” idea Ppt19 25 Follow up from PS8a Q2 (diff rxn eqn) CaO(s) + 3 C(s) CaC2(s) + CO(g); 5 mol CO x 1) If 5 mol CO is formed: DH = 465 kJ 3 mol C = 15 mol C 1 mol CO How many moles of C react? 465 kJ = 2325 kJ What is the DH of the rxn? 5 mol CO x 1 mol CO 2) If 8 mol C is to react: 8 mol C x # of moles of CaO needed? Amt of energy absorbed? Ppt19 1 mol CaO = 8 / 3 mol CaO 3 mol C 8 mol C x 465 kJ = 1240 kJ 3 mol C 26 Stoichiometry with energy! (Example) #2 on Handout Sheet: How much heat (in kJ) is evolved or absorbed in the reaction of 233.0 g of carbon with enough CaO to produce calcium carbide? CaO(s) + 3 C(s) CaC2(s) + CO(g); DH = 464.8 kJ (b) Is the process exothermic or endothermic? Reminder: If there’s a “process”, q flow is ultimately caused by that process, with the amount being dependent on how much process occurs) Ppt19 27 Another Example # 3 on Handout Sheet: 85.8 kJ of energy is evolved (i.e., released) at constant pressure when 3.56 g of P4 is burned according to: P4(s) + 5 O2(g) → P4O10(s) What is the ΔH for the (thermo)chemical equation? Q7 on PS8b uses “Stoichiometry with energy” ideas/skills. Try it now! Ppt19 28 Return to Calorimetry • Recall: qsys = - qsurr • If there IS a process in the system (but not in the surroundings), This reduces to DHsys = -(Csurr x msurr x DTsurr) (chemical or physical change in system) DHsys is “caused” by the process in the system (“stoichiometry with energy”), but we can determine its value experimentally in a particular situation by measuring the T change of the surroundings. Ppt19 29 Example—Calorimetry (case with a physical or chemical change) #4 on Handout Sheet: Instant cold packs contain solid NH4NO3 and a pouch of water. When the pack is squeezed, the pouch breaks and the solid dissolves, lowering the temperature because of the endothermic process: NH4NO3(s) NH4NO3(aq); DH = +25.7 kJ What is the final temperature in a squeezed cold pack that contains 50.0 g of NH4NO3 dissolved in 125 mL of water? Assume the specific heat capacity of the dissolved NH4NO3 is negligible compared to water, an initial temperature of 25.0 C, and no heat transfer between the cold pack and the environment. dwater ~ 1.0 g/mL Q11 and Q12 on PS8b use similar ideas/skills. Try them now! Ppt19 30 Exp 14 Part B • Dissolve some solid in some water • In an insulated cup • System is the solid (plus the small amount of water molecules that interact with the dissolved FUs of the solid) – Process that occurs in sys is “dissolution” • Assume (excess) H2O is the surroundings • DHsys = - qsurr becomes: DHdiss = - qwater • qwater = Cwater x mwater x DTwater (assume Twater=Tsol’n) Ppt19 31 Figure 6.9 Ppt19 32 Reminder (DE vs DH) (Slide 11, recopied here) • Internal Energy (Esys) = sum of all KE & PE of particles • Enthalpy (H) is a property of a system whose change is similar to the change in E for typical chemical processes The formal definition of H is not important. Its change is. • At constant P, the difference between DE and DH is the amount of work (w) done on (or by) the system: DH = DE – w (P const.) For typical chemical processes, w << DE, and DH DE Since DE = q + w, q = DE – w …. = DH! (IF P is constant) More about “work” later Ppt19 33 Return to Internal Energy (E)—Heat and Work Matter here. • There are two ways to increase the (internal) energy of a system: – Have heat (q) flow into it (qsys > 0) – Have the surroundings do work (w) on it (w > 0; chemists’ convention) In equation form: DEsys = q + w (chemists) • Of course, if heat flows out of the system, or if the system does work, Esys decreases Ppt19 34 Return to Internal Energy (E)—Heat and Work Matter here. • Recall that qp = DH, so at constant P DEsys = q + w becomes DEsys = DHsys + w and thus DHsys = DEsys - w (as noted earlier) If work is small, DH is approximately equal to DE Ppt19 35 Figure 6.7 If the system expands against an external pressure (i.e., piston moves upward), DVsys is positive and “w” is negative (system does work on surroundings). w = -PDV Ppt19 36 Example #5 on Handout Sheet: Assume that a particular reaction produces 244 kJ of heat and that 35 kJ of PV (expansion/contraction) work is done on the system. What are the values of DE and DH for the system? For the surroundings? Q13 & Q14 on PS8b use similar ideas/skills. Try them now! Ppt19 37 IV. Hess’s Law • Value of H (or E) for a system depends ONLY on the state of the system (i.e., the P, T, moles of substances, states of substances, etc.) • It doesn’t matter how you got to that state • Called a “state function” • The change in H in going from State 1 to State 2 does not depend on how you get there (i.e., “path”). “Hess’s Law”: • DHoverall= DH1 + DH2 + DH3 + etc. (1,2,3 are processes that “add up” to the overall) Ppt19 38 Example 1 3 H2 + N2 2 NH3 ; DH = ??? What is the DH for the above equation if we know the following? 2 H2 + N2 N2H4 ; DH = 95.4 kJ H2 + N2H4 2 NH3 ; DH = -187.6 kJ Answer: The sum of these two, because the sum of these two equals the overall process! Ppt19 39 Ppt19 40 Generalized “Procedure” for Creating a set of equations that sum to the equation of interest (“target”) • See handout sheet examples and boardwork • Apply the following ideas (PS8a, Q2!): 1) If you reverse an equation, the sign of DH opposite becomes the __________. 2) If you multiply an equation through by a number x, the DH becomes ____ x times the original value. Q’s 15-17 on PS8b use similar ideas/skills. Try them now! Ppt19 41 Ppt19 42 “Special” Example Find DH for CH4(g) + 2 O2(g) → CO2(g) + 2 H2O(l) given: C(s) + 2 H2(g) → CH4(g); O2(g) → O2(g); DH1 = -74.8 kJ DH2 = ? 0 kJ C(s) + O2(g) → CO2(g); DH3 = -393.5 kJ H2(g) + ½ O2(g) → H2O(l); DH4 = -285.8 kJ Ppt19 b/c nothing changed! 43 “Special” Example Find DH for CH4(g) + 2 O2(g) → CO2(g) + 2 H2O(l) CH4(g) → C(s) + 2 H2(g) ; DH1’ = - DH1 2 O2(g) → 2 O2(g); DH2’ = -2DH2 ------------------------------------------------------------------------------------------------------------------ DH3’ = C(s) + O2(g) → CO2(g); 2 H2(g) + O2(g) → 2 H2O(l); DH3 DH4’ = 2DH4 Elements! Ppt19 44 “Special” Example Find DH for CH4(g) + 2 O2(g) → CO2(g) + 2 H2O(l) given: C(s) + 2 H2(g) → CH4(g); O2(g) → O2(g); DH1 = DH2 = C(s) + O2(g) → CO2(g); DH3 = H2(g) + ½ O2(g) → H2O(l); DH4 = Ppt19 45 Q: Is there an “easy” way to calculate the DHeqn without doing calorimetry? • Yes and no • Calorimetry does have to be done, but it can be done ahead of time and not on the reaction of interest. Obtain an effective value of “H” for one mole of every substance (via calorimetry) Effective “H” is actually called the “standard enthalpy of formation” of a substance X: DH°f(X) • Use these “H” values of reactants and products (substances) to calculate DHrxn for any chemical equation! Ppt19 46 How to calculate DH°eqn (from tabulated data) For an equation: a A + b B c C + d D; DH°eqn DH°eqn = “H” of all products” - “H” of all reactants “Hfinal” “Hinitial” = [c DH°f(C) + d DH°f(D)] - [a DH°f(A) + b DH°f(B)] mol C x kJ mol C From Tro: Ppt19 47 Ppt19 48 CH3Cl(g) Find DH for : CH4(g) + Cl2(g) → CH3Cl(g) + HCl(g) Find DH for : CH4(g) + 2 O2(g) → CO2(g) + 2 H2O(l) -86.3 49 How to calculate DH°eqn EXAMPLE Find DH for : CH4(g) + 2 O2(g) → CO2(g) + 2 H2O(l) Ppt19 50 What’s really going on with the prior equation/procedure • Imagine making ELEMENTS from reactants and then turning ELEMENTS into products. (Generalized “path” for any reaction!!) • Determine DH for making a substance from its ELEMENTS (called DHformation) • Tabulate these “DHf’s” for ALL SUBSTANCES • Use them to calculate DHrxn for any chemical equation! Ppt19 51 NOTES • The DH°f of a substance is the: – Enthalpy change associated with forming one mole of a substance from its elements – As such, the value for any element is zero* • The DH°f for a substance depends on the physical state of the substance in question (because it takes or releases energy to change a substance’s state) * In its standard state Ppt19 52 Figure 6.8 Pathway for the Combustion of Methane Ppt19 53 Figure 6.9 Schematic Diagram of Energy Changes Ppt19 54 Figure 6.11 (Tro): Ppt19 55 Figure 6.10 A Pathway for the Combustion of Ammonia I showed on the board how to use data from the Table to find DH for this process. Can generalize: DHoverall = Sum(n x DHf [P’s]) - Sum(n x DHf [R’s]) 56