AP Chemistry Unit 5 Thermodynamics Study Guide PDF

AP CHEMISTRY UNIT 5 TEST: THERMODYNAMICS TEST INFO & REVIEW ASSIGNMENT __________ Format: 4-6 Free Response Questions 30-35 MC questions Material Covered: Thermo (80%), Other: (20%) Stoichiometry, Bonding, Atomic, and Reactions Study: practice FRQ’s, no...

AP CHEMISTRY UNIT 5 TEST: THERMODYNAMICS TEST INFO & REVIEW ASSIGNMENT __________ Format: 4-6 Free Response Questions 30-35 MC questions Material Covered: Thermo (80%), Other: (20%) Stoichiometry, Bonding, Atomic, and Reactions Study: practice FRQ’s, notes, homeworks, quizzes, reference sheets, and your Chemical Reactions/Atomic/Bonding Review Assignments. Thermo - Day 1 Notes 1) Heat vs. Temperature Heat Temperature Transfer of thermal energy from one system to A measure of average KE of particles within a another substance Measured in Joules (J) or Calories (cal) Ex. Farenheit, Celsius, Kelvin Note: When average KE doubles, Kelvin temperature increases K is proportional to avg KE 2) Boltzmann distribution The probability that a system will be in a certain state Displays the distribution of KE for a set of molecules Distribution @ each temp ○ Some particles collide and are deflected off @ higher speed while others are nearly stopped Molecules reach certain Ea before reacting ○ Molecules need to collide w/ correct orientation and energy ≥ Ea ○ Ea = min energy needed to from the transition complex Transition complex: where bonds start to bond/break Ea = Activation energy 3) Activation energy Not enough Ea = no reaction ○ Ex. ○ Matches sit around w/o lighting on fire ○ Bunsen burner needs spark to react bc need Ea Note: Sunlight can break down substances that are stored in brown bottles 4) Heat transfer between 2 sys. 2 sys @ diff. temps and that are in contact with one another will exchange energy (heat) via. conduction until both are at thermal equilibrium. 5) Conservation of Energy Energy can’t be created or destroyed Energy of sys changes during ○ Chem rxn ○ Phase change ○ Change in temp. 6) Endo vs Exo Endothermic system Exothermic system Heat is absorbed [by system] Heat is released [by system] Bonds in product contains more PE than Bonds in product contains less PE than reactants reactants ○ Bc energy absorbed fr ○ Released energy means products surroundings is stored as chemical are in a lower energy state (more PE stable) Product contains less KE than reactants Product contains less KE than reactants ○ Absorbed energy breaks bonds or ○ Increase motion of particles, changes states raising temp of surroundings and product 6) Less PE = bond energy of products is greater than reactants Note: Stronger bonds in the products = inc bond energy = dec potential energy of the system. 7) Enthalpy: Total internal energy of a system (sum of potential and kinetic) Extensive Property: Dep on amt ○ Ex.Mass, Volume, … Intensive Property: Doesn’t dep on amt ○ Ex. Luster, Temp. ΔH = ΔHproducts – ΔHreactants ΔH = - [exo] | ΔH = + [endo] 8) When valence e- get closer to the nucleus, the PE decreases. According to Coulomb’s law, the closer distance results in greater bond energy because of the coulombic attraction between the bonding nuclei and bonding e-. 9) Determine ΔH of a rxn/process Bond Enthalpies Calorimetry Hess’ Law Standard Enthalpies formation 9) Avg Bond Enthalpy PE vs Internuclear Distance Graph Average amt of energy required to break a bond (look on table) Energy always released during formation of bond ○ PE decreases as atoms move closer ○ PE increases as atoms repel further or if too close 10)Find ΔH using bond energies. Thermo- Day 2 Notes 11) Specific Heat Specific heat: The amount of heat energy required to raise the temperature of one gram of a substance by 1°C (or 1 K). Measures a substance’s ability to absorb/retain heat 12) Specific Heat of Water Water has a higher specific heat than most substances because its IMFs are require immense energy to break. This enables temperature regulation of the ocean that is necessary for living stability among marine life and moderate weather/climate conditions, especially along the coast. 13) Calorimetry Eq. q = mCΔT q: Total heat transfer (J) | m: mass (g) | C: specifc heat (J/Kg*K) | ΔT: change in temp 14) Finding Molar Enthalpy −𝑞 ΔHrxn = 𝑚𝑜𝑙 15) Coffee Cup Calorimetry Lab System: substance in water/acid Surroundings: water/acid, cup, room, universe 16) Sys vs Surroundings Sys Actual chem rxn/process happening Surroundings  Entire universe outside rxn 17) How would you find the molar enthalpy of neutralization when 1M of HCl and 1M of NaOH are mixed? Explain step by step how to calculate each of the following. (i) The number of moles of water formed during the experiment Moles of HCl = 1M · L Moles NaOH = 1M · L Assume 1:1 ration ○ Moles of H2O = moles of HCl or NaOH Not 1:1 ratio ○ Use stoichiometric ratio 𝑚𝑜𝑙𝑒 𝑤𝑎𝑡𝑒𝑟 𝑓𝑟 𝑏𝑎𝑙𝑎𝑛𝑐𝑒𝑑 𝑒𝑞 ○ Moles of H2O = mol LR · ( 𝑚𝑜𝑙𝑒 𝐿𝑅 𝑓𝑟 𝑏𝑎𝑙𝑎𝑛𝑐𝑒𝑑 𝑒𝑞 ) (ii) The value of the molar enthalpy of neutralization, ∆Hneut, for the reaction between HCl(aq) and NaOH(aq) 𝑞 ∆Hneut = 𝑚𝑜𝑙𝑒𝑠 𝑜𝑓 𝑤𝑎𝑡𝑒𝑟 𝑓𝑜𝑟𝑚𝑒𝑑 (d) The student repeats the experiment with the same equal volumes as before, but this time uses 2.0 M HCl and 2.0 M NaOH. (i) Indicate whether the value of q increases, decreases, or stays the same when compared to the first experiment. Justify your prediction. Moles of reactant doubled, so amt of heat released (q) increases. Heat is directly proportional to amt of reactants. (ii) Indicate whether the value of the molar enthalpy of neutralization, ∆Hneut, increases, decreases, or stays the same when compared to the first experiment. Justify your prediction. ∆Hneut remains the same because it is already the ER. The dispersion of heat is neutralized. (e) Suppose that a significant amount of heat were lost to the air during the experiment. What effect would this have on the calculated value of the molar enthalpy of neutralization, ∆Hneut? Justify your answer. q is lower than the actual heat released by the neutralization rxn because not all heat fromthe rxn is captured by the calorimeter. The calculated value of ∆Hneut will be lower than it should be. 18) What is Hess’s Law Hess's Law states that the total Δ𝐻rxn = sum of the enthalpy changes of the steps into which the reaction can be divided. Note: Enthalpy is a state function 18) State Function Property of the sys that depends on current state rather than the approach to that state.  Note: State function doesn’t change even if process does Depends on [chage between] final and initial state 19) Finding ∆𝐻 using ∆𝐻f° ∆𝐻f° = Σn∆𝐻products – Σn∆𝐻reactants Standard state enthalpies for monoatomic elements is 0 20) How much heat is released when 4 grams of H2O2 decomposes into water and oxygen gas? 2H2O2 (l) → H2O(l) + O2(g) 4𝑔 34𝑔 = 0.12 mol H2O2 ∆𝐻 = -187.8 Heat Released = (0.12)(-187.8) = -22.54 kJ/mol Thermo-Day 3 Notes 21) TFP vs NTFP TFP (spontaneous) Assistance from outside sys needed to Proceeds w/o assistance from outside sys. induce change Ex. @ 25℃ Ex. ○ H2O evaporates ○ H2O doesn’t freeze at 75℃ ○ Iron rusts ○ Fe2O3 + 3C → 2Fe + 3CO (oppo of ○ NaCl dissolves rusting) A process that is NTFP in one direction, is NTFP (non spontaneous) TFP in the opposite direction Note: Thermodynamics cannot determine the speed of a reaction. 23)Entropy (S) A Measure of disorder in a system Units for entropy (S): JK-1mol-1 Entropy increases w/ more disorder and Entropy decreases w/ more order, less microstates microstates water Water being frozen Melting ice Salt crystals forming Toasting bread Water vapor condensing Salt dissolving Buildling a lego tower Making Popcorn Cooling jello liquid Microstates: microscopic arrangements and energies of atoms/molecules of macroscopic sys. 24) Identifying ∆𝑆 + ∆𝑆 = favored  Dissolving ionic compounds ○ Dissolved ions have greater entropy than lattice Vaporization Melting Rxn where products have same phase as reactants, but contain more particles (decomposition) Making most solutions Adding heat ○ inc distribution of KE ○ Entropy = 0 when K = 0 Inc. vol of gas 26) Exceptions for favorable ∆𝑆 Small highly charged cations (Al3+, Fe3+, Cr3+, Be2+) attract and reduce # of microstates of H2O molecules, reducing entropy of H2O ○ Entropy of H2O may be reduced to the point where entropy of overall sol becomes less than the components before mixing Making solutions w/ liquids and gases ○ Rapid and chaotic mvmts of gas particles are greatly reduced by molecules in sol. 27) Laws of Thermodynamics 1. Energy can’t be created or destroyed 2. Defines spontaneity and acknowledges that entropy increases over time 3. At 0K, there is no movement 26) How do you calculate ΔS? ΔSuni = ΔSsys + ΔSsurr. 26)Endo vs Exo Entropy (surroundings) Endo Exo SSurr dec Ssurr inc ○ Heat is absorbed by sys ○ Heat is released by sys ○ Less microstates in surroundings ○ Greater microstates in surroundings 27)Gibbs Free Energy Eq. ΔG° = ΔH - TΔS ΔG = Free Energy (kJ/mol) | ΔH = Enthalpy (kJ/mol) | T = temperature (K) | ΔS = Entropy (J/K) ΔG° = mac amt of free energy remaining that can be used to do work ΔH° = energy transferred as heat TΔS = energy used to create disorder 28) TFP = ΔG0 ΔH ΔS TFP – + always TFP + – never NTFP – – TFP @ low temps NTFP @ high temps + + TFP @ high temps NTFP @ low temps 29) ΔG > 0 External energy possibly used to recharge battery Photons can supply energy req to remove e- Electrolytic cells need battery 30) Calculating ΔG 1. Plug in givens into equation Calculate ΔH Calculate ΔS w/ entropy values 2. Energy Coupling 31) Find TFP temp ∆𝐻 T= ∆𝑆 = ____ K