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Thermochemistry

Thermochemistry. kinetics. Equilibrium. Endothermic vs. Exothermic Rxns. Heat = enthalpy, Δ H. Joules, or KJ. break bond (reactants). requires energy (endothermic). form bond (products). releases energy (exothermic). Avg. bond Enthalpy:.

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Thermochemistry

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  1. Thermochemistry kinetics Equilibrium

  2. Endothermic vs. Exothermic Rxns Heat = enthalpy, ΔH Joules, or KJ break bond (reactants) requires energy (endothermic) form bond (products) releases energy (exothermic) Avg. bond Enthalpy: The energy needed to break 1 mole of covalent bonds between atoms in the gas state

  3. potential energy in the system is released as heat to surroundings exothermic rxn Ex: combustion: CH4 + O2→ CO2 + H2O + heat Ex: combustion: CH4 + O2→ CO2 + H2O Ex: neutralization: HCl + NaOH→ NaCl + H2O + Δ gets warmer! -ΔH more energy is released to form the products than it to took to break up the reactants Implies stronger bonds in products

  4. endothermic rxn Energy absorbed from surroundings is stored up as potential energy in system +ΔH feel colder Ex: NH4NO3 (s) + heat → NH4+ (aq) + NO3- (aq) more energy is required to break up the reactants than was released to form the products Implies stronger bonds in reactants

  5. Calculating enthalpy change,ΔH stoichiometry Given the following change: KOH (s) → KOH(aq) ΔH = -43 KJ/mol What is the enthalpy change when 14.0 grams of KOH are dissolved in water. 1 mol 14.0 g -43 KJ 0.2496 mol 56.1 g 1 mol -10.7 KJ

  6. q = mCΔT calorimetry OR ΔH = mCΔT temp change mass specific heat capacity specific heat capacity: amt of heat needed to raise the temp of 1 gram of a substance by 1°C 46.2 grams of a metal are heated to 95.4°C then placed in a calorimeter containing 75.0 grams of water at 19.6° C . The final temperature of the water and metal is 21.8°C. Calculate the specific heat of the metal. Cwater = 4.18 J/g°C

  7. heat lost by M = heat gained by H2O q = mCΔT q = mCΔT M H2O m C ΔT = m C ΔT M M M H2O H2O H2O (46.2 g)(C) (95.4-21.8°) = (75.0 g)(4.18 J/g°C)(21.8-19.6) 73.6 2.2°C 3400.32 3400.32 g °C C = 0.203 J/g°C

  8. Hess’s Law When several rxns can be added to give one overall rxn, the ΔH for the overall rxn is the sum of the ΔH’s for the individual rxns. If one of the individual rxns needs to be reversed, reverse the sign of ΔH If one of the individual rxns needs to be multiplied by some factor, multiply its ΔH by that same factor

  9. What is the value of ΔH for: 2 H2O2 (l) → 2 H2O(l) + O2 (g) rev, x2 H2 (g) + O2(g) → H2O2(l) ΔH = -187.6 KJ 2 H2 (g) + O2(g) → 2 H2O(l) ΔH = -571.6 KJ keep 2 H2O2(l) → 2 H2 (g) + 2 O2 (g) ΔH = 375.2 KJ 2 H2O2 (l) → 2 H2O(l) + O2 (g) ΔH = -196.4 KJ exo

  10. Ex: Calculate ΔH for: 2 N2(g) + 5 O2(g) → 2 N2O5 rev, x2 rev, x2 H2(g) + ½ O2(g) → H2O(l) ΔH = -285.8 KJ/mol rev, x2 rev, x2 N2O5 (g) + H2O(l) → 2 HNO3(l) ΔH = -76.6 KJ/mol ½ N2(g) + 3/2 O2(g) + ½ H2(g) → HNO3(l) ΔH = -174.1 KJ/mol x 4 x 4 4 HNO3 → 2 H2O(l) + 2 N2O5ΔH = 153.2 KJ/mol 5 2 N2 + 6 O2 + 2 H2 → 4 HNO3 ΔH = -696.4 KJ/mol 2 H2O → 2 H2 + O2ΔH = 570.4 KJ/mol 2 N2(g) + 5 O2(g) → 2 N2O5 ΔH = 27.2 KJ/mol

  11. Ex: Calculate ΔH for: 3 V2O3 → V2O5 + 4 VO rev rev 4 VO + O2→ 2 V2O3ΔH = -753 KJ/mol x ½ x ½ 2 V2O3 + O2 → 4 VO2ΔH = -452 KJ/mol x ½ x ½ 4 VO2 + O2 → 2 V2O5ΔH = -243 KJ/mol 2 VO2 + ½ O2 → V2O5ΔH = -121.5 KJ/mol 2 V2O3→ 4 VO + O2ΔH = + 753 KJ/mol V2O3 + ½ O2 → 2 VO2ΔH = -226 KJ/mol 3 V2O3 → V2O5 + 4 VO ΔH = 406 KJ/mol

  12. Enthalpy Change using Bond Energies ΔHrxn = Σ B.E. reactants: + products: - OR ΔH = B.E. (react) – B.E. (prod) Calculate ΔH for the following rxn: CH4 + 2 Cl2 + 2 F2→ CF2Cl2 + 2 HF + 2 HCl I bond energy (KJ/mol) C-H 413 Cl-Cl 239 F-F 154 C-F 485 C-Cl 339 H-F 565 H-Cl 427 -C- + 2 Cl-Cl + 2 F-F → I F I F-C-Cl + 2 H-Cl + 2 H-F I Cl

  13. B.E. (react): C-H = (413)(4) = 1652 Cl-Cl = (239)(2) = 478 2438 F-F = (154)(2) = 308 B.E. (prod): C-F = (485)(2) = 970 C-Cl = (339)(2) = 678 3632 H-F = (565)(2) = 1130 H-Cl = (427)(2) = 854 ΔH = 2438 – 3632 = -1194 KJ

  14. kinetics study of reaction rates Rate of rxn: change in concentration over time unit: mol dm3 ·sec Ex: A + 2 B → C + 3D or mol dm-3 sec In terms of the disappearance of A or B: reactants ↓ B disappears twice as fast as A In terms of the appearance of C or D: products ↑ D appears 3x as fast as C

  15. ways to measure the rate of a rxn: mass or volume changes for gaseous reactants or products 2. Change in pH for rxns involving acids or bases 3. Change in conductivity measurements for rxns involving electrolytes 4. Use of a spectrophotometer for rxns involving color changes

  16. HCl + CaCO3→ CO2 + H2O + CaCl2 H2O2→ H2O + O2

  17. collision theory In order for atoms/molecules to react, they must collide; collisions must be effective particles must have sufficient energy: Activation Energy, Ea 2. particles must collide in the correct orientation anything that results in more effective collisions speeds up the rxn

  18. Factors that affect the rate of a rxn 1. Nature of the reactants ions in sol’n react faster than solids molecules w/ weaker bonds react faster than those w/ stronger bonds 2. Surface area (solids only) smaller the particle size, greater the surface area, faster the rxn. 3. Concentration ↑ conc. ↑ rate

  19. 4. Pressure (gases only) ↑ pressure by ↓ volume of container ↑ rate 5. temperature ↑ temp ↑ rate b/c a greater % of molecules will have the minimum Ea to react Activation Energy, Ea: the minimum amount of energy needed to bring about a rxn.

  20. Ea Maxwell-Boltzmann distribution Low T Ea high T

  21. chemical that speeds up rxn w/o being consumed 6. Catalyst works by lowering the activation energy, Ea Maxwell-Boltzmann distribution uncatalyzedrxn catalyzed rxn Fraction of molecules that react more molecules have minimum Ea Ea too low; no rxn Ea Ea

  22. Effect of a catalyst exo endo

  23. Homogeneous catalyst: catalyst is in same phase as reactants Ex: sulfuric acid catalyst (solution) Heterogeneous catalyst: catalyst is in different phase than reactants Fe Ex: Haber process: 3 H2 + N2→ 2 NH3 V2O5 Ex: Contact Process: SO3 + H2O → H2SO4 Ex: decomposition of hydrogen peroxide: MnO2 2 H2O2→ 2 H2O + O2

  24. Reaction Mechanisms Reaction Mechanism: The series of steps by which a rxn occurs rate determining step ( RDS ): The slowest step in the mechanism; the overall rxn cannot go any faster than this step.

  25. Chemical Equilibrium reversible rxns only A + B ↔ C + D rate of forward reaction equals the rate of the reverse reaction in a closed system concentrations are constant; not equal!

  26. Equilibrium Constant, Keq indicates the “extent” of reaction Large Keq means products are favored more product than reactant > 1 extremely large: rxn goes to completion small Keq means reactants are favored more reactant than product < 1 Temperature the only thing that affects the value of Keq

  27. EquilibriumConstant Expression a A + b B↔ c C + d D Kc = [C]c [D]d [A]a [B]b examples Kc = [NH 3 ]2 [H 2 ]3 [N 2] 3 H2 (g) + N2 (g)↔ 2 NH3 (g) 4 NH3 (g) + 6 NO(g) ↔ 5 N2 (g) + 6 H2O (g) Kc = [N2 ]5 [H2O]6 [NH3]4 [NO]6

  28. Le Chatelier’s Principle when a stress is applied to a reaction at equilibrium, the reaction shifts (left or right) to relieve the stress and establish a new equilibrium change concentration Increasing conc. (adding more) of a substance shifts rxn in opposite direction 3 H2 (g) + N2 (g) ↔ 2 NH3

  29. decreasing conc. shifts rxn in same direction (to make more) 3 H2 (g) + N2 (g) ↔ 2 NH3 Change temp Increasing temp (add heat) shifts rxn to consume the heat favors endothermic 3 H2 (g) + N2 (g) ↔ 2 NH3 + heat Decreasing temp (remove heat) shifts rxn to “replace” the heat (favors exothermic)

  30. Effect of temp on Kc exothermic rxn: 3 H2 + N2 ↔ 2 NH3 + heat heating shifts rxn left, makes more reactant and uses up product Kc = [P] [R] Kc decreases cooling ↑ K endothermic rxn: 2 HI + heat → H2 + I2 heating shifts rxn right, makes more product and uses up reactant cooling ↓ K Kc increases

  31. Gases only! Change pressure Increasing pressure (by making container smaller) shifts rxn to the side with fewer moles of gas vice-versa ↑ P, shift right 3 H2 (g) + N2 (g) ↔ 2 NH3 ↓ P, shift left 4 moles 2 moles C(s) + O2(g)↔ 2 CO (g) ↑ P, shift left 1 mole ↓ P, shift right 2 moles

  32. will not shift equilibrium; Catalyst? equilib. reached more quickly 2 NO2(g) ↔ N2O4 (g) brown colorless ΔH = - 125 KJ/mol What is the effect of increasing the temp on The amount of NO2 ? increase The color of the mixture ? darker brown The value of Kc? decrease How will the position of equilibrium be affected by reducing the volume of the container? shift right

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