1. THERMODYNAMICS REVIEW

2. Energy Ability to do work Units– Joules (J), we will use “kJ” Can be converted to different types Energy change results from forming and breaking chemical bonds in reactions

3. Heat (q) Energy transfer between a system and the surroundings due to a temperature change Transfer is instantaneous from high----low temperature until thermal equilibrium Temperature— Measure of heat, “hot/cold”

4. Heat (q) continued Kinetic theory of heat Heat increase resulting in temperature change causes an increase in the average motion of particles within the system. Increase in heat results in Energy transfer Increase in both potential and kinetic energies

5. Thermodynamics 101 First Law of Thermodynamics Energy is conserved in a reaction (it cannot be created or destroyed)---sound familiar??? Math representation: ΔE total = ΔE sys + ΔE surr = 0 Δ= “change in” ΔΕ= positive (+), energy gained by system ΔΕ= negative (-), energy lost by system Total energy = sum of the energy of each part in a chemical reaction

6. 1 st Law of Thermodynamics (Conservation of Energy) Energy cannot be created or destroyed. With physical and chemical changes, energy can be transferred or converted. Total energy = Σenergy of its components ΔU = q + w, ΔE total = ΔE sys + ΔE surr = 0

7. How do we find the change in energy/heat transfer that occurs in chemical reactions??? CALORIMETRY

8. Calorimetry Experimentally “measuring” heat transfer for a chemical reaction or chemical compound Calorimeter Instrument used to determine the heat transfer of a chemical reaction Determines how much energy is in food Observing temperature change within water around a reaction container ** assume a closed system, isolated container No matter, no heat/energy lost Constant volume

9. Specific Heat Capacity Amount of heat required to increase the temperature of 1g of a chemical substance by 1°C Units--- J/g  °K Unique to each chemical substance Al (s) = 0.901J/g  °K H 2 O (l) = 4.18 J/g  °K

10. q = smΔT

11. “Coffee Cup” calorimeter (cont.) q chemical = - q water

12. Example 2: Using the following data, determine the metal’s specific heat. Metal mass = 25.0g Water mass = 20.0g Temperature of large water sample = 95°C Initial temperature in calorimeter = 24.5°C Final temperature in calorimeter = 47.2°C Specific heat of water = 1.00 cal/g  °C OR 4.184 J/g  °K (KNOW!!!!)

13. ENTHALPY

14. Enthalpy (H) Measures 2 things in a chemical reaction: 1)Energy change 2)Amount of work done to or by chemical reaction 2 types of chemical reactions: 1)Exothermic—heat released to the surroundings, getting rid of heat, -ΔΗ 2)Endothermic—heat absorbed from surroundings, bringing heat in, +ΔΗ **Enthalpy of reaction—heat from a chemical reaction which is given off or absorbed, units = kJ/mol Enthalpy of reaction Heat from a chemical reaction which is given off or absorbed At constant pressure Units = kJ/mol

16. Example 2: Calculate the ΔH for the following reaction when 12.8 grams of hydrogen gas combine with excess chlorine gas to produce hydrochloric acid. H 2 + Cl 2  2HCl ΔH = -184.6 kJ

17. Hess’ Law Enthalpy change for a chemical reaction is the same whether it occurs in multiple steps or one step ΔH rxn = ΣΔH A+B+C (sum of ΔH for each step) Allows us to break a chemical reaction down into multiple steps to calculate ΔH Add the enthalpies of the steps for the enthalpy for the overall chemical reaction

18. Guidelines for using Hess’ Law Must use data and combine each step in a way that gives the chemical reaction with the unknown ΔH Set up steps so chemical compounds not in the final reaction are cancelled Reverse a reaction if necessary and change the sign on ΔH Check for correct mole ratios

19. Example 1: H 2 O (l)  H 2 O (g) ΔH° = ? Based on the following: H 2 + ½ O 2  H 2 O (l) ΔH° = -285.83 kJ/mol H 2 + ½ O 2  H 2 O (g) ΔH° = -241.82 kJ/mol

20. Enthalpy of Formation (ΔH f °) Enthalpy for the reaction forming 1 mole of a chemical compound from its elements in a thermodynamically stable state. Elements present in “most thermodynamically stable state” 25°C°, 1atm

21. Apply Hess’ Law---- Really ΔH f (products) - ΔH f (reactants) Calculate ΔH rxn based on enthalpy of formation (ΔH f ) aA + bB  cC + dD ΔH° =[c (ΔH f °) C + d(ΔH f °) D ] - [a (ΔH f °) A + b (ΔH f °) B ]

22. How does a chemical reaction have energy? BOND ENTHALPIES

23. Bond Energy Energy required to make/break a chemical bond Endothermic reactions Products have more energy than reactants More energy to BREAK bonds Exothermic reactions Reactants have more energy than products More energy to FORM bonds

24. Bond Enthalpy Focuses on the energy/heat between products and reactants as it relates to chemical bonding Amount of energy absorbed to break a chemical bond---amount of energy released to form a bond. Multiple chemical bonds take more energy to break and release more energy at formation Amount of energy absorbed = amount of energy released to break chemical bond to form a chemical bond

25. Calculating ΔH rxn. by bond enthalpies (4 th method) Least accurate method ΔH = ΣBE (bonds broken) - ΣBE (bonds formed)

26. Example 1: Using average bond enthalpy data, calcaulate ΔH for the following reaction. CH 4 + 2O 2  CO 2 + 2H 2 O ΔH = ? BondAverage Bond Enthalpy C-H413 kJ/mol O=O495 kJ/mol C-O358 kJ/mol C=O799 kJ/mol O-H467 kJ/mol

27. Methods for Calculating Enthalpy-- 1)Stoichiometric Calculations using Balanced Chemical Equation 2)Calorimetry (lab based method) 3)Hess’s Law 4)Enthalpy of Formation 5)Bond Enthalpies **Which method is the LEAST accurate?

28. ENTROPY

29. Spontaneous vs. Nonspontaneous 1)Spontaneous Process Occurs WITHOUT help outside of the system, natural Many are exothermic—favors energy release to create an energy reduction after a chemical reaction Ex. Rusting iron with O 2 and H 2 O, cold coffee in a mug Some are endothermic Ex. Evaporation of water/boiling, NaCl dissolving in water

30. Spontaneous vs. Nonspontaneous 2) Nonspontaneous Process REQUIRES help outside system to perform chemical reaction, gets aid from environment Ex. Water cannot freeze at standard conditions (25°C, 1atm), cannot boil at 25°C **Chemical processes that are spontaneous have a nonspontaneous process in reverse **

31. Entropy (S) Measure of a system’s disorder Disorder is more favorable than order ΔS = S (products) - S (reactants) ΔS is (+) with increased disorder State function Only dependent on initial and final states of a reaction Ex. Evaporation, dissolving, dirty house

32. Thermodynamic Laws 1 st Law of Thermodynamics Energy cannot be created or destroyed 2 nd Law of Thermodynamics The entropy of the universe is always increasing. Naturally favors a disordered state

33. 3 rd Law of Thermodynamics The entropy (ΔS) of a perfect crystal is 0 at a temperature of absolute zero (0°K). No particle motion at all in crystal structure All motion stops

34. How do we determine if a chemical reaction is spontaneous? 1)Change in entropy (ΔS) 2)Gibbs Free Energy (ΔG)

35. Change in entropy (ΔS) For a chemical reaction to be spontaneous (ΔS T > 0), there MUST be an increase in system’s entropy (Δs sys > 0) and the reaction MUST be exothermic (Δs surr > 0). Exothermic reactions are favored, NOT endothermic reactions. Exothermic (ΔH 0) Endothermic (ΔH > 0, ΔS < 0) ΔS T = Δs sys + Δs surr If ΔS T > 0, then the chemical reaction is spontaneous

36. Example 1: Will entropy increase or decrease for the following? a)N 2 (g) + 3H 2 (g)  2NH 3 (g) b)2KClO 3 (s)  2KCl (s) + 3O 2 (g) c)CO (g) + H 2 O (g)   CO 2 (g) + H 2 (g) d)C 12 H 22 O 11 (s)  C 12 H 22 O 11

37. How do we calculate the entropy change (ΔS) in a chemical reaction? Same method as using the enthalpies of formation to calculate ΔH and use the same table. aA + bB  cC + dD ΔS° =[c (ΔS° C ) + d(ΔS° D )] - [a (ΔS° A ) + b (ΔS° B )]

38. Example 2: Calculate ΔS° for the following reaction at 25°C…. 4HCl (g) + O 2 (g)  2Cl 2 (g) + 2H 2 O (g)