1. Thermochemistry Study of the transfer of energy in chemical reactions

2. Heat and Temperature What’s the difference?? Temperature (C 0 ): measures the average kinetic energy of particles. Instrument: thermometer Units: Celcius, Kelvin, F

3. Heat Heat (Q): Q= m ∆t Cp (Total Thermal Energy of a system) Instrument: calorimeter p.519 Units: Joules, Cal

4. Heat of Reaction (∆H) The net energy released or absorbed in a chemical reaction. The net energy released or absorbed in a chemical reaction. Thermochemical equation: includes the amount of energy released/absorbed. Thermochemical equation: includes the amount of energy released/absorbed. Write the thermochemical equation of the formation of water if 483.6 kJ of energy are produced from 2 moles of hydrogen. 2H 2 (g) + O 2 (g)  2H 2 O (l) + 486.3 kJ (exo) Always include states of matter!

5. Enthalpy (heat) Changes (ΔH) The amount of energy absorbed or lost by a system. How to calculate ΔH: How to calculate ΔH: A) from a graph: Heat products – Heat reactants A) from a graph: Heat products – Heat reactants If ΔH = negative: exothermic (Energy Lost) If ΔH = positive: endothermic (Energy absorbed)

6. B) ∆H can be calculated using: Hess’s Law :The overall ΔH is equal to the sum of the ΔH for individual steps. Molar heat of formation: ∆H f is the net heat released or absorbed when one mole of a compound is formed by the combination of its elements. (synthesis) ∆H f values (Appendix A-14) are based on one mole of product. -∆H f indicates a stable compound. Molar heat of combustion: ∆H c is the heat released by the complete combustion of one mole of a substance. ∆H c values (Appendix A-5) are based on one mole of reactant.

7. Calculate the ΔH rxn for: 2SO 2 + O 2  2SO 3 2SO 2 + O 2  2SO 3 Equation must be balanced. Equation must be balanced. Table A-14 (ΔH f ) Table A-14 (ΔH f ) ΔH f SO 2 = (-296.8)kJ/mol ΔH f SO 2 = (-296.8)kJ/mol ΔH f SO 3 = (-395.7)kJ/mol ΔH f SO 3 = (-395.7)kJ/mol ∆H rxn = (∆H f products – (∆H f reactants) ∆H rxn = (∆H f products – (∆H f reactants) (-791.4) - (-593.6) (-791.4) - (-593.6) Multiply both SO 2 and SO 3 by 2 because we have 2 moles of each. -593.6 kJ -791.4 kJ Answer: -197.8 KJ

8. Sample problems(continued) 2) Calculate the heat of reaction for: 2) Calculate the heat of reaction for: 2 H 2 O 2 2H 2 O + O 2 (must be balanced) 2 H 2 O 2 2H 2 O + O 2 (must be balanced) Find ∆H f table A-14 Find ∆H f table A-14 ∆H f H 2 O 2 (-187.8 kJ/mol) (2) = -375.6 kJ ∆H f H 2 O 2 (-187.8 kJ/mol) (2) = -375.6 kJ ∆H f O 2 (0.00 kJ/mol) ∆H f O 2 (0.00 kJ/mol) ∆H f H 2 O (-285.8 kJ/mol) (2) = -571.6 kJ ∆H f H 2 O (-285.8 kJ/mol) (2) = -571.6 kJ ∆H rxn = ∆H f products - ∆H f reactants ∆H rxn = ∆H f products - ∆H f reactants (-571.6) - (-375.6) (-571.6) - (-375.6) ANSWER: - 196.0 kJ ANSWER: - 196.0 kJ

9. Calculating ∆H from ∆H c values. Calculating ∆H from ∆H c values. 1. Calculate the heat of formation for: 1. Calculate the heat of formation for: C (s) + 2H 2(g) CH 4(g) C (s) + 2H 2(g) CH 4(g) Make sure the equation is balanced. Make sure the equation is balanced. Use ∆H c values of C, H 2 and methane (tbl. A-5) Use ∆H c values of C, H 2 and methane (tbl. A-5) NOTE: ∆H c values are based on one mole of NOTE: ∆H c values are based on one mole of REACTANT. Therefore: REACTANT. Therefore: ∆H f = ∆H c of REACTANTS – (∆H c PRODUCTS) C = -393.5 kJ/mol CH 4 = -890.8 kJ/mol H = (-285.8 kJ) (2) = (-965.1 kJ) – (-890.8kJ) ANSWER: ∆H f = -74.3 kJ

10. 17.2 Two Driving Forces of Reactions…Allow one to predict if a reaction will occur spontaneously. Enthalpy (ΔH): exothermic reactions favored (- ∆H) (products at lower energy) Enthalpy (ΔH): exothermic reactions favored (- ∆H) (products at lower energy) Entropy (ΔS): measures disorder of a system. (+∆S)High disorder favored. Entropy (ΔS): measures disorder of a system. (+∆S)High disorder favored. What does high disorder look like? Increase in randomness (disorder): What does high disorder look like? Increase in randomness (disorder):  Solid to liquid to gas  Lesser # particles to greater #particles  Solid dissolving

11. Possible reactions: RESULTS: 1.Exothermic and Entropy increase GO 2.Endothermic and Entropy decrease NO GO 3.Exothermic and Entropy decrease ? 4.Endothermic and Entropy increase ? #3 & #4 may or may not “Go” depending on which of the two driving forces is dominant. #3 example: water freezes. In this case, enthalpy(∆H) is the dominant factor. #4 example : Ice melts. Therefore, the entropy (∆S) is the dominant factor.

12. Gibb’s Free Energy (G) Predicts whether ∆H or ∆S will dominate in a reaction Natural processes proceed in the direction that lowers the free energy (G) of a system. Natural processes proceed in the direction that lowers the free energy (G) of a system. Only ΔG can be measured: ∆G=∆H-(T∆S) Only ΔG can be measured: ∆G=∆H-(T∆S) … Predicts whether a reaction will be spontaneous (at constant T & P). Negative ΔG = spontaneous reaction Negative ΔG = spontaneous reaction Positive ΔG = nonspontaneous reaction 

13. Sample problem: Calculate the free energy change for the following reaction at 25 0 C: Ca (s) + 2H 2 O (l) Ca(OH) 2 + H 2 Data: ∆H = -411.6 kJ ∆S = 31.8 J/mol.K ∆G = ∆H – (T∆S) -411.6 - (298 x.0318 kJ/mol.K) ANSWER: ∆G = -421.1 kJ

14. Summary Favorable conditions for a spontaneous reaction:  Free Energy (ΔG) = Negative  Enthalpy (ΔH) = Negative  Entropy (ΔS) = Positive ΔG = ΔH – TΔS  When enthalpy and entropy are at odds, temperature determines whether the reaction will proceed: ΔG = ΔH – TΔS

15. 17.3 The Reaction Process Collision Theory: Explanation of reactions as a result of collisions. –Particles must collide –Particles must collide in the correct spatial orientation –Collision must be energetic enough to disrupt bonds

16. Reaction Pathways Activation Energy: minimum energy required to transform reactants  activated complex. Activation Energy: minimum energy required to transform reactants  activated complex. Activated Complex: transitional structure formed from an effective collision. Activated Complex: transitional structure formed from an effective collision. A= Energy of Reactants B= Activation Energy C= Energy of Activated Complex D= Energy of Products Is this an exothermic or endothermic reaction?

17. 17.4 Reaction Rates and Kinetics Rate-Influencing Factors: Anything that will influence collision frequency and/or efficiency. Rate-Influencing Factors: Anything that will influence collision frequency and/or efficiency. –Nature of Reactants –Surface Area (increased surface area) –Temperature (increased temperature) –Concentration (increased conc. of reactants) –Catalyst (addition of)

18. Rate Laws …Relate reaction rate to the [concentration] of reactants, at a given temperature. As concentration increases, rate increases. 1) Elementary Reactions (single step) 2A + 3B A 2 B 3 2A + 3B A 2 B 3 Rate Law: R = k[A] 2 [B] 3 Rate Law: R = k[A] 2 [B] 3 3D E + F 3D E + F Rate Law: R = k [D] 3 Rate Law: R = k [D] 3

19. 2) Multi-step Reactions Rate is determined from the slowest step. example: Net equation: NO 2 + CO NO + CO 2 step 1: 2NO 2 NO 3 + NO (slow) step 2: NO 3 + CO NO 2 + CO 2 (fast) Rate Law? ANSWER: Rate Law = k[NO 2 ] 2

20. Rate law determination by experimental data : In a one step reaction: Doubling [A]: the rate increases by 2X Doubling [B]: the rate increases by 4X Reduce [B] to 1/3: the rate decreases to 1/9 What is the rate law? ANSWER: Rate= k [A] [B] 2