1. 50 mL 100 o 100 mL 25 o 150 mL (4.184 J/ o C g) q = C x  T x mass q 2 = (4.184 J/ o C g) x q 1 = - q 2 (T f - 100) x (50) = T f = 50 o C q 1 =x (T f - 100)x (50 g) T f Heat transfer experiments (T f - 25) x(100 g) -(T f - 25) x (100)

2. Enthalpy of reaction  H rxn = q rxn coffee cup calorimeter 10.5 g KBr T f = 21 o Calculate  H rxn KBr(s)  K + (aq) + Br - (aq) q system = =  H rxn in 125g waterat 24 o - q surroundings

3.  H rxn q system = - q surroundings =  H rxn q surroundings =C x q surroundings = (4.184 J/g o C)(21(10.5 g = -1756 J q system = - q surroundings = +1756 J =  H rxn H is extensive  H rxn = 1756 J = 167 J/g= 19873 J/mol TT x mass - 24 o C)+ 125 g) a) endothermic b) exothermic 10.5 g KBr

4.  E = q + w  E = q - P  V At constant V, Bomb calorimeter q rxn = q system = -q calorimeter q calorimeter = C (J / o C) x  E = q v  T ( o C)

5. Constant Volume calorimetry 2Fe (s) + 3/2 O 2 (g)  Fe 2 O 3 (s) 11.2 g Fe(s), 1 atm O 2 C calorimeter = 2.58 kJ/ o C  T calorimeter = + 31.9 o C q rxn = -q calorimeter =  E rxn = - (2.58 kJ/ o C) = - 82.2 kJ  E rxn = - 822 kJ/mol Fe 2 O 3 / 0.1 mol Fe 2 O 3 (31.9 o C)

6. Thermite reaction H is an 2Al(s) +Fe 2 O 3 (s) Hess’ Law extensive,State function  Al 2 O 3 (s) + 2Fe(l)

7. 2Al(s) +Fe 2 O 3 (s)  Al 2 O 3 (s) + 2Fe(l) 2Al(s)  H = -1676 kJ/mol  H = - 822 kJ/mol 2Al(s) -854 kJ/mol Fe(s)  Fe(l) +15 kJ/mol _______________________________ __________ 2Al(s) +Fe 2 O 3 (s)  Al 2 O 3 (s) + 2Fe(l)  H rxn = -824 kJ/mol 2 ( ) 2 2 Fe(s) + 3/2 O 2 (g)  Fe 2 O 3 (s) Fe 2 O 3 (s)  2Fe(s) + 3/2 O 2 (g) _______________________________ __________ + + 3/2 O 2 (g) + Fe 2 O 3 (s)  Al 2 O 3 (s) + 2Fe(s)  Al 2 O 3 (s)

8. Hess’ Law Always end up with exactly the same reactants and products If you reverse a reaction, reverse the sign of  H If you change the stoichiometry, change  H

9. Heats of formation,  H o f  H = heat lost or gained by a reaction “ o ” = standard conditions: all solutes 1M all gases 1 atm “ f ” = formation reaction: 1mol product from elements in standard states for elements in standard states,  H o f = 0

10. 2Al(s) +Fe 2 O 3 (s)  Al 2 O 3 (s) + 2Fe(l) reactants products elements 2 Al(s) Al 2 O 3 (s) Fe 2 O 3 2 Fe(s) 3/2 O 2 (g) 2 Fe (l) HofHof  H o f Al 2 O 3 (s) + 2  H o f Fe (l) Fe 2 O 3 Al(s)

11. 2Al(s) +Fe 2 O 3 (s)  Al 2 O 3 (s) + 2Fe(l) reactants products elements 2 Al(s) Al 2 O 3 (s) Fe 2 O 3 2 Fe(s) 3/2 O 2 (g) 2 Fe (l) Fe 2 O 3  H o f Al 2 O 3 (s) + 2  H o f Fe (l)  H rxn =n  H o f products - - HofHof n  H o f reactants HofHof - HofHof Al(s)

12. 2Al(s) +Fe 2 O 3 (s)  Al 2 O 3 (s) + 2Fe(l)  H rxn = = -824 kJ n  H o f products - n  H o f reactants [  H o f Al 2 O 3 (s) + - [(-1676) -  H o f Fe(l)] 2 [  H o f Fe 2 O 3 (s)+  H o f Al(s)] 2 + (15)]2[(-822)+ 0]kJ

13. Bond Energies chemical reactions =bond breakage and bond formation bond energies energy required to break bond bond breakagea) endothermic b) exothermic (raise P.E.) bond formation(lower P.E.)exothermic positive

14. Bond energies CH 4 (g)  H rxn = C-H 413 kJ O=O 495 kJ C=O 799 kJ O-H 467 kJ  H rxn = bonds broken - bonds formed [ (C-H) + (O=O)] [ (C=O)+ (O-H)] = -824 kJ  H o f products-  H o f reactants =- 802 kJ - 4 2 24 + 2O 2 (g)  CO 2 (g) + 2H 2 O (g)

15. q v v.s. q p q v =  E q p =  H H = E + PV  H =  E +  PV if  n = 0 2Fe (s) + 3/2 O 2 (g)  Fe 2 O 3 (s)  n =  H =  H = -826 kJ/mol =  E +  nRT  H =  E (0 - 3/2)= - 3/2 - 822 kJ/mol + (- 3/2)(8.314 x 10 -3 kJ)(298)