3. Branch of chemistry that studies the relationship between the chemical action and the amount of heeeet absorbed or generated.

4. Thermochemistry Terminology  Heeet (not to be confused with temp.)  Transfer of energy to an object of higher energy to an object of lower energy.  Symbolized by ( q )  Heeet is stoichiometric which means more stuff in reaction more heeet involved.

5. ( Continued )  System is the concentrated object  Surroundings are everything around the system  Enthalpy  Total heeet of a system  Symbolized by ( H )  State function: Only initial and final conditions matter not how you get thurr

6. Endothermic or Exothermic?  Endothermic: Requires input of heeet from surrounding for reaction to take place  The system feels to cool to touch  ∆H > 0  Exothermic: Releases heeet into the surroundings as the process occurs  The system feels to hot to touch  ∆H < 0

7. 3 Ways to determine Enthalpy change (∆H) of a reaction 1.Calorimetry 2.Hess’s Law 3.Standard Enthalpies of Formation

8. Calorimetry:  Measurement of heeet flow  Coffee cup Calorimeter C x M x ∆T=q Specific heeet capacity ( j/g°C) Mass ( g ) Change in Temp. ( C° ) Heeet   “ BOMB “ Calorimeter q= C x ∆T Heeet Specific heeet capacity ( j/g°C) Change in Temp. ( C° )

9. Calorimetry Example  H+(aq) + OH-(aq) → H2O(l)  The temperature of 110 g of water rises from 25.0°C to 26.2°C when 0.10 mol of H+ is reacted with 0.10 mol of OH-.  Calculate q of the water  Calculate ∆H

11. Hess’s law  Hess’s Law states that the heat of a whole reaction is equivalent to the sum of it’s steps.  For example: C + O 2  CO 2 This occurs as 2 steps C + ½O 2  CO  H  = – 110.5 kJ CO + ½O 2  CO 2  H  = – 283.0 kJ C + CO + O 2  CO + CO 2  H  = – 393.5 kJ I.e. C + O 2  CO 2  H  = – 393.5 kJ  Hess’s law allows us to add equations.  We add all reactants, products, &  H  values.

12. Hess’s law: Example We may need to manipulate equations further: 2Fe + 1.5O 2  Fe 2 O 3  H  =?, given Fe 2 O 3 + 3CO  2Fe + 3CO 2  H  = – 26.74 kJ CO + ½ O 2  CO 2  H  = – 282.96 kJ CO + ½ O 2  CO 2  H  = – 282.96 kJ 1: Align equations based on reactants/products. 2: Multiply based on final reaction. 3: Add equations. 2Fe + 1.5O 2  Fe 2 O 3 2Fe + 1.5O 2  Fe 2 O 3 3CO + 1.5 O 2  3CO 2  H  = – 848.88 kJ 3CO + 1.5 O 2  3CO 2  H  = – 848.88 kJ 2Fe + 3CO 2  Fe 2 O 3 + 3CO  H  = + 26.74 kJ 2Fe + 3CO 2  Fe 2 O 3 + 3CO  H  = + 26.74 kJ CO + ½ O 2  CO 2  H  = – 282.96 kJ CO + ½ O 2  CO 2  H  = – 282.96 kJ  H  = – 822.14 kJ

13. Standard Enthalpies of Formation  Standard conditions: Most stable form of the substance  1atm and 25°C ( 298K )  Standard Enthalpy, ∆H°, is enthalpy measured when everything is measured in standard state

14. Multiple Choice  1.) Which of the following is NOT a characteristic of an exothermic reaction? A.Reaction feels warm B.System gains energy C.Enthalpy change of reaction is negative

15. B 2 H 6 + 6H 2 0  6H 2 + 2H 3 BO 3  2.) ∆H=? KJ/mol A.-3604 KJ/mol B.-772 KJ/mol C.3604 KJ/mol D.772 KJ/mol B 2 H 6 : ∆H°= +36 KJ/mol H 2 0: ∆H°= -242 KJ/mol H 3 OBO 3 : ∆H°= -1094 KJ/mol

16.  3.) Which of the ∆H’s is exothermic? A.563 B.0 C.-375 D.989