2.  Work (W) is a force acting over a distance  Universe is the system and surroundings  System is the part of the universe you are focusing on (molecules, atoms or ions undergoing a change)  Surrounding is everything else  State function is a property or system that depends only on initial and final conditions

3.  Energy (E) is the capacity to do work or transfer heat (q)  Two forms of energy are Kinetic Energy and Potential energy  The Law of Conservation of Energy states that energy can not be created or destroyed  Internal Energy is the sum of kinetic energy and potential energy in a system.  E = q + W

4.  First: Energy cannot be created or destroyed or the energy of the universe is constant  Second: In any spontaneous process there is an increase in entropy of the universe. The entropy (disorder) of the universe is increasing  Third: The entropy of a perfect crystal at 0K is zero

5.  Enthalpy (  H) is a state function (independent of the path) and is related to heat   H =  H (products) –  H (reactants)   H forward  H reverse   H depends on the state (phase) of the substances   H is an extensive property – it depends on the number of moles present

6. EXOTHERMICENDOTHERMIC heat flows out of systemheat flows into system heat is a productheat is a reactant system has less energysystem has more energy ΔH is negative to indicate direction of heat flow ΔH is positive to indicate direction of heat flow energy released making bonds > than energy absorb to break bonds energy absorb to break bonds > than energy released making bonds product’s bond energy > reactant’s bond energy reactant’s bond energy > product’s bond energy

9.  H system +  H surroundings = 0  H system = -  H surroundings HH H

10.  The enthalpy of a reaction does not depend on the path, only initial and final conditions. It is a state function  The enthalpy of a reaction is the sum of the enthalpies of simpler reactions  If a reaction is reversed, the sign of the enthalpy is changed  If the coefficients in a balanced reaction are multiplied by an integer, the enthalpy is multiplied by the same integer

11.  Nitrogen and oxygen react to produce nitrogen dioxide  N 2 (g) + O 2 (g)  2NO (g)  H = 180 KJ  2NO (g) + O 2 (g)  NO 2 (g)  H = -112 kJ  N 2 (g) + O 2 (g)  2NO 2 (g)  H = 68 kJ  Notice that the compounds on both sides of the yield sign cancel out.  The heat of the reaction is the sum of the two steps.

12. Reaction  H  2B (s) + 3 / 2 O 2 (g)  B 2 O 3 (s) -1273 kJ  B 2 H 6 (g) + 3O 2 (g)  B 2 O 3 (s) + 3 H 2 O(g) -2035 kJ  H 2 (g) + ½ O 2 (g)  H 2 O(l) - 286 kJ  H 2 O(l) + H 2 O (g)44kJ  2B(s) + 3H 2 (g)  B 2 H 6 (g)

13.  Label the reactions A, B, C …  Determine how to add the reactions to obtain the desired overall reaction  Multiply reactions by integers to make coefficients work (also multiply  H)  Flip reactions to make intermediate compounds cancel (change the sign of 

14. 2B(s) + 3H 2 (g)  B 2 H 6 (g) A. 2B (s) + 3 / 2 O 2 (g)  B 2 O 3 (s) B. B 2 H 6 (g) + 3O 2 (g)  B 2 O 3 (s) + 3 H 2 O(g) C. H 2 (g) + ½ O 2 (g)  H 2 O(l) D. H 2 O(l) + H 2 O (g) A: 2B (s) + 3 / 2 O 2 (g)  B 2 O 3 (s) -B: B 2 O 3 (s) + 3H 2 O (g)  B 2 H 6 (s) + 3O 2 (g) 3C: 3 H 2 (g) + 3 / 2 O 2 (g)  3H 2 O(l) 3D: 3H 2 O(l)  3H 2 O (g) 2B(s) + 3H 2 (g)  B 2 H 6 (g)

15.  You can use the heat of formation to determine the heat of a reaction.  This is a short cut which used the property of state functions (like Hess’s Law)  Simply determine the heat of formation  H f for each substance from the provided charts  Multiply the coefficient by the heat for each substance  Determine the sum of the heat of formation for the reactants and then the products  Subtract the heat of formation for the reactants from the products

16. The heat of formation for an element in its standard state is 0kJ. Calculate the change in enthalpy for: 4NH 3 (g) + 7O 2 (g)  4NO 2 (g) + 6H 2 O(l) using the ΔH o f in the appendix

17. NH 3 (g) = 7O 2 (g) = NO 2 (g) = H 2 O(l) =