 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

 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

 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

 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

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

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

 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

 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.

Reaction  H  2B (s) + 3 / 2 O 2 (g)  B 2 O 3 (s) kJ  B 2 H 6 (g) + 3O 2 (g)  B 2 O 3 (s) + 3 H 2 O(g) kJ  H 2 (g) + ½ O 2 (g)  H 2 O(l) kJ  H 2 O(l) + H 2 O (g)44kJ  2B(s) + 3H 2 (g)  B 2 H 6 (g)

 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 

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)

 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

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

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