2.  Chemical Equilibrium occurs when opposing reactions are proceeding at equal rates.  When the forward reaction equals the reverse reaction.  It results in the formation of an equilibrium mixture of the reactants and products of the reaction.  The composition mixture does not change with time. Neither does the concentration once equilibrium is reached!!

5.  N 2 (g) + 3H 2 (g) ↔ 2NH 3 (g)  This reaction involves the presence of a catalyst, a pressure of several hundred atmospheres and a temperature of several hundred degrees Celsius.  This equilibrium mixture can be reached regardless of where one starts.  This RXN is exothermic

6.  Homogeneous: all substance are in the same state.  Heterogeneous: substance are in different states.

7.  Position of a heterogeneous reaction does not depend on the amount of pure solid and liquid present !! WHY?  Because concentrations don’t change during the course of a RXN

8.  The relationship between the concentrations of the reactants and the products of a system in equilibrium is given by the law of mass action.  aA + bB ↔ dD + eE  K c = [D] d [E] e ← products [A] a [B] b ← reactants Kc = equilibrium constant

10.  N 2 O 4 ( g ) ↔ 2NO 2 ( g )  Kc = = = 0.211 [NO 2 ] 2 [N 2 O 4 ] [0.0172] 2 [0.00140]

11.  When the equilibrium system consists of gases, it is convenient to express the concentrations of reactants and products in terms of gas pressures:  Kp = (P D ) d (P E ) e (P A ) a (P B ) b

12.  CO(g) + Cl 2 (g) ↔ COCl 2 (g)  Kc = = 4.56 x 10 9  For the equilibrium constant to be so large, the numerator must be much larger than the denominator. Therefore, the equilibrium concentration of COCl 2 must be greater than that of CO or Cl 2. [COCl 2 ] [CO][Cl 2 ]

13.  A large value for the equilibrium constant indicates that the mixture contains more products than reactants and therefore lies towards the product side of the equation.  A small value for the equilibrium constant means the mixture contains less products than reactants and therefore lies toward the reactant side.

15.  If the concentrations of all species in an equilibrium are known, the equilibrium-constant expression K can be used to calculate the value of the equilibrium constant.  The changes in the concentrations of reactants and products in the process of achieving equilibrium are governed by the stoichiometry of the equation.

16. What is the Reaction Quotient? KQ Equilibrium ConstantReaction Quotient Expression is ratio of products to reactants with balanced equation coefficients as powers Only includes gases and solutions To solve for K, plug in concentrations at equilibrium Expression is ratio of products to reactants with balanced equation coefficients as powers Only includes gases and solutions To solve for Q, plug in concentrations at any time

17. The Difference Between K and Q What exactly is the difference? 2 H 2 (g) + O 2 (g)  2 H 2 O (g) The expressions for K and Q are the same. To solve for “K”, plug in concentrations at equilibrium only. To solve for “Q”, plug in concentrations right now.

18. Using Reaction Quotient [products now] [reactants now] = Q [products at equilibrium] [reactants at equilibrium] =K Q = K[now] = [equilibrium] System is at equilibrium Q > K [Products now] too large [Reactants now] too small System will make more reactants to reach equilibrium Q < K [Products now] too small [Reactants now] too large System will make more products to reach equilibrium

19. Let’s Practice For N 2 (g) + O 2 (g)  2 NO (g) if [N 2 ] = 0.81 M, [O 2 ] = 0.75 M and [NO] = 0.030 M, is the reaction at equilibrium if K = 0.0025? If not, which way will it go to reach equilibrium?

20. Q = 0.0015 Q < K Reaction is not at equilibrium More products will need to be made (and also thereby reducing reactants) to have Q = K Reaction will go to the right to reach equilibrium Let’s Practice For N 2 (g) + O 2 (g)  2 NO (g) if [N 2 ] = 0.81 M, [O 2 ] = 0.75 M and [NO] = 0.030 M, is the reaction at equilibrium if K = 0.0025? If not, which way will it go to reach equilibrium?

21.  Knowing the value of K makes it possible to calculate the equilibrium amounts of the reactants and products, often by solving an equation where the unknown is the change in a partial pressure or concentration.

22. Predicting the direction of a reaction by comparing Q and K.

23. ◦ If a system at equilibrium is disturbed by a change in temperature, pressure or concentration of one of the components, the system will shift its equilibrium position to counteract the effect of the disturbance. ◦ OPPOSITE DIRECTION/ AWAY FROM ◦ IF SOMETHING IS REMOVED WE GO TOWARDS.

24.  Volume goes up- pressure is down  Pressure is up – volume is down

25. The effect of adding H 2 to an equilibrium mixture of N 2, H 2 and NH 3.

26.  If a chemical system is at equilibrium and the concentration of a substance is increased (either product or reactant), the system responds by consuming some of the substance.  If some of the concentration is decreased, the system will respond by producing some of the substance.  N 2 + 3H 2 ↔ 2NH 3

27.  At constant temperature, reducing the volume of a gaseous equilibrium mixture causes the system to shift in the direction that reduces the number of moles of gas.  The system will always favor the side with fewer moles of a gas in order to re-equilibrate.

29.  When the temperature of a system in equilibrium is increased, the system reacts as though a reactant were added to an endothermic reaction or a product was added to an exothermic reaction.  It will shift in the direction to consume the excess reactant or product, which is heat.  Endothermic:reactants + heat ↔ products  Exothermic:reactants ↔ products + heat

31.  Adding a catalyst increases the rate at which equilibrium is achieved, but it does not change the composition of the equilibrium mixture.