UNIT 17 Review, Part III Rate, Thermodynamics, Equilibrium, & Electrochemistry

C. Johannesson A. Collision Theory n Reaction rate depends on the collisions between reacting particles. n Successful collisions occur if the particles...  collide with each other  have the correct orientation  have enough kinetic energy to break bonds (activation energy)

B. Factors That Increase Rxn Rate n Surface Area (more SA = more opportunities for collisions) n Concentration (more concentrated = more frequent collisions) n Temperature (hotter = faster moving particles)  This means more frequent collisions AND  More activation energy n Catalysts (lower activation energy)

B. Exothermic Reaction n A reaction that releases (gives off) energy n products have lower energy than reactants n Energy is a PRODUCT 2H 2 (l) + O 2 (l)  2H 2 O(g) + energy  H (energy released)

C. Endothermic Reaction n A reaction that absorbs (takes in) energy n reactants have lower energy than products n Energy is a reactant 2Al 2 O 3 + energy  4Al + 3O 2  H (energy absorbed)

A. Reaction Pathway Diagram n Catalyst: speeds up the reaction by lowering the E a (activation energy) (catalyzed reaction in red) NOTE: E reactants, E products, &  H are unchanged by a catalyst Catalyst only affects E a

Equilibrium  Most reactions are reversible  2 Na + Cl 2  2 NaCl kJ  This is ___thermic  The REVERSE reaction:  2 NaCl kJ  2 Na + Cl 2  Is endothermic (absorbs energy) exo (releases energy)

Equilibrium  A reversible reaction becomes an equilibrium reaction if the forward and reverse reaction occur at the same rate.  Fe(OH) 3 ⇌ Fe OH -  The equilibrium arrow: ⇌  Indicates that the forward and reverse reactions will compete until they are occurring at the same rate.

Equilibrium  If the reverse reaction (iron (III) hydroxide forming from its ions) happens at the same rate as the forward reaction, the reaction will never “finish.” The concentrations stop changing because product and reactant are being made and used up at the same rate.

Le Chatelier’s Principle  A reaction at equilibrium, when “stressed,” will react to relieve the stress.  (If you mess with it, it will work to return to its equilibrium ratio.)

Le Chatelier: Concentration 2SO 2 (g) + O 2 (g) ⇌ 2 SO 3 (g)  If the SO 2 or O 2 concentration was increased: (i.e., we added more reactants)  Equilibrium will shift right to decrease the concentration of reactants.  (Some of the reactant we added gets used up and more SO 3 (product) is produced.)

Le Chatelier: Concentration 2SO 2 (g) + O 2 (g) ⇌ 2 SO 3 (g)  If the SO 3 concentration was increased: (i.e., we added more products)  Equilibrium will shift left to decrease the concentration of products.  (Some of the product we added gets used up and more SO 2 and O 2 (reactants) are made.)

Le Chatelier: Concentration 2SO 2 (g) + O 2 (g) ⇌ 2 SO 3 (g)  If the SO 2 concentration was decreased: (i.e., we removed reactants)  Equilibrium will shift left to replace the removed reactants.  Some SO 3 will get turned into SO 2 & O 2.  The amount of SO 3 decreases. The amount of O 2 increases.

Le Chatelier: Concentration 2SO 2 (g) + O 2 (g) ⇌ 2 SO 3 (g)  If the SO 3 concentration was decreased: (i.e., we removed products)  Equilibrium will shift right to replace the removed product.  Some SO 2 and O 2 will get turned into SO 3.  The amount of SO 2 and O 2 decreases.

Le Chatelier: Temperature N 2 (g) + 3 H 2 (g) ⇌ 2 NH 3 (g) + 92 kJ  The reaction is exothermic. (Heat is a product.)  An increase in temperature shifts this reaction to the left.  A decrease in temperature shifts this reaction to the right.

Le Chatelier: Temperature 2 NH 3 (g) + 92 kJ ⇌ N 2 (g) + 3 H 2 (g)  This reaction is endothermic. (Heat is a reactant.) It is the OPPOSITE of the previous reaction!  An increase in temperature shifts this reaction to the right.  A decrease in temperature shifts this reaction to the left.

+ 2 e - Loss of Electrons is Oxidation 2 e - + Gain of Electrons is Reduction

Zn (0  +2) Cu (+2  0) Zn (0  +2)

Zn  Zn e - 2e - + Cu 2+  Cu e-

Zn  Zn e - 2e - + Cu 2+  Cu KCl Cl - K+K+

Given a reduction half-reaction AND an oxidation: Cu e -  Cu Sn 2+  Sn e - Find their potentials (V) 0.34 V V The oxidation is written opposite from the version on the chart. So I “flip” it and change the sign on its voltage. Now ADD V 0.49 V

Given two reduction half- reactions: Cu e -  Cu Zn e -  Zn Find their potentials (V) 0.34 V V The more positive voltage will be the reduction. (Cu) The less positive voltage will change its sign and run as an oxidation. (Zn)

Given two reduction half- reactions: Cu e -  Cu Zn  Zn e V V 1.10 V Cu 2+ + Zn  Zn 2+ Cu Always flip the less positive half-reaction (and change the sign on its voltage). Then add the voltages.