Final Study Guide Chem 2 Chapter 13 13 3 The Rate Law of Reaction Rate Law shows the dependence of the rate of a reaction on the concentration of different reactants aA cC dD Rate Law k A m k rate constant specific for each reaction m order of the reaction with respect to A NOT based on stoichiometric coefficients If m 0 Rate k A 0 k Zero Order Reaction Rate is independent of concentration of A M s If m 1 Rate k A 1 k A First Order Reaction Rate depends on concentration of A If A doubles rate doubles s 1 If m 2 Rate k A 2 Second Order Reaction Rate depends on concentration of A If A doubles rate will be 4 times faster M 1s 1 Ex A M 1 10 2 20 3 40 Initial Rate M s 1 015 2 030 3 060 Rate k A m Rate 2 Rate 1 k A 2 k s cancel out 015 030 20 10 m 2 2m ln2 mln2 m ln2 ln2 m 1 First order reaction Rate constant Rate k A 1 015 M s k 10M k 015 M S 10 M k 15 s m A 1 m Reaction Order for Multiple Reactants aA bB cC dD Rate k A m B n m order of reaction w respect to A n order of reaction w respect to B Order of reaction m n Ex NO2 CO NO CO2 NO2 M 1 10 2 20 3 20 4 40 CO M 1 10 2 10 3 20 4 10 initial Rate M s 1 0021 2 0082 3 0083 4 033 Rate Law Rate k NO2 m CO n m n Determine m Choose 2 experiments where NO2 is changed and CO is not rate 2 rate 1 k NO2 2 k and CO s both cross out b c the same 0082 0021 20m 10m 4 2m log4 mlog2 m log4 log2 2 Second Order Reaction Do the same steps to get n 0 Order of Reaction 2 0 2 second order reaction n k NO2 1 m CO 1 m CO 2 n Rate Constant Exp 1 0021 M s k 10 M 2 10M 0 k 0021 M s 01 M2 k 0 21 M 1s 1 Effect of Concentration on the Rate Rate k A 2 B 3rd Order A is doubled 2 2 4 rate will be 4 times faster B is doubled A unchanged 2 1 2 rate is doubled A is tripled B doubled 3 2 9 2 1 9 x 2 18 rate is 18 times faster If both concentrations change multiply by each other 13 4 Integrated Rate Law Zero Order A time kt A 0 given A 0 initial concentration Linear Decreasing Straight Line Slope k y int A 0 First Order ln A t kt ln A 0 given Linear Decreasing Straight Line will be curved before ln is put in Slope k y int ln A 0 Second Order 1 A t kt 1 A 0 given Linear Increasing Straight Line Slope k y int 1 A 0 The Half Life of a Reaction t Time required for concentration of reactant to fall to of its initial value Zero Order t A 0 2k given know which half life goes with which First Order t 693 k given know which half life goes with which order Does not depend on initial concentration of reactants Second Order t 1 k A 0 given know which half life goes with which order Depends on initial concentration 13 6 Reaction Mechanism Reaction Mechanism is a series of individual chemical steps by which an overall chemical reaction occurs Ex Step 1 is the slow step and step 2 is the fast step Step 1 k1 H2 g ICl g HI g HCl g Step 2 k2 HI g ICl g HCl g I2 g Overall RXN H2 g 2ICl g 2HCl g I2 g Intermediates Molecules ions that appear in elementary steps but do not show up in overall rxn In this example HI is an intermediate Elementary Steps Each step taken to finally reach the overall reaction Rate Law is derived from the slowest step in the case Step 1 or k1 Rate Law k1 H2 ICl The order of reaction with respect to each reactant is equal to the stoichiometric coefficient of that reactant in the slow step m 1 n 1 Total Order 2nd Energy Diagram for a two step Mechanism Step 1 Ea1 is larger slower step rate determining step Step 2 Ea2 is smaller faster step of transition states of reaction steps Chapter 14 14 2 The concept of Dynamic Equilibrium Dynamic Equilibrium For a chemical reaction the condition in which the rate of the forward reaction to the right equals the rate of the backwards reaction to the left 14 3 The equilibrium Constant K aA bB cC dD K C c D d A a B b Only gaseous and aqueous species are considered in the expression of K Significance of the Equilibrium Constant K K 1 large Products are favored so FORWARD reaction move right More Products than Reactants K 1 small Reactants are favored so BACKWARD reaction move left More reactants than products K 1 Neither reactants or products are favored Relationships Between Equilibrium Constant Chemical Equation K Forward 1 K Backward K1 Kn Koverall K1 K2 Ex N2 3H2 2NH3 K 3 7 108 25 celsius NH3 N2 3 2 H2 K1 It is flipped so take reciprocal of K It is ed so K1 2 K1 1 K1 2 1 3 7 108 K1 5 2 10 5 Ex2 A 2B 3C K A 2B 2D C K1 C D K2 A 2B 2D C K1 2D 2C 1 K2 Soooo K k1 1 K2 2 2 Kc Products Reactants Kc C c D d A a Kp Pc CPd D Pa A Kp Kc RT n 14 4 Expressing the Equilibrium Constant in Terms of Pressure aA cC dD n Sum of stoichiometric coefficients of products sum of stoichiometric coefficients of reactants c d a ONLY COUNT IF AQUEOUS OR GASEOUS R 08206 Latm molK 14 5 Heterogeneous Equilibrium 2CO g CO2 g C g Kc CO2 C CO 2 CO2 g H2O l H aq HCO3 K H HCO3 CO2 aq 14 6 Calculating the Equilibrium Constant A g 2B g Initial Conc Of A 1 00 M When equilibrium is reached conc Of A 025 M Find K A 2 B I 1 00 M 0 M C E 025 M 2 025 M 75 M 50M K B 2 A K 50 2 75 33 14 7 The reaction Quotient Q aA bB cC dD Qc C c D d A a B b Qp PC d PA c PD aPB b K 1 reactants favored backwards reaction moves left Q K rxn is equilibrium Q K rxn proceeds to the left backwards to reach equilibrium reactants favored Q K rxn proceeds to the right forwards to reach equilibrium products favored 14 9 Le Chatelier s Principle …
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