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Module 2

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by: Simone

Module 2 Chem 1160



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Chapter 13 & 14
General Chemistry II
Dr. Bell
Study Guide
General Chemistry 2
50 ?




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1 review
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"Yes please! Looking forward to the next set!"

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This 7 page Study Guide was uploaded by Simone on Wednesday February 3, 2016. The Study Guide belongs to Chem 1160 at a university taught by Dr. Bell in Winter 2016. Since its upload, it has received 41 views.


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Date Created: 02/03/16
Module 2 (T est Review/Study Guide)  Reaction rates are measurements of how FAST a reaction occurs, not the distance. It is used to compare the rates of two reactants/products in the same reaction, or to determine the average rate of reaction.  Formula: aA+bB→cC+dD +¿ aq)→3Br aq)+3H O(l) 2−3 2 ¿  Ex. −¿ ( )+BrO ( )+6H 5 Br Br } Br} ¿ ¿ Δ ¿ Δ ¿ o −1 = −1 ¿ ¿ 5 3 −4 1 o ( )1.66∗10 = 3 3( ) m o 9.96∗10−5 =x s Reactant (Change∈Amount ofProduct)  Rate of Reaction: Rateof Reaction= TimeTaken fortheChange  Rate Law and Reaction Order: Used when the rate is dependent on concentration of reactants rate=k[A] [B]  Formula: where “k” is the rate constant, “A” and “B” are the concentrations in mole, and “a” and “b” are the order of operation (leading values)  The order determines how the rate will be affected.  With given a table of values such as; TRAIL [A] [B] [C] INTIAL (M) (M) (M) RATE -4 1 .40 .40 .40 1.2*10 2 .40 .40 1.20 3.6*10 -4 3 .80 .40 .40 4.8*10 -4 -4 4 .80 .80 .40 4.8*10  The rate law of this is; rate= k[A][B][C] k=[A] [C]  The order in respect with each 2, 0, and 1 determined by [A] – (Trail 1 M by Trail 4) raised to the power of n. (n=moles) and the Initial rate trial 1 by initial rate trial 4. Getting a difference of 2 from the trial M’s and 4 from the initial rate. Subtract the two values and you get 2. 2 is equal to your n. M −1  Units of k are; zero order- s first order- s second 1 -1 -1 2- -1 order- M∗s (M s ) third order- M s  To get the units you start off at zero order units and you -1 multiple by either 1/M or M  Integrated Rate Law: used when time is involved, and you are given the initial and final concentrations and initial and final time, and you solve for the missing variable. (if initial is not given assume it is 1.)  Zero order- o Rate law= rate=k[A] =k 0 2 o the linear plot is [A] vs. time o a straight line equation of [A] =[A] -tt 0 [A]0 o a half life equation of 2k  First Order- o Rate Law= rate= k[A] 1 o Linear Plot= Ln[A] vs time o Straight line equation= Ln[A] =Ln[A] tkt 0 ln(2) .693 o Half life= k = k  Second Order- o Rate law= rate= k[A] 2 1 vs.time o Linear Plot = [ ] 1 = 1 +kt o Straight line equation= [ t [ ]0 1 o Half life equation k[ A0  Arrhenius Equation; rate constant is affected by rate and temperature and to calculate activation energy.  As temperature increases, more molecules have enough energy to react and to reach the transition state, activation energy being met increase rate of reaction and k increase −Ea/RT  Formulas k=Ae 3  A is the frequency factor, T is in Kelvins, and R= 8.314 J/mol*K o Ln k=ln A-E /Ra o Graph ln k vs 1/T ln k1= Ea(1 − 1) o (k2 R T 2 T 1 o A catalyst speeds up a reaction by decreasing the Activation energy, and increasing the k. o Activation energy does not increase the rate of reaction.  Reaction Mechanisms; series of individual steps by which a reaction occurs o Rules;  The elementary steps must add up to give overall balanced equation for the reaction  The rate law for the rate-determined step must agree with the experimentally determined rate law.  A catalyst speeds up the reaction.  Ex. Given OO +O2(fast) O 3O2O (slo2) Balanced Equation- O  Chemical Equilibrium:  Dynamic Equilibrium- refers to when the forward reaction is equal to the reverse rate of reactions (concentrations and reactants are not always equal to each other) 4  K<1- low concentration- reverse reaction is favored high concentration reactant and low concentration of products, to the left.  K>1- high concentration- forward reaction is favored, high concentration of products & low concentration of reactants, to the right. (SLOW)  K=1 products and reactants have equal concentrations.  In an equilibrium expression; solids and liquids are excluded c d K = [C] [D]  Formula: c [A] [B] Kc is equilibrium constant, In brackets are concentrations in M, and the lowercase letters are numbers in front of each substance.  Relationship between K and K c p n  K =p (Rc)  n=(moles of gaseous product)- (moles of gaseous reactant)  When chemical reaction is o Multiplied by a factor, K is raised to the multiplication factor. (K ) n o Inversed, K is inverted (K or 1/K) o Added to the other reactions, multiplied together to find final K (K 1K *2 ….3K)  Reaction quotient refers to the stoichiometric ratios, not in equilibrium  Q<K- products/forward reactions are favored, going to the right.  Q>K- reactants/reverse reactions are favored, going to the left. 5  Q=K- reaction is at equilibrium, equal.  Use an ICE table to find K  Ex. Assume zero Initial if no initial provided. H 2g)+ I2(g)= 2HI(g) I q r p C -x -x +2x E 2 q-x r-x p+2x p+2 x¿  ¿ ¿  ¿Chatelier sPrinciple  describes how the system shifts, to minimize disturbance, damage control o Change, Occurrence, Effect on Equilibrium, and Effect on K o Addition of Reactant, Added Reactant consumed, shift to the right, no change on K. o Addition of Product, added product consumed, shift to the left, no change on K. o Decrease Volume & Increase Pressure, pressure decreases, shift to fewer gas molecules, no change on K. o Increase Volume & Decrease Pressure, Pressure decreases, shift to more gas molecules, no change on K. o Increase temperature, heat is consumed, shift in endothermic direction, change on K. direction is right. o Decrease temperature, heat is generated, shift in exothermic direction, change on K. direction is left. 6 7


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