Chemistry Exam 2 Study Guide
Chemistry Exam 2 Study Guide Chem 032
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This 0 page Study Guide was uploaded by Amy Vaughn on Monday March 14, 2016. The Study Guide belongs to Chem 032 at University of Vermont taught by Dr. Ruggles in Spring 2016. Since its upload, it has received 186 views. For similar materials see General Chemistry 2 in Chemistry at University of Vermont.
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Date Created: 03/14/16
Chemistry Exam 2 Study Guide Quick note you are going to do well if you do the textbook problems All of Ruggles questions come from the book The multiple choice is the only part that he makes up which both question in the practice exams will be gone over in this study guide Do the problems and get the practice Chapter 13 Mechanisms the end of Chapter 13 we cannot predict mechanism thus it will be given in the problem Rules for Mechanisms 1 Rate is determined by the slowest step called the determining step 2 Stoichiometry and order are the same in mechanisms not for overall rxn 3 Rate has to be in terms of true starting materials this can be found in overall rxn The slow step will have the highest activation energy be able to find the rate law of a reaction when given the mechanism very important is keep true starting materials in rate law look at 104 in chapter 13 for example Ratek3NH2Cl NH3 however NH2C1 is an intermediate so it has to be substituted by true starting materials OCl Z 3 0H Z K1NH3 OC139k2NH2Cl OH39 then do some algebra NHzCl k2 g k1NH3Z z Then substitute this for NHzCl in the rate law and combine k1 k2 k3 into just k and due to chapter 15 stuff get rid of OH and finally get rateNH32OCl39 Chapter 14 when rate forward rate backward then there is an equilibrium this doesn t mean Ks are equal products starting materials describes where a rxn is when establishing an equilibrium when at equilibrium use Keq QgtKeq will shift left more starting materials QltKeq will shift right more products Rules for Writing Equilibrium Equations 1 Stoichiometry and order are directly related Stoichiometry is the order 2 Solids DO NOT appear in the equation 3 Pure liquids DO NOT appear in equation like H201 C 4 r Keq L C Dd from the equation aAbB H cCdD Z Z Kppressure k in atms KCConcentration k in molarity KpKcRTAn Rules for finding K by using concentration 1 Balance the equilibrium 2 Equilibrium equation 3 Make an ICE table 4 Using the equablibriums found in the ICE table substitute concentrations into equilibrium equation to find K Assumptions we use assumptions when some really crappy algebra comes up in a Kc equation and we don t want to do the quadratic formula or a cube root Two assumptions used 1 When Keq gt 1X104 and the initial products are favored when this happens then the corresponding X becomes 0 because the change is so small that the X in the equation Example to understand what this X concept is K4 18 X106 and using an ICE table find the product equilibrium to be 642X and the two starting materials to be X Thus 64 2x 418X106 xx do to the assumptions X becomes on top associated with the 64 products becomes 0 thus the new equation becomes 418X106 V with simple algebra the answer becomes X0001 15 2 Keqlt1X10394 and the initial starting materials are favored when this happens then the corresponding X becomes zero for the starting materials in the equation Example K299X10397 and by using the ICE table products equal X2 and starting material is 075X Thus 2 x 299X10397 0175 x this causes a quadratic so an assumption can be used thus the 2nd assumption says the X associated with the starting material goes to 0 because the change is 2 X just so small that it doesn t matter so the equation becomes 299X10397 0175 so X equals 2287x10 4 When doing assumptions you HAVE to check 0175 2287x10 4 0175 X100998 above 95 is a good assumption LeChatelionwhen a stress is applied to an equilibrium the equilibrium will shift to alleviate the stress these can include volume pressure concentration and temperature Volume Pressure Increaseshift to side with more mols increaseshift to side with less mols Decreaseshift to side with less mols decreaseshift to productsside w more mols Concentration Remove gas bubbles away shift to produce more of bubbled away compound Precipitate coming out of solution products are favored Add increase in concentration will shift to opposite side of the increased concentration Temperatures write a AH on starting material or product side Exothermic AH is a product Increaseshift to starting material side Decreaseshift to product side Endothermic AH is a starting material Increaseshift to product side Decrease shift to starting material Chapter 15 acids Ka and bases Kb and neutrality neutrality is different at different temperatures Parent base conjugate acid Parent aid conjugate base To identify an acid H is in the front conjugate base is negative Relative strengths Acids Inorganic acids have no carbon HCl HBr H3PO4 these are generally stronger Oxyacid stronger the more oxygen it has has a more stable base If same of oxygens then more electronegative central atom is the stronger one Organic acids have carbon CH3C02H CC13C02H end in COZH Inductive effect created by the substitution of electronegative elements close to C02H the more in number and the closer they are the stronger the acid Exceptions HF is actually weak and CF3C02H is actually strong Relative strengths Bases Strongmetal OH Weak amine acids R3N greater the inductive effect the weaker the base The pH of anything is actually the log Example determine the pH of a 00158 M HCl solution pHlogH30 log 00158 18 In the case of a strong acid or base the products are greatly favored so much that is kind of stops being an equilibrium For the next part the practice exams will be needed reasons for the answers will be given 12 on 2015 exam a Smallest pKa which means large Ka strongest acid want most electronegative most of and closest proximity CH3CBrFC02H is the strongest because of the proximity b Largest Kb which means strongest base want least electronegative least of inductive effect CH32NH has the most donation of carbon c Strongest acid more oxygens more electronegative element HZSO4 because S is more electronegative then P d Weakest base more acid like CBr3NH2 strong electronegative central negative e Least stable conjugate base think of the parent least stable conjugate base comes from the most stable parent acid which the weakest acid it wants to go back to the acid form 0139 is the least stable because it has the weakest parent f Most stable conjugate acid looking for strongest parent base since strongest is the least stable since the conjugate acid wants to stay the acid CC13CH2NH3 has the strongest base parent Strongestmost stable conjugate Stableweakest doesn t want to convert back Unstablereactive 15 on 2015 exam a Largest pKa weakest acid CH3CBrC02H because Br has a lower electronegatiVity than F and there is only one b Largest Kb strongest base CH3C0239 least electronegative least inductive effect c Weakest acid CH32NH2 has the strongest parent base d Weakest base CN0333N lots of electronegatiVity e Least stable conjugate bas CH3C0239 this is the only base i Most stable acid the weakest acid H2TeO4 Te is the least electronegative Practice Practice Practice that is how you will do well
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