Study Guide for Exam !!
Study Guide for Exam !! CHEM 1200
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Popular in Chemistry
This 5 page Study Guide was uploaded by Alexi Martin on Thursday March 31, 2016. The Study Guide belongs to CHEM 1200 at Rensselaer Polytechnic Institute taught by Dr. Alexander Ma in Spring 2016. Since its upload, it has received 54 views. For similar materials see Chemistry II in Chemistry at Rensselaer Polytechnic Institute.
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Date Created: 03/31/16
Exam II Study Guide Question distribution: Easy Medium Hard Total Ch 16 1 7 2 10 Ch 17 1 4 1 6 Ch 18 2 4 3 9 Total 4 15 6 25 *number of quantative Questions=18* (highlighted things indicate things that need to be known or hints) Basic outline: Chapter 15 Chemical Kinetics reaction rates rate laws integrated rate laws half lives effect of temperature (Arrhenius) reaction mechanisms catalysts vs intermediaries Chapter 17 Acids and Bases - strong acids/bases - pH and pOH - weak acids o acid ionization constant, Ka, pKa o strength of the acid - finding pH of strong acids and weak acids - weak bases o base ionization constans Kb and pKb o strength of the base - Finding pH of strong base and weak bases - Acid base propertieies ions and salts - Ionization of polyprotic acids - Ka1>>> Ka1 = 1 ionization - Acid strength and molecular structure o Binary acids (HX) Xhalogens o Oxyacids - Lewis acids/bases (e pairs) - Organic acids and bases pay attention 1 Chapter 18 Ionic Equilibrium Buffers Buffer solutions and their pH (Henderson Hasselbach equation) Buffer effectiveness pH=pKa plus or minus 1 buffer capacity Titrations and pH cures strong acids and strong bases weak acids and strong bases strong acids and weak bases diprotic acids, Ka1,Ka2 solubility equilibrium AB> A+ +B Ksp Ksp=[A+][B] precipitation complex ion equilibria In Detail descriptions: Chapter 15 Chemical Kinetics 1 –reaction rates 2A+3B> C rate= d[C]/dt= d[A]/2dt= d[B]/dt 1 – rate law - 1 order rate=k[A] or rate=k’[B] nd - 2 order rate=k[A]^2 or rate=k’[B]^2 - 0order rate=k - in general rate= k[A]^n[B^m] for nA+mB> C double concentration A, rate 2x double B rate 3x 2 – integrated rate law y=mx+b - 0 order [A]t= kt+[A]0 line of [A] vs t is downward sloping and linear - First order ln[A]t = kt+ln[A]0 line of ln[A] vs t is downward sloping and linear - 2 order 1/[A]t = kt+1/[A]0 line of 1/[A] vs t is upward sloping and linear no derivations intial A final A or calculate K 2 –half lives - know all equations for half life th - 0 t1/2=[A]0/2k - 1 t1/2=0.693/k - 2 1/k[A]0 2 – effect of temperature - Arrhenius equation =k=Ae^ Ea/RT - lnK= Ea/R(1/T)+ln A or ln(k2/k1)= Ea/R [(1/T2)(1/T1)] 2 reaction mechanisms elementary vs multistep rate determining step (the slowest) intermediaries final reaction cannot have intermediaries catalysts: (beginning and end) cancel, cannot be in overall reaction Chapter 17 Acids and Bases strong acids and bases complete dissociation HCl+H2O>H3O++Cl pH NaOH+H2O>Na++OH pOH pH+pOH=pKa=14 pH= log[H3O+], pOH= log[OH] weak bases yields incomplete dissociation B+H2O>BH++OH kB=[BH+][OH]/[B] pOH 14pOH=pH weak acids yields incomplete dissociation HA+H2O>A+H3O+ Ka=[A][H3O+]/[HA] pH use ICE 1 –acid base properties of ions and salts (depends on the conjugate ions of acids or bases) strong acid= conjugate base ions anions are neutral [Cl Br I NO2 SO42] strong base= conjugate acid are cations are neutral [Na+, K+,Ca2+] weak acid=conjugate base anions are basic [F CH3COO] weak base=conjugate acid cationa are acidic [NH4+] 1 – ionization of polyprotic acids pH= ? Ka1>>Ka2>>>Ka3 [H3O+] after 1 dissociation is all you need - acid strength and molecular structure o binary acid increases down a column (HI,HBr,HCl,HF) o larger anion, larger area charge spread out more stable yields a strong acid o I>Br>Cl>F stability F>OH>NH2 o HF; H2O;NH2 HF>H2O>NH3 oxyacids strength depends on oxidation # of central atoms more bonds higher oxidation number, higher the strength more O attached to central atoms HClO4>HClO3>HClO2>HClO 1organic acids lab discussion for similar questions - stability of anion – charge 3 - electronegativity of atom unless down - Resonance stabilization RSH>ROH RS>RO - Inductive effect (CF3) o e withdrawing groups –stabilizing o e donating groups destabilizing o hybrid orbitals sp>sp2>sp3 - more ‘s character’ make stable the – charge Chapter 18 Ionic Equilibrium 2+ Buffers - buffer solutions and their pH - Henderson Hasselbach equation - pH=pKa+log[base]/[acid] M or mol - 1:1 buffer pH=pKa most effective buffer - pOH=pKb+log[acid]/[base] 1:1 buffer pOH=pKb - Calculate pH or pOH of solution after a strong acid or base added effectiveness of buffer pH is plus or minus one of the pKa [base]:[acid] nature from 0.1 to 10 pH=pKa most effective buffer capacity stoich calculatenew initial concentration # of mol of H+ or OH 2 titration and pH curve strong acid/strong base titration OH vs H3O+/total V pH or POH 7 equivalence point= pKa= 4.5 no equilibrium no ICE limiting reagent weak acids and weak bases I Befire Titration pH=pH of weak acid ICE II Before Equivalence buffer stoich table intial [A] and conjugate basepH=pka(1/2 equivalence) henderson hasselbach pH for buffer III @ equilibrium buffer capacity is filled pH> pKa ICE IV excess OH mol OH/total volume=pOH 14pOH=pH pH> [OH] excess stoich solubility AnBm>mA^m++nB^n+ example: PbCl2>Pb2++Cl kSp=[Pb2+][Cl]^2 s=n+m(Ksp/N^n*m^m)^1/2 4 common ion effect limiting reagent add more ion push to right add more solid move to the left 1 –compare Q to Ksp A<Ksp none Q=Ksp precipitate Q>Ksp precipitate Selective precipitation *Kf and ligands* 5
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