CEM142 1nd EditionExam # 1 Study Guide Lectures: 1 - 9Lecture 1 -overview of syllabus (available online)-overview of schedule (available on inside cover of lecture notes book or online)-chemical kinetics: the rates and mechanisms of chemical reactions-rate = (# moles) / time = [concentration]/time = [A]/time-positive rate indicates that A is a product (it is appearing)-negative rate indicates that A is a reactant (it is disappearing)-concentration = moles/liter-1 mole = 6.02 x 1023 molecules-to find the normalized rate, divide by the coefficient of the reactant or product-rate law: rate vs concentration of the reactant-rate = k[reactant 1]m[reactant 2]nLecture 2-relationship between concentration and time: as a reaction proceeds, the concentrations of reactants and products change depending on the reaction order-reaction order: values for m and n in the equation learned in last lecture-review: rate = k[reactant 1]m[reactant 2]n-note: [ ] indicates a concentration of the substance within the brackets-integrated rate equations-ln([R]t/[R]0) = -kt from time 0 to t where [R]0 is the concentration at time 0 and [R]t is the concentration at time t-half life: the time taken for the concentration to reduce to one-half of its original value-when [R]t = ½ [R]0Order Rate law Integrated rate law Linear form Half-life Successivehalf-livesLinear plot0 Rate=k [R]=[R]0-kt [R]=[R]0-kt t1/2=[R]0/(2k) Halves [R] vs t slope=-k1 Rate=k[R] [R]=[R]0e-ktln[R]=ln[R]0-kt t1/2=0.693/k Constant ln[R] vs t slope=-k2 Rate=k[R]2[R]=[R]0/(1+[R]0kt) 1/[R]=(1/[R])+kt t1/2=1/(k[R]0) doubles 1/[R] vs t slope=kLecture 3-collision theory: a reaction is a rearrangement of atoms; bonds between atoms must be brokenand new bonds between atoms formed-molecules must collide with one another-sufficient energy must be available to break bonds that need to be broken-orientation of the collision must be appropriate-temperature increases the frequency of collisions and increases the energy in the collision (very important)-arrhenius equation-k = Ae-Ea/RT-lnk = lnA – (Ea/RT)-catalysis: provides an alternative route for the reaction with a lower activation energy; the reaction rate is increased-homogeneous catalysis: the catalyst is in the same phase as the reactants-heterogeneous catalysis: the catalyst is in a different phase than the reactantsLecture 4-reaction mechanisms: the sequence of bond breaking and bond making, the rearrangement of the atoms that is the chemical reaction; the reaction may be a single step or a sequence of steps-unimolecular: involves a single molecule-bimolecular: collision of two molecules-termolecular: simultaneous collision of three molecules-rate equations can be written based upon the stoichiometry of an elementary step-rate equations cannot be written based upon the overall reaction stoichiometry-example: NO2 + CO  NO + CO2step 1: NO2 + NO2  NO + NO3step 2: NO3 + CO  NO2 + CO2overall: NO2 + CO  NO + CO2-the overall equation is the same as the original equation-you can cancel out terms that are on both sides of the arrow to simplify the step 1 and step 2 equations to find the overall equationLecture 5-rate forward = k[reactants]-rate reverse = k’[products]-when equilibrium is reached…-K = k/k’ = [products]/[reactants]-deltaGo determines the value of K-deltaG = deltaGo + RTlnQ-deltaGo = -RTlnK-if K is larger than 1, the products are favored, there are more products than reactants-if K is less than 1, the reactants are favored, there are more reactants than products-things to watch out for with K-K is a unitless value-K depends on how we choose the stoichiometry of the reaction-you can find two different values for K using the same equation if you choose to balance the equation using different coefficients-changing the stoichiometry of the reaction changes K-reaction quotient: describes the composition of a system, not necessarily at equilibrium-Q = [products]/[reactants]-K and Q are expressed in partial pressures-Kp = K(RT)deltan-deltan is the change in the number of moles of gas when reactants are converted to productsLecture 6-review from last lecture-K = [products]/[reactants]-Q = [products]/[reactants] -same equation as for K, but used with a reaction that is not at equilibrium-electrolytes: something that can yield ions-strong electrolytes: complete dissociation-NaCl  Na+ + Cl--weak electrolytes: only dissociate partially-CH3COOH  CH3COO- + H+-extremely weak electrolyte: non-electrolyte-CH4  CH3- + H+ (reaction rarely/never occurs)-most important electrolyte: H2O-H2O  H+ + OH--H2O + H2O  H3O+ + OH--K = [H3O+][OH-]/[H2O]2-Kw=equilibrium constant of water = [H3O+][OH-] = 10-14-with pure water at 25 degrees Celsius… [H3O+] = [OH-]-pH-pH = -log[H3O+]-pOH = -log[OH-]-pKw = pH + pOH-strong acids:-HCl, HBr, HI, HNO3, H2SO4, HClO3, HClO4-strong bases: -NaOH, KOH-weak acids: -HCN, HF, H3PO4, H2CO3-weak bases:-NH3Lecture 7-brief review from last lecture-H2O + H2O  H3O+ + OH--equilibrium heavily favors the left side (reactants)-Kw = [H3O+][OH-] = 10-14 for pure water at 25 degrees Celsius-pH = -log[H3O+]-acids-HA + H2O  A- + H3O+-an acid donates its H+ to water-example: HCl + H2O  Cl- + H3O+-bases-B + H2O  BH+ + OH--a base accepts H+ from water-example: NH3 + H2O  NH4+ + OH--strong acids and bases: they are completely dissociated; equilibrium is pushed 100 percent to the right-weak acids and bases: they are partially dissociated; equilibrium is not pushed 100 percent to the right-percent ionization-% ionization = [A-]/total concentration x 100-the more you dilute a solution, the percent ionization will increaseLecture 8-Arrhenius definition of acids and bases-acid: releases H+-base: releases OH--Bronsted-Lowry definition (best definition)-acid: H+ donor-base: H+ acceptor-acid: HA (acid) + H2O (base)  A- (base) + H3O+ (acid)-there are 2 couples called conjugate acid-base pairs-HA/A- and H3O+/H2O-note: acids are typically listed on the left and bases are typically listed on the right-amphiprotic: molecules that can act as an acid or base, depending on the molecule that they are reacting with-main one: H2O-as an acid: H2O/OH- or -as a base: H3O+/H2O-HSO3--as an acid: HSO3-/SO32- or -as a base: H2SO3/HSO3--NH3-as an acid: NH3/NH2- or -as a base: NH4+/NH3-a strong acid has a weak conjugated base-a weak acid has a strong conjugated base-Ka x Kb = Kw = 10-14-if Ka is large, Kb is small-if Ka is small, Kb is largeLecture 9-review:-acid: H+ donor-HA +