Chem 122 First Exam Study Guide
Chem 122 First Exam Study Guide Chem 122
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Popular in Chemistry
This 14 page Study Guide was uploaded by Alisha Notetaker on Monday February 8, 2016. The Study Guide belongs to Chem 122 at University of North Dakota taught by Dr. Harmon Abrahamson in Spring 2016. Since its upload, it has received 92 views. For similar materials see General Chemistry 2 in Chemistry at University of North Dakota.
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Date Created: 02/08/16
Ch. 10 Intermolecular forces: attractive forces between molecules -determined by differences in electronegativity -types: 1. ion-dipole: between an ion and the partial charge on the end of a polar molecule 2. dipole-dipole: when polar molecules are close together 3. hydrogen bonding: hydrogen must be directly bonded to a small atom of high electronegativity (N,O,F) -have higher boiling points -more energy is needed to break the intermolecular forces 4. London dispersion forces: very weak attraction formed by the uneven electron cloud in a molecule increase with polarizability increase with both size and molecular weight Hydrogen bond > Ion-dipole > Dipole-dipole > London dispersion forces Electronegativity: atom’s ability to attract electrons to itself when it’s bonded to another atom -higher electronegativity=higher ability to attract electrons -trend on periodic table: increases going across to the right, and decreases going down Bond polarity: measure of how unevenly the electrons in a bond are shared and is indicated by the difference between the electronegativity of each atom involved -larger difference=polar=contains dipole ionic dipole exists when two equal but opposite charges are separated -smaller difference=more covalent bond character Covalent bond: electrons are shared Ionic bond: electrons are transferred Electronegativity differences can determine type of bond: Nonpolar covalent < 0.5 Polar covalent 0.5 - 2.0 Ionic > 2.0 Nonpolar: when the bond dipoles are equal in magnitude and opposite in direction, then there is no molecular dipole Polar: when bond dipoles are present, but are not equal, then there is a molecular dipole Electric fields: can orient molecules based upon their molecular dipoles and partial charges (+) partial charge will align toward negative end (-) partial charge will align toward positive end Nonspontaneous: reaction does not occur Spontaneous: reaction does occur -irreversible -a process in which the energy content of the system decreases tends to occur spontaneously Equations: -Gibbs free energy: −∆G=∆ H−T ∆S Exothermic: - + temp increases Endothermic: + + temp decreases -ΔH is the enthalpy -T is the temp in Kelvin -ΔS is the change in entropy -spontaneous: decrease in H (-ΔH) Increase in S (+ΔS) -nonspontaneous: increase in H (+ΔH) Decrease in S (-ΔS) Ch. 11 Solution: when one substance disperses uniformly throughout another what forms is a homogenous mixture: solvent and solute Solvent: substance present Solute: substance dissolved in solvent Solvent + Solute Solution ∆ H soln∆H solvent soluteH mixture ∆ H ¿ -the energy gained ( soln needs to be greater than the energy required in order for the intermolecular forces to be broken in the solvent and solute -if ∆ H mix∆H solv∆H sol , then −∆ H soln favorable ∆ H <∆H +∆H +∆ H -if mix solv sol , then soln unfavorable Equilibrium: the forward reaction rate is equal to the reverse reaction rate Saturated: a solution in equilibrium with undissolved solute Solubility: amount of solute in a saturated solution -supersaturated: excess solid present -saturated: equilibrium -unsaturated: no solid present the stronger the interactions are between solute and solvent molecules, the greater the solubility LIKE DISSOLVE LIKE attractions that can lead to h-bonding will increase the solubility of a solute the presence of a carboxyl group will increase the strength of h-bonding polar –OH groups readily form h-bonds in water as length of carbon chain increases, the solubility decreases Not energetically favorable: breaking a strong interaction (h-bonding) to form a weak interaction (London dispersion) Miscible: when two solvents mix in all proportions to be miscible in water, there must be strong enough interactions to overcome the nonpolar portions of the molecule Endothermic: energy entering a system and have a positive enthalpy Exothermic: energy released from the system and have a negative enthalpy Hydration: when water molecules cluster around solute particles Equations: -Henry’s Law: Sg=kP g Sg - is the solubility in mol/L -k is Henry’s constant in mol/L*atm - P g is the pressure Example: How many grams of CO2 dissolve in 250mL of bottled water under 1900mm Hg at 25˚C? 10−2 (k=3.1* mol/L*atm) S gkP g −2 mol 1atm mol 3.1* 10 (1900 * ) = 0.0775 L∗atm 760mm Hg L −2molCO2 44.0gCO2 7.75* 10 * 0.250L * = 0.8525g CO2 L 1molCO2 Expressing Solution Concentration: massof concentration 1. % by mass = totalmassof solution100 2. % by volume molesof solute 3. Molarity (M) = Litersof solution -dependent on temp molesof solute 4. Molality (m) = kgof solvent -independent of temp molesof component 5. Mole fraction (X) = totalmolesof allcomponents wieght∨volumeof solute ∗10 6 6. Parts per million (ppm) = weight∨volumeof solution 1gsolute 1mgsolute -1 ppm = 6 = 10 gsolution 1 L solution ¿ 1gsolute 7. Parts per billion (ppb):1ppb 10 solution Example: 86% by mass of nitric acid. What is the Molarity and Molality? (Density=1.425 g/mol) (g/mol = g/mL) 86 g HNO3 Mass % = 86g HNO3 + 14g H2O 14g H20 * 100 g solution 1kg 1000 g=0.0140kg 1 mol 86g HNO3 * = 1.365 mol HNO3 100g soln * 63.02g 1mL ∗1L 1.425g 1000mL =0.07042L 1.365 1.365 Molarity = 0.07042=19.4 M Molality = 0.0140=97.5 m Colligative Properties: 1. Vapor pressure lowering: P =X P˚ X = molesof A Raoult’s Law: A A A A moles A+molesB P - A is the vapor pressure of solution - XA is the mole factor of solvent P˚A - is the vapor pressure of pure solvent -as concentration increases, vapor pressure decreases -if contains volatile solutes: Ptotal AP =B P˚ AX PA B B Example 1: 0.340 mol nonvolatile dissolved in 3.70 mol H20. ( P˚ Af wateris23.8 ) 3.70mol =0.916 Mole fraction = 4.04mol Vapor pressure = 0.916 * 23.8 = 21.8 torr P˚ A Example 2: 1.80 mol Cyclohexane ( = 97.6 torr) and 2.00 mol Acetone ( P˚ B = 229.5 torr) 1.8mol 2.0 mol X A =0.474 XB= =0.526 3.8mol 3.8mol P = 0.474∗97.6 )+(0.526∗229.5 =166.98torr total ∆ T bk b 2. Boiling point elevation: -k is the molal boiling point constant -m is the molality k Example: 50% by mass C2H6O2 ( b =0.512˚C/m) 1kgC2H 6O2 ∗1000g 1kgH 20 m = 1kg ∗1mol =16.1mol/kg 62.08g ∆ T b = 0.512 * 16.1 = 8.24˚C boiling point = 100 + 8.24 = 108.24˚C ∆ T =k m 3. Freezing point elevation: f f -k is the molal freezing point constant -m is the molality k Example: 50% by mass C2H6O2 ( f =1.86 ˚C/m) 1kgC2H 6O2 1kgH 20 ∗1000g ∗1mol m = 1kg 62.08g =16.1mol/kg ∆ Tf = 1.86 * 16.1 = 29.95˚C freezing point = 0 - 29.95 = -29.95˚C molesof particles∈solution Van’t Huff Factor: i = molesof solutedissolved -used in both boiling and freezing point if the solute is ionic Example: How many grams of KNO3 have to be added to 275mL H2O in order for it to freeze at -14.5˚C? ( kf =1.86˚C/m) ∆T =k ∗m∗i f f 2mol solute particles i = 1 molsolute dissolved2 14.5 ∆ Tf = 0 – 14.5 = -14.5˚C m = 1.86∗2 =3.9mol/kg 1kg 275mL H2O = 275g * 1000 g=0.275kg H 2O x mol=1.0725mol 3.90 mol/kg = 1.0725 mol = 0.275kg 101.1 g =108g KNO3 1mol m 4. Osmosis: π= RT π=MRT v - π is the pressure required to prevent osmosis -M is the molarity -R is 0.08206 L*atm / k*mol -T is the temp Ch. 12 Reaction rates: the change in concentration of a reactant or product per unit of time -measured in M/sec A + 2B C + 2D -reactant concentration will decrease, product concentration will increase −∆[A] −1 ∆[B] ∆[C] 1 ∆[D] = = = Rate= ∆t 2 ∆ t ∆t 2 ∆t Instantaneous rate: rate at a particular time. Obtained from the slope at a point Initial rate: rate at t=0 Rate law: expresses how the rate depends on the concentration A + B C + D k[A] [B]y Rate law= -x and y are the order of reaction - x and y can only be determined by experimental data, not the stoichiometric coefficient -orders: 0 order means changing the concentration will not change the rate 1 order means that if concentration doubles, the rate doubles. If concentration triples, then the rate triples. And so on. 2 order means that if the concentration doubles, then the rate quadruples. If the concentration triples, then the rate increases by a factor of 9. And so on. x2 -units of k: −1 1 order: s −1 −1 2ndorder : M s rd −2 −1 3 order: M s [N2O5] Initial rate Example 1: 2 N2O5 4 NO2 + O2 0.9 M −4 5.4* 10 M/ x Rate=k [N 2O5] sec 0.45 M 10−4 2.7* x −4 k[N 2O5] =5.4∗10 M/sec k[N 2O5] =2.7∗10 −4 −4 0.9 5.4∗10 x 0.45= −4 2 =2 xln2 = ln 2 x=1 2.7∗10 X=1 is a first order reaction 1 −4 −4 −1 rate=k [N 2O5] 5.4* 10 =k[0.9] k = 6.0 * 10 s 6.0∗10 s [N 2O5] 1 Rate law = Ex [ [ Initial p −¿ rate −¿ ¿ 2−¿ ¿ −¿+I 3 S2O 8 I ¿ Example 2: −¿→2SO 4 2−¿+3I ¿ ] ¿ 1 0.15M 0.21 1.14 S2O 8 M M/s 2 0.45M 0.21 3.41 M M/s 3 0.45M 0.12 1.95 2−¿ M M/s S2O 8¿ ¿ −¿¿ Rate= I ¿ ¿ ¿ k¿ 0.45M 0.15M ¿ ¿ 0.21M 3.41M/s 3 =3 0.21M = 1.14M/s xln3 = ln 3 x=1 ¿ ¿ exp2 =¿ exp1 0.45M 0.45M ¿ ¿ 0.21M y 1.75 =1.75 yln1.75 = ln1.75 x=1 0.12M ¿ ¿ exp2 exp3 =¿ 2−¿ S O ¿ 2 8 ¿ −¿ k[0.15] [0.21] =1.14M/s 1.14 s−1 I k = 0.15∗0.21 = 36.2 ¿ ¿ ¿ k¿ 2−¿ ¿ S2O 8 ¿ −¿ Rate law = 36.2 M/s I¿ ¿ ¿ ¿ ¿ [A] Half-life: ln 0=kt 0.693 = kt1 [A] 2 Example: certain 1 order has half-life of 40 min. What is rate constant? How much time is required for the reaction to be 75% complete? kt −1 0.693 = 2 0.693 = k * 40 k = 0.0173 min [A]0 A ln =0.0173t At t=0, 100% 75% complete [¿¿t] = 25% 0.25 [A] ¿ ln 1 =0.0173t ln4 0.25 t = 0.0173 t = 80 minutes Integrated rate laws: relates concentration to reaction time −kt [A] 0 For 1 order: [A] = [A]0e ln =kt [A] -rearranged in y = mx + b form: Order Integrated rate law Graph slope 0 [A] [A] v. t -k [A] = -kt + 0 1 Ln [A] = -kt + ln Ln [A] v. t -k [A]0 2 1 1 1 k =kt+ v. t ln[A] [A]0 [A] Example 1: 0 order, [A] = 7.00 * −2 −2 10 M after 125 sec. and [A] = 1.00 * 10 M after 300 sec. What is k? What rise was the initial concentration? K = m = run Rise = (1.00 * 10 ¿ – (7.00 * 10−2 ) = -0.06 M Run= 300 – 125 = 175 sec 0.06M −4 -k = -m = - ( ¿ = 3.43 * 10 M/sec 175s [A] e−kt 10−2 10−4 [A]0 [A] = 0 7.00 * M = -3.43* (125) + [A]0 =0.113 M st −2 −3 Example 2: 1 order, [A] = 8.00 * 10 M after 50.0 sec. and [A] = 7.50 * 10 M after 65.0 sec. What is k? Make a data table: rise = (-4.893) – (-2.501) = -2.392 run = 65 – 50 = [A] Ln t 15 sec [A] 8.00 * - 5 −2.392 −1 10−2 2.50 0 -k = -m = - ( 15 ) = 0.159 s 1 7.50 * - 6 nd 10−3 4.89 5 Example 3: 2 order, [A] = 0.860M after 285 sec. and [A] = 3 −2 5.10 * 10 M after 750 sec. What is k? Make a data table: 1 1 [A] 1 t Rise = ( 5.10∗10 −2 ) – ( 0.860 ) = 18.45 [A] −1 0.860 285 M 1 0.860 Run = 750 – 285 – 465 sec 5.10 * 1 750 −2 −2 10 5.10∗10 18.45 −2 −1 −1 K = m = =3.96∗10 M s 465 The Collision Model: molecules must collide to react, and the number of collisions is proportional to concentration and temp -increased concentration = increased molecules = increased probability of collision -increased temp = increased velocity = increased frequency of collisions For a reaction to occur, the collision energy must be greater than the activation energy Collision E > E A If temp increases, there is a much larger fraction of molecules that could react The Orientation Factor: the probability of a particular high energy collision resulting in a reaction depends on the molecular orientation during the reaction Example: 2BrNO Br2 + 2NO Transition state theory: focuses on hypothesized intermediate species called an activated complex which forms during an energetic collision. Exists very briefly and dissociates either back to reactants or to products EA Example 1: If = 171 kJ/mol for forward reaction and ΔH = -13 kJ/mol E What is A for the reverse reaction? For exothermic reaction, the ΔH = (R – P). E A for the forward reaction = (intermediate – R). EA for the reverse reaction = ( EA of forward + ΔH). Example 2: If the E A for A B is 150 kJ/mol. And EA for B A is 100 kJ/mol. Then what is ΔH? For endothermic reaction, ΔH = (B – A). E A for the forward reaction = (intermediate – A). E A for the reverses reaction = (intermediate – B). −EA −E 1 RT A( )+ln A Arrhenius Equation: k=Ae ln k = R T slope = −E A R -A is the frequency factor -R is the gas constant (8.314 J/mol) If we have two data points, then the equation will be: k E ln 2= A( 1 − 1 ) k R T T 1 1 2 Example: Calculate rate constant at 25˚C for hydrolysis of sucrose given that it’s equal 1∗10 M s −1 −1 EA to at 37˚C. R = 8.314 J/mol*K = 108 kJ/mol 108000 J/mol (25˚C = 308 K) (27˚C = 310 K) −3 1∗103 1080001 1 ln 1∗10 = 108000 ( 1 − 1 ) ln k 8.31308−310 k 8.314 308 310 e 1 =e 1 1∗10 −3 −5 1∗10 −3 0.272 = e12990∗(2.09∗10 ) =e k1 k1 −3 1∗10 −4 −1 −1 k 1 k1=7.63∗10 M s 1.31 Reaction Mechanisms: many times, a reaction doesn’t occur in one step. And it may not show us the other steps involved. -those hidden steps are known as the elementary steps -together, the steps make up a mechanism -the slowest step is known as the rate determining step (rds) Example: NO2 + CO NO + CO2 rate law = k[NO2] 2 Takes 2 steps: 1. NO2 + NO2 NO3 + CO2 (based on rate law) (rds) 2. NO3 + CO CO2 + NO2 NO2 + CO NO + CO2 -the parts crossed out are known as the intermediates Molecularity: how molecules collide in a reaction -biomolecular: collision of 2 molecules -unimolecular: involves only one molecule in the reactants -termolecular: very rare. Collision of 3 molecules Order and rate law: -overall order of elementary steps depends on molecularity - rate law can be determined by the number of each type of molecule in the reactants of the elementary steps Example: 1. 3X E + F 2. E + 2M F + N 3X + 2M 2F + N -intermediate: E -rate law for step 1 = k [X]3 k[E] [M] 2 -rate law for step 2 = -Steady state approximation: solve for concentration of intermediate by assuming that an equilibrium is established in the fast step which precedes the slowest step Example: 1. Cl2↔2Cl (fast) 2. Cl+CHCl ↔H3l+CCl 3 (slow) Cl+CCl ↔CCl 3. 3 4 (fast) Cl2+CHCl ↔3HCl+CCl 4 CHCl Predicted rate law = [¿¿3] ???? (but Cl is an intermediate) k[Cl]¿ Cl2↔2Cl Cl k [Cl ] Rate of 2 decomposition: 1 2 (put them equal to each other because it’s in equilibrium) Cl 2 k−1Cl] 2 Rate of formation: (solve for Cl) Cl [¿¿2] k1[Cl2] k−1Cl] 2 [Cl]2 = = k1 (take square root of k ¿ −1 both sides) Cl k 1 [ ]= 1[Cl 2 [Cl] = k 2 (plug into rate law) √ k−1 [¿¿2] ¿ Cl CHCl [¿ ¿3] Rate law = k 1 [¿¿2] ¿ ¿ Catalyst: speeds up a reaction by reducing the activation energy for both forward and reverse reactions
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