AP Chemistry Exam Review

AP Chemistry Exam Review

+ AP Chemistry Exam Review + Big Idea #3 Chemical Reactions + Changes in matter involve the rearrangement and/or reorganizations of atoms and/or the transfer of electrons. +Types of Chemical Reactions Synthesis A + B AB Source Decomposition AB A + B Video Single Displacement A + BC AC + B Double Displacement AB + CD AD + CB Images from: Wilbraham, Antony C. Pearson Chemistry. Boston, MA: Pearson, 2012 Print.

LO 3.1: Students can translate among macroscopic observations of change, chemical equations, and particle views. +Types of Chemical Reactions C. Pearson Chemistry. Boston, MA: Pearson, 2012. Print. Combustion CxHx + O2 CO2 + H2O Oxidation-Reduction A + + e- A B B + e- Source Acid-Base (Neutralization) HA + BOH H2O + BA Video Precipitation AB (aq) + CD (aq) AD (aq) + CB (s) LO 3.1: Students can translate among macroscopic observations of change, chemical equations, and particle views. Source +Balanced Equations Complete Molecular: Complete Ionic : Net Ionic : AgNO3 (aq) + KCl (aq) AgCl (s) + KNO3 (aq)

Ag+(aq) + NO3-(aq) + K+(aq) + Cl-(aq) AgCl (s) + K+(aq) + NO3(aq) Video Quizlet Ag+(aq) + Cl-(aq) AgCl (s) Spectator ions should not be included in your balanced equations. Remember, the point of a Net Ionic Reaction is to show only those ions that are involved in the reaction. Chemists are able to substitute reactants containing the same species to create the intended product. You only need to memorize that compounds with nitrate, ammonium, halides and alkali LO 3.2: The student can translate an observed chemical change into a metals are soluble. balanced chemical equation and justify the choice of equation type (molecular, ionic, or net ionic) in terms of utility for the given circumstances. +Making Predictions Source Solid copper carbonate is heated strongly: CuCO3 (s) CuOClick (s) + reveals CO2 (g)

answer and explanation. What evidence of a chemical change would be observed with this reaction? Video One would observe a color changeanswer and evolution of a gas Click reveals and explanation. What is the percent yield of CO2 if you had originally heated 10.0g CuCO 3 and captured 3.2g CO2 ? Step 1: Find the Theoretical Yeild Click reveals answer and explanation. 10.0g CuCO3 x(1mol/123.555g) x (1mol CO2 /1 molCuCO 3 ) X 44.01gCO2/mol = 3.562 gCO2 Step 2: Find Percent Yield (3.2 g / 3.562 g) * 100 = 89.8 % 90% with correct sig figs How could you improve your percent yield? and explanation. -reheat the solid, Click to see reveals if there is answer any further mass loss

-make sure you have pure CuCO3 LO 3.3: The student is able to use stoichiometric calculations to predict the results of performing a reaction in the laboratory and/or to analyze deviations from the expected results. +Limiting Reactants D.A. Source Al2S3 + 6 H2O ---> 2Al(OH)3 + 3 H2S 15.00 g aluminum sulfide and 10.00 g water react a) Identify the Limiting Reactant 15.00g Al2S3 x (1mol/ 150.158 g) x (6mol H2O/1mol Al2S3) x (18g/mol H20 ) = 10.782 g H20 needed Click reveals answer and explanation. Video Sim pHet 10g H20 x (1mol/ 18.015 g) x (1 mol Al2S3 / 6mol H2O) x (150.158 g/mol) = 13.892g Al2S3 needed b) What the maximum mass ofmore H 2S than which formed from these reagents? H20 isislimiting, because we need wecan

werebe given Theoretical Yield 10.00 g H20 x (1mol/ 18.015 g) x (3/6) x (34.0809 g/mol ) = 9.459 g H2S produced Click reveals answer and explanation. c) How much excess reactant is left in the container? 15.00 g 13.892 g = 1.11g Al2S3 Click reveals answer and explanation. **Dimensional Analysis is not the only way to solve these problems. You can also use BCA tables (modified ICE charts), which may save time on the exam LO 3.4: The student is able to relate quantities (measured mass of substances, volumes of solutions, or volumes and pressures of gases) to identify stoichiometric relationships for a reaction, including situations involving limiting reactants and situations in which the reaction has not gone to completion. Source +Limiting Reactants BCA Table 15.00 g aluminum sulfide and 10.00 g water react according to the following equation: Al2S3 + 6 H2O ---> 2Al(OH)3 + 3 H2S a) Video Identify the Limiting Reactant 15.00g Al2S3 x (1mol/ 150.158 g) = .100mol 10g H20 x (1mol/ 18.015 g) = .555 Before Complete the table using the molar relationships

Al2S3 .0999 6 H2O 2Al(OH)3 .5551 0 Click reveals Change answer and explanation. -.0925 -.5551 + .1850 Water is the limiting reactant. After .0074 0 .1850 3 H2S 0 + .277 5 .2775 b) What is the maximum mass of H2S which can be formed from these reagents? 0.2775 mol H2S x (34.0809 g/mol ) = 9.459 g H2S produced Click reveals answer and explanation.

c) How much excess reactant is left in the container? .0074mol Al2S3 x 150.158 g/mol = 1.11g Al2S3 Click reveals answer and explanation. LO 3.4: The student is able to relate quantities (measured mass of substances, volumes of solutions, or volumes and pressures of gases) to identify stoichiometric relationships for a reaction, including situations involving limiting reactants and situations in which the reaction has not gone to completion. +Experimental Design Source Synthesis A sample of pure Cu is heated in excess pure oxygen. Design an experiment to determine quantitatively whether the product is CuO or Cu2O. Video Find the mass of the copper. Heat in oxygen to a constant new mass. Subtract to find the mass of oxygen that combined with the copper. Click reveals basic steps Compare the moles of oxygen atoms to the moles of original copper atoms to determine the formula. Decomposition CaCO3(s) CaO(s) + CO2(g) Design a plan to prove experimentally that this reaction illustrates conservation of mass. Find the mass of calcium carbonate and seal it in a rigid container. Evacuate the container of remaining gas. Heat the container and take pressure readings (this will be the pressure exerted by the CO PV=nRT, calculate

2). Using Click reveals basic stepsthe moles of carbon dioxide gas present in the container and compare it to the molar relationships afforded by the balanced chemical equation. LO3.5: The student is able to design a plan in order to collect data on the synthesis or decomposition of a compound to confirm the conservation of matter and the law of definite proportions. Source +Data Analysis When tin is treated with concentrated nitric acid, and the resulting mixture is strongly heated, the only remaining product is an oxide of tin. A student wishes to find out whether it is SnO or SnO2. Mass Mass Mass Mass Mass of pure tin 5.200 grams. of dry crucible 18.650 g of crucible + oxide after first heating after second heating 25. 253 g after third heating 25. 252 g Video 25.500 g How can you use this data, and the law of conservation of mass, to determine the formula of the product? 1) Determine the number of moles of tin. 5.200/118.7 = 0.0438 moles. Sn

2) Subtract the mass of the crucible from the mass after the third heating. 25.252-18.650 = 6.602 g SnOx 3) Subtract the mass of tin from the mass of oxide to get the mass of oxygen. 1.402 grams of oxygen. Click reveals answer and explanation. 6.602-5.200 = 4) Calculate the moles of oxygen atoms, and divide by the moles of tin atoms to get the formula ratio. 1.402 g/16.00 g/mol of atoms = 0.0876 moles. 0.0876/0.0438 = 2.00 The formula must be SnO . 2 LO 3.6: The student is able to use data from synthesis or decomposition of a compound to confirm the conservation of matter and the law of definite +Bronsted-Lowery Acids & Bases Source According to Bronsted-Lowery (B.L.) an acid is a "proton donor" and a base is a "proton acceptor. The proton here is shown as a hydrogen. Video The acids conjugate base is the anion. The bases conjugate acid now has the proton (hydrogen ion). Quizlet

Amphoteric nature of water Water acts as both an acid & a base. H2O H+ + OH2H2O H3O+ + OH- LO 3.7: The student is able to identify compounds as Bronsted-Lowry acids, bases and/or conjugate acid-base pairs, using proton-transfer reactions to justify the identification. Redox Reactions + When an electron is transferred, it is called a redox reaction. When something is reduced, the RED part of redox, it gains electrons. You may have a difficult time with this definition because when something is reduced, it usually means that it is losing something. In this case, it is a reduction in charge. Remember, electrons are negatively charged so if something is being reduced, it's getting more negatively charged by receiving more electrons. The other reaction that is coupled with this is called oxidation--the "OX" part of redox. Whenever something is reduced, the electron it gains has to come from somewhere. The oxidation is the loss of an electron, so if an atom is oxidized it loses its electron to another atom. And these are always coupled reactions. If one molecule is oxidized, another molecule must be reduced and vice versa: the electron must go somewhere. OILRIG LO 3.8: The student is able to identify redox reactions and justify the identification in terms of electron transfer Source Video +Redox Titrations

Source A redox titration (also called an oxidation-reduction titration) can accurately determine the concentration of an unknown analyte by measuring it against a standardized titrant. A common example is the redox titration of a standardized solution of potassium permanganate (KMnO4) against an analyte containing an unknown concentration of iron (II) ions (Fe2+). The balanced reaction in acidic solution is as follows: MnO4- + 5Fe2+ + 8H+ 5Fe3+ + Mn2+ + 4H2O In this case, the use of KMnO4 as a titrant is particularly useful, because it can act as its own indicator; this is due to the fact that the KMnO4 solution is bright purple, while the Fe2+ solution is colorless. It is therefore possible to see when the titration has reached its endpoint, because LO 3.9: The student is able to design and/or the solution will remain slightly purple from the interpret the results of an experiment a redox titration unreactedinvolving KMnO Video

Evidence of Chemical Change + Source Note: it is a common misconception that boiling water makes O2 Video and H22 gas. Notice that Video the water molecule stays Video intact as the water boils. Chemical Changes: Physical Changes: Production of a gas:Covalent bonds are not may produce similar visible evidence (i.e. this physical 2KClO3 (s) + heat broken 2KCl (s) +with 3O2 (g) boiling water creates bubbles, but bonds are not broken and reformed. No new change- only substances are made. Formation of a precipitate: intermolecular AgNO3) (aq) + KCl (aq) attractions AgCl (s)+ 2KNO 3 (aq) (hydrogen bonds) between water Change in color: molecules. Two white solids react to produce a mixture of a yellow and a white solid when shaken forcefully!

Pb(NO3)2 (s) + 2KI (s) PbI2 (s)+ 2KNO3 (s) Production of heat*: 2 Mg (s) + O2 (s) 2MgO (s) + heat *can also Evaluate include thethe absorption of heat of a process as a physical, chemical, or LO 3.10: classification ambiguous change based on both macroscopic observations and the distinction between rearrangement of covalent interactions and noncovalent +Energy Changes Chemical reactions involve the formation of new products Bonds between atoms or ions in the reactants must be BROKEN (the enthalpy of the system is increasing ENDOTHERMIC process) Bonds are then FORMED between atoms or ions to make the producsts of the reaction. (the enthalpy of hte system is decreasing...EXOTHERMIC process) LO 3.11: Source Video Video The student is able to interpret observations regarding macroscopic energy changes associated with a reaction or process to generate a relevant symbolic and/or graphical representation of the energy changes. Source

+ Galvani c Cell Potentia l Video Video Click reveals answer and explanation. LO 3.12: Make qualitative or quantitative predictions about galvanic or electrolytic reactions based on half-cell reactions and potentials and/or +Redox Reactions and Half Cells Source Video Video LO 3.13: The student can analyze data regarding galvanic or electrolytic cells to identify properties of the underlying redox reactions + Big Idea #4 Kinetics +Factors Affecting Reaction Rate Factors that Affect Reaction Rate Collision theory states that reactants must collide in the correct orientation and with enough energy

for the molecules to react; changing the number of collisions will affect the reaction rate Rate is the change in concentration over time [A] / tA] / t Source State of reactants Rate increases as state changes from Video solid gas as increased molecular movement allows for more opportunity for collision Greater surface area of solids will increase rate as more reactant is exposed and able participate in collisions Temperature - more kinetic energy leads to more successful collisions between molecules Concentration more reactants more collisions

Use of a catalyst affect the mechanism of reaction leading to faster rate LO 4.1: The student is able to design and/or interpret the results of an experiment regarding the factors (i.e., temperature, concentration, surface area) that may influence the rate of a reaction. Source +Determining Rate Order Rate law for a reaction has the form: rate = k [A]m[B]n (only reactants are part of the rate law) Exponents (m, n, etc. ) are determined from examining data, not coefficients: forWhen A + [A] B isC When [A] is doubled, the rate do not change, so the reaction is zero order with respect to A Trial Initial [A] (mol/L)

Initial [B] (mol/L) Initial Rate (mol/ (Ls) 1 0.100 0.100 0.002 2 0.200 0.100 0.002 3 0.200 0.200 0.004 When [B] is doubled, the rate doubles, so the reaction is

first order with respect to B Video The overall rate expression for the reaction is rate = k [B] k is the rate constant and is determined Plot to create a straightexperimentally line graph: by plugging in data into the rate expression Zeroth Order [A] / Time First Order ln[A] / Time Second Order The first and second 1/[A] / time order integrated rate laws can be found on the Kinetics section of the AP Equations Sheet LO 4.2: The student is able to analyze concentration vs. time data to determine the rate law for a zeroth-, first-, or second-order reaction. Source +Half-life (First Order) Time needed for the concentration of

reactant to reach half its initial value The first order half life equation is derived from the first order integrated rate law Time to reach half concentration is dependent on k, not initial concentration Half life remains constant in a first order reaction Video Example: when t1/2= 30 sec, the concentration is halved each 30 seconds Initial Conditions seconds (12 molecules) After 30 seconds (6 molecules) After 60 (3 molecules) LO 4.3: The student is able to connect the half-life of a reaction to the rate constant of a firstorder reaction and justify the use of this relation in terms of the reaction being a first-order Source

+Reaction Mechanisms X2 + Y2 X2Y2 rate = k[A] / tX2] A reaction and its experimentally determined rate law are represented above. A chemist proposes two different possible mechanisms for the reaction, which Video are given below. Mechanism 1 X2 2X X + Y2 XY2 X + XY2 X2Y2 Mechanism 2 (slow) (fast) (fast) X2 X + Y2 X + XY X2Y + Y 2X XY + Y X2Y X2Y2 (slow) (fast) (fast) (fast)

Based on the information above, which of the mechanisms is/are consistent with the rate law? List the intermediates in each mechanism: Answer: Both are consistent. In both mechanisms, the molecularity of the slow, rate determining step is consistent with the rate law. Furthermore, the sum of the elementary steps for both mechanisms gives the overall balanced equation for the reaction. Intermediates in mechanism 1: X, XY2. Intermediates in mechanism 2: X, XY, Y, X2Y LO 4.7: Evaluate alternative explanations, as expressed by reaction mechanisms, to determine which are consistent with data regarding the overall rate of a reaction, and data that can be used to infer the presence of a reaction intermediate. + Reaction Mechanisms The rate law for a reaction is found to be Rate = k[A] 2[B]. What is the intermediate? Which of the following mechanisms gives this rate law? I. II. Video Answer: E is the intermediate. Only A + B E (fast) Mechanism II is consistent with the rate E + B C + D (slow) law. Whenever a fast equilibrium step A + B E E + A C III. A + A E E + B C A. B. C. D. Source

I II III Two of these producing an intermediate precedes the slow rate determining step and we want (fast) + D (slow) to remove the intermediate from the rate law, we can solve for the concentration of the intermediate by assuming that an (slow) equilibrium is established in the fast + D (fast) step. The concentration of the intermediate in the rate determining slow step can be replaced with an expression derived from the equilibrium constant [E] =Keq[A][B]. This substitution gives us the desired rate law: rate = k[A]2[B] LO 4.7 The student is able to evaluate alternative explanations, as expressed by reaction mechanisms, to determine which are consistent with data regarding the overall rate of a reaction, and data that can be used to infer the presence of a reaction intermediate. + Reaction Mechanisms and Energy Profiles Practice Problem Draw and label axes for the energy profiles below. Match the curves with the appropriate description. LO 4.7 Cont. 2015 Reaction pathway B Reaction pathway Potential Energy

F Reaction pathway A Reaction pathway Potential Energy Potential Energy C Potential Energy E Potential Energy D Potential Energy A. exothermic reaction with a 2 step D.endothermic reaction with a 2 mechanism where the first step step mechanism where the first is slow. step is slow. B. endothermic reaction with a 2 E. exothermic reaction with a 1 step step mechanism where the mechanism. second step is slow. C. exothermic reaction with a 2 step F. endothermic reaction with a 1 mechanism where the second step mechanism. step is slow.

Reaction pathway Reaction pathway Dena K. Leggett, PhD Advanced Chemistry Teacher Allen High School Copyright +Catalysts Source a. A catalyst can stabilize a transition state, lowering the activation energy. b. A catalyst can participate in the formation of a new reaction intermediate, providing a new reaction pathway. Video The rate of the Haber process for the synthesis of ammonia is increased by the use of a heterogeneous catalyst which provides a lower energy pathway. N2(g) + 2H2 (g) iron-based catalyst + 2NH3 (g) Iron based catalyst LO 4.8 The student can translate among reaction energy profile representations, particulate representations, and symbolic representations (chemical equations) of a chemical reaction occurring in the presence and absence of a +Catalysts catalysts provide alternative mechanisms with lower activation energy a. In acid-base catalysis, a reactant either gains or loses a proton, changing the rate of the reaction. b. In surface catalysis, either a new reaction intermediate is formed or the probability of successful collisions is increased. Source

Video c. In Enzyme catalysis enzymes bind to reactants in a way that lowers the activation energy. Other enzymes react to form new reaction intermediates. Homogeneous catalysis of the decomposition of H2O2 LO 4.9 The student is able to explain changes in reaction rates arising from the use of acid-base catalysts, surface catalysts, or enzyme catalysts, including selecting appropriate mechanisms with or without the catalyst present. + Big Idea #5 Thermochemistry +Bond Energy, Length & Bond strength is determined by the distance between the atoms Strength in a molecule and bond order. Multiple bonds shorten the Source distance & increase the force of attraction between atoms in a molecule. Video Bond Energy is ENDOTHERMIC the energy needed to break the 3 Factors bond. 1) Size: H-Cl

is smaller than H-Br 2) Polarity: HCl is more polar than HC 3) Bond order Lowest PE =Bond (length) C=C Energy involves more LO: 5.1 The student is able to create or use graphical representations in order to connect the eis shorter dependence of potential energy to the distance between atoms and factors, such as bond order than C-C. and polarity, which influence the interaction strength. +Maxwell Boltzmann Distributions Temperature is a measure of the average Kinetic Energy of a sample of substance. Particles with larger mass will have a lower velocity but the same Average KE at the same Temperature.

Kinetic Energy is directly proportional to the temperature of particles in a substance. (if you double the Kelvin Temp you double the KE) The M-B Distribution shows that the distribution of KE becomes greater at higher temperature. The areas under the curve are equal and therefore the number of molecules is constant Increasing Temperature (KE) increases the number of particles with the Activation Energy necessary to react. Activation Energy is not changed with temperature but may be changed with a catalyst. Source Video LO 5.2: The student is able to relate Temp to motions of particles in particulate representations including velocity , and/ or via KE and distributions of KE of the +Thermodynamic vocabulary Universe: The sum of the system and surroundings

System: The species we want to study Surroundings: the environment outside the system Endothermic: Heat flows to the system from the surroundings (surroundings temperature drops-i.e. beaker feels cold) Exothermic: Heat flows from the system to the surroundings. (surroundings temperature rises-i.e. beaker feels hot) Source Video LO 5.3: The student can generate explanations or make predictions about the transfer of thermal energy between systems based on this transfer being due to a kinetic energy transfer between systems arising from molecular collisions . +Heat Transfer Kinetic energy transferred between particles of varying temperature is heat energy. Heat flows from particles of higher energy (hot) to those of lower energy (cold) when particles collide.

When the temperature of both particles are equal the substances are in thermal equilibrium. Not all particles will absorb or release the same amount of heat per gram. Specific Heat Capacity is a measure of the amount of heat energy in Joules that is absorbed to raise the temperature of 1 gram of a substance by 1 degree Kelvin. Heat transfer can be measured q=mcpT Source Video LO 5.3: The student can generate explanations or make predictions about the transfer of thermal energy between systems based on this transfer being due to a kinetic energy transfer between systems arising from molecular collisions. +Conservation of Energy 1st Law of Thermodynamics: Energy is conserved

Temperature is a measure of the average Kinetic energy of particles in a substance Energy can be transferred as Work or Heat E = q+w Work = -PV (this is the work a gas does on the surroundings i:e the volume expanding a piston) a gas does no work in a vacuum. Source Video LO 5.4: The student is able to use conservation of energy to relate the magnitude of the energy changes occurring in two or more interacting systems, including identification of the systems, the type (heat vs. work), or the direction of the energy flow. Source +Conservation of Energy Video Expansion/ Compression of a gas

Volume increases, work is done by the gas Volume decreases, work is done on the gas LO 5.4: The student is able to use conservation of energy to relate the magnitude of the energy changes occurring in two or more interacting systems, including identification of the systems, the type (heat vs. work), or the direction of the energy flow. + Conservation of Energy when Mixing Energy is transferred between systems in contact with one another Energy lost by one system is gained by the other so that total energy is always conserved. -Q lost by system = +Q gained by surroundings

Source Video For example : When room temperature water T1 (system) is mixed with cold water T2 (surroundings), the final temperature T3 will be inbetween. Q1 + Q 2student = 0 and energy is conservation conserved of energy to relate the LO 5.5: The is able to use magnitudes of the energy changes when two non reacting substances are mixed or brought into contact with one another. Calorimetry: an experimental technique used to determine +the heat transferred in a chemical system. System can be a Source chemical reaction or physical process. Can use Calorimetry to solve for Heat Capacity of a calorimeter (C),, specific heat of a substance, (c), and Hvap, fus, Hrxn.Hvap, Hvap, fus, Hrxn.fus, Hvap, fus, Hrxn.Hrxn.

The data handling and math: Law of Conservation of Energy: Q system + Qsurroundings = 0 Qsystem = - Qsurroundings where System = reaction, Surroundings = calorimeter SO: Q rxn = - Q calorimeter Heat Transfer due to Temperature Change in the Calorimeter: Q= CHvap, fus, Hrxn.T, or Q= mc Hvap, fus, Hrxn.T where Q in J, C in J/K, m in g, c in J/g-K, Hvap, fus, Hrxn.T in K Q rxn = - Q calorimeter = - CT if the calorimeter Heat Capacity is Known, or can be determined. Q rxn = - Q calorimeter = - mcT for reactions in solution.

When calculating H, must take into account the H, must take into account the mass of reactant that caused Q rxn. Video Exampl e problem in video LO 5.5: The student is able to use conservation of energy to relate the magnitudes of the energy changes when two non-reacting substances are brought into contact with one another. Chemical Systems undergo 3 main processes that change their energy: heating/cooling, phase transitions, and chemical reactions. + Source 1. Heat Transfer due to Temperature Change: (kJ) Q= mcTT m= mass (g), cT= specTificT heat cTapacTity (J/g-C), T= Temp. cThange in C Q is + for Heating, - for cTooling 2. Heat Transfer due to Phase Change: (kJ/mol ) Video Q= H phase cThange Q phase cThange = + for H fusion, H vaporizing, H subliming, - for H freezing, H cTondensing, H deposition 3. Q for a chemical reaction at constant pressure = H, must take into account the H rxn When calculating H rxn from Q, remember H rxn must H rxn from Q, remember H rxn from Q, remember H rxn must H rxn must agree with the stoichiometric coefficients in the reaction.

Units of H rxn from Q, remember H rxn must H rxn are kJ/mol rxn. 4. When a gas expands or contracts in a chemical reaction, energy is transferred in the form of PressureVolume work. W= -PV (l-atm) V (l-atm) V (l-atm) Gas Expands Does work on surroundings (system loses energy) Gas ContracTts Work done on the gas (system gains energy) No cThange in volume, no work done. LO 5.6: The student is able to use calculations or estimations to relate energy changes associated with heating/cooling a substance to the heat capacity, relate the energy changes associated with a phase transition to the enthalpy of fusion/vaporization, relate energy changes associated with a chemical reaction to the enthalpy of the reaction, and relate the energy changes to PV work. Calorimetry: an experimental technique used to determine the heat transferred in a chemical system. System can be a chemical reaction or physical process. + Source Video LO 5.7 The student is able to design and/or interpret the results of an experiment in which calorimetry is used to determine the change in enthalpy of a chemical process. (heating/cooling, phase transition, or chemical reaction) at constant pressure . The net energy change during a reaction is the sum of the energy required to break the reactant bonds and the energy released in forming the product bonds. The net energy change may be positive for endothermic reactions where energy is required, or negative for exothermic reactions where energy is released. + Any bond that cTan be formed cTan be broken. These

procTesses are in opposition. (their enthalpy cThanges are equal in magnitude, opposite sign) H bonds breaking ENDOTHERMIC (+) H bonds forming EXOTHERMIC (-) To find Hrxn, apply Hesss Law: Hrxn = H bonds breaking (+) + H bonds forming (-) To calculate or estimate V (l-atm) Hrxn from Bond Energy: 1. Draw the Lewis StrucTture. Dont forget about double and triple bonds! 2. Add up H bonds breaking. Its + (kJ) 3. Add up H bonds forming. Its - (kJ). 4. Add the two terms. Units are kJ/mol rxn. To calculate V (l-atm) Hrxn from a table of standard

enthalpies of formation: V (l-atm) Hrxn = Hf producTts - Hf reacTtants . Source Video If a reacTtion is EXOTHERMIC, there is a net release in energy, sincTe weaker bonds break and stronger bonds form. ProducTt has higher kineticT energy and lower potential energy than reacTtant. If a reacTtion is ENDOTHERMIC, there is a net absorption of energy, sincTe stronger bonds break, and weaker bonds form. ProducTt has lower kineticT energy, and higher potential energy than reacTtant . LO 5.8: The student is able to draw qualitative and quantitative connections between the reaction enthalpy and the energies involved in the breaking and formation of chemical bonds. Source +Electrostatic forces exist between molecules as well as between atoms or ions, and breaking these intermolecular interactions requires energy. The Stronger the IMF the more energy required to break it, the Higher the Boiling Point, the Lower the Vapor Pressure. Intermolecular Forces Listed from weakest to strongest. Thus the boiling points and vapor pressure of molecTular substancTes cTan be ordered based on IMF strength: Video 1. Dispersion (Induced Dipole- Induced Dipole): Caused by distortion of elecTtron cTloud. The larger the elecTtron cTloud, and the more surfacTe area, the more polarizable the cTloud, the stronger the dispersion forcTe. Thus the boiling point trend in halogens is I2 >Br2>Cl2> F2 and n-butane (30.2 C) has a

higher boiling point than isobutane (-11 C). All substances have dispersion forces, as all electron clouds distort. Nonpolar molecTules and atoms have only dispersion forcTes, as they have no permanent dipoles. 2. Dipole- Induced Dipole: OcTcTurs between a polar molecTule (HCl) and a nonpolar molecTule. (Cl 2) The nonpolar molecTules cTloud distorts when affecTted by a dipole. 3. Dipole-Dipole: OcTcTurs between 2 polar molecTules. (HCl-HCl) 4. Hydrogen Bond: An extreme cTase of Dipole Dipole. OcTcTurs between molecTules cTontaining a H cTovalently bonded to F,O, or N. The bond ocTcTurs between the lone pair of F, O, or N, and the H whicTh is attacThed to one of those elements. Weaker Weaker IMF, IMF, Lower Lower Boiling, Boiling, Higher Higher Vapor Vapor Pressure Pressure Stronger Stronger IMF, IMF, Higher Boiling, Higher Boiling, Lower

Lower Vapor Vapor Pressure Pressure LO 5.9: Make claims and/or predictions regarding relative magnitudes of the forces acting within collections of interacting molecules based on the distribution of electrons within the molecules and the types of intermolecular forces through which the + Interstates- Between States IMF- Between Molecules Inter vs Intra Chemical vs. Source Source Source Source Video Physical Chemical vs. Physical Changes A physical change doesnt produce a new substance. Phase changes are the most common. It involves IMF changes. A chemical change produces new substances. Bonds are broken and new bonds are formed! The Intramolecular forces are changed.

Strong IMF= High BP, High MP, High viscosity, high surface tension, low vapor pressure! LO 5.10: The student can support the claim about whether a process is a chemical or physical change (or may be classified as both) based on whether the process involves changes in intramolecular versus intermolecular +IMF and Biological/Large Molecules Sourc e Sourc e Sourc e Video LO 5.11: The student is able to identify the noncovalent interactions within and between large molecules, and/or connect the shape and function of the large molecule to the presence and magnitude of these interactions. EntropyEmbrace the Chaos! + Entropy Changes that result in a + S: Increasing moles Increasing temperature Increasing volume Solid to liquid to gas Forming more complicated molecules. (More moles of electrons) Source Source

Video LO 5.12: The student is able to use representations and models to predict the sign and relative magnitude of the entropy change associated with chemical or physical processes. Predicting How Reactions Will + Go Video # 1 Video # 2 Source Entropy is typically given in J/K so you MUST convert to kJ! LO 5.13: The student is able to predict whether or not a physical or chemical process is thermodynamically favored by determination of (either quantitatively or qualitatively) the signs of both delta H and delta S, and calculation or estimation of delta G when needed. + Source Source Video Video Video How can I calculate G?G? GG

LO 5.14: Determine whether a chemical or physical process is thermodynamically favorable by calculating the change in standard Gibbs Coupling + Reactions erm h t o Ex i on t c a Re ic Source Video #1 Choo Choo ic m r e oth End tion c Rea LO: 5.15 The student is able to explain the application the coupling of favorable with unfavorable reactions to cause processes that are not

+ Coupled Reactions and LeChatelier Source Video LO: 5.16 The student can use LeChatelier's principle to make qualitative predictions for systems in which coupled reactions that share a common intermediate drive formation of a product. + COUPLING OF REACTIONS Reactions with known values can be COUPLED to Video produce a new reaction. Coupled Reactions and K Source Source When we couple reactions (or half-reactions), we Source may multiple by a factor or reverse the reaction before we couple or add them together. Manipulation Equilibrium constants (K) Free energy,

enthalpy or entropy Standard Reduction potential (Eo) Multiply by factor Raise K to power of factor Multiply by factor NOTHING! Flip/reverse rxn Inverse of K Change sign Change sign Add Reactions Multiply values Add Values Add values LO 5.17: The student can make quantitative predictions for systems involving coupled reactions that share a common intermediate, based on the equilibrium constant for the combined reaction.

+ K? Is it thermo, kinetics, or V (l-atm) Go and the EquilibriumFAVORABLE Constant THERMODYNAMICALLY The free energy at non-standard, non-equilibrium states is given by: Gibbs Equilibrium Description Standard Kinetics tells us how FAST a reaction will constant Free Energy o Video proceed. V (l-atm) G = V (l-atm) G + RTln(Q) Source Equilibrium tells us how FAR a reaction At Equilibrium, V (l-atm) G = 0 and Q becomes K Source o Neither reactant

nor G?G = 0 K=1 proceeds. V (l-atm) product Go = -RTln(K) favored Thermodynamics tells us whether or not the Thermodynamically (aka spontaneous) means is FAVORABLE at a given G?Goa< 0 K > 1 reactionfavorable Product favored reaction is PV (l-atm) RODUCTtemperature. favored at a given temperature. (thermodynamically favorable) K > 1, ln(K)V (l-atm) is positive, V (l-atm) Go is negative H A reaction that is thermodynamically UNFAVORABLE (aka non-spontaneous) reactions o Thermodynamically G?G > 0 will K < 1

Reactant favorable will form more products at run in reverse when set up with standard conditions (1M/1atm of ALL (thermodynamically thefavored reaction may be so SLOW reactantsequilibrium and products) BUT BUT can be made to proceed forward under different o that few products form in a reasonable As K decreases, V (l-atm) conditions. G becomes positiveunfavorable) " more amount of time. L 5.18: Explain why a thermodynamically favored chemical reaction may not produce large amounts of product (based on consideration of both initial conditions and kinetic effects), or why a thermodynamically unfavored

chemical reaction can produce large amounts of product for certain sets of

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