Chem 102 Final Exam PDF

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Review guide for cumulative final exam...

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CHEM 102 FINAL EXAM -

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Chapter 9: Gases - Gas Pressure - Ideal Gas Law - Stoichiometry - Effusion/Diffusion - Kinetic molecular theory - Non-ideal gas behavior Chapter 10: Liquids and Solids - IMFs - Properties of liquids - Phase transitions - Phase diagrams - Solid state of matter Chapter 11: Solutions - Dissolution process - Electrolytes - Solubility - Colligative properties - Colloids (is this on the exam??) Chapter 12: Kinetics - Chemical reaction rates - Factors that affect rxn rates - Rate laws - Integrated rate laws - Collision theory - Mechanisms - Catalysis Chapter 13: Equilibrium - Chemical equilibria - Equilibrium constants - Le Chatelier's principle - Equilibrium calculations Chapter 14: Acids and bases - Bronsted lowry - Ph and pOH - Relative strengths of acids and bases - Hydrolysis of salt solutions - Polyprotic acids - Buffers - Titrations Chapter 15: Solubility - Precipitation and dissolution - Lewis acids and bases - Multiple equilibria Chapter 16: Thermodynamics

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- Spontaneous rxns - Entropy - 2nd and 3rd laws - Gibbs free energy Chapter 17: Electrochemistry - Redox rxns - Galvanic cells - Standard reduction potential - Nernst equation - Batteries and fuel cells, Corrosion, electrolysis (also are these on the exam???)

Final Exam Equation Sheet R = 8.3145 J/K mol

R = 0.08206 L atm/mol K

1 J = 1 kg m2/s2

[A]t = -kt + [A]0 t1/2 = 0.693/k

t1/2 = [A]0/2k

t1/2 = 1/[k [A]0]

k = A e-Ea/RT

ln K = ln A – Ea/RT

∆G˚ = ∆H˚ - T ∆S˚

ΔG = ΔG˚ + RT ln Q

∆G = ∆H - T ∆S

∆S˚ = qrev / T

∆G˚ = -RT ln K ∆S˚univ = ∆S˚sys + ∆S˚surr = 0

∆S˚univ = ∆S˚sys + ∆S˚surr > 0

Kw = 1.0 x 10-14 (at 25 °C)

PV = nRT

urms = √(3RT/MM)

PA = XA PTotal

Cg = kHPg

Π=MRTi

Eo = Eored(cathode) – Eored(anode) → Ecathode – Eanode = Ered + Eox F = 9.6485 x 104 C/mol e-

At 298K: ∆G = −nFE

1 J = 1 C 1V

CHEM 102 Final Exam Concept Review IMFs - Distinguish the relative strengths of different types of intermolecular forces 1. London Dispersion forces: fourth strongest (weakest) 2. Dipole-dipole interactions: third strongest 3. Hydrogen bonding: second strongest 4. Ion-Ion: strongest ● ●

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Unbranched molecules have stronger IMFs than branched molecules London dispersion is in all particles; bigger molar mass — larger dispersion forces because of the number of electrons with the formation of the temporary dipole; have more points of contact for electron buildup; larger the molar mass the greater the tendency to distort the electron cloud (polarizability) Equilibrium: ICE tables ○ At equilibrium, the rates of the forward and reverse reactions are always the same ○ But the concentrations of products and reactants are not necessarily the same

- Determine the factors that affect the strength of London dispersion forces London dispersion forces are temporary and present in all molecules. Larger atoms are stronger because they have a larger electron cloud. They occur when atoms collide. Compact atoms are weaker than atoms with a lot of surface area. - Determine whether a molecule has a permanent dipole and the relative strengths of dipoles Dipole attractions occur in molecules with a permanent dipole. They are polar molecules that are attracted to each other by electrostatic interactions. - Be familiar with what compounds will hydrogen bond and how many hydrogen bonds can form Hydrogen bonds can only bond between a hydrogen and a highly electronegative ion such as oxygen, nitrogen, and fluorine, F.orce O.f N.ature - Sort a set of compounds by increasing boiling points Boiling point is determined by the polarity of the molecule. The more polar the molecule, the higher its boiling point. (Water has a high boiling point) Strong attractions also cause high boiling points. -What determines a molecule’s polarity? In order to be polar, a molecule must: - Have at least one polar covalent bond - Have a molecular structure so that the sum of the vectors doesn’t cancel -If you have stronger IMFs, will you have a higher or lower vapor pressure? - Lower (harder to disrupt IMFs to convert molecules from liquid to vapor, therefore you have a lower pressure of vapor → said molecules will still be liquid) Kinetics - Define the rate of reaction in terms of concentrations of reactants/products over time The reaction rate is the change in concentration of a reactant or product with time. There are three different types of rates: 1. Average reaction rate- The reaction rate of the overall reaction.

2. Instantaneous rate- The slope of a line tangent to the curve o any point of a concentration vs. time reaction plot. 3. Initial rate- The rate of the reaction at the beginning of the reaction. - Relate the rate of appearance of products to the rate of disappearance of reactants using stoichiometry 2 H2O2(aq) ⟶ 2 H2O(l) + O2(g) ●

In this reaction, the stoichiometry of H2O2 is equal to that of water; they interact in a 2:2 (or 1:1) ratio. ● Therefore, Reaction Rate = ● The rate of disappearance of hydrogen peroxide is equal to the rate of appearance of water. - Explain why the initial rate of reaction is the best indicator of rate defined by instantaneous rates It is the best indicator of the rate because all reactions slow down over time. - Explain how physical state, temperature, concentrations, and presence of a catalyst affect the rate of a reaction 1. Physical state: a. In order to react, molecules must come in contact with each other. b. The more homogenous the mixture of reactants, the faster the molecules can react. c. The phase can affect rate. Gasses react faster than liquids and even faster than solids. Sometimes aqueous solutions react even faster. 2. Temperature: a. At higher temperatures, reactant molecules have more kinetic energy and therefore move faster and collide more often with greater energy. 3. Concentration reactants: a. The increase of concentration increases the likelihood of reactant molecules colliding. 4. Presence of a Catalyst a. Catalysts speed up reactions by changing the mechanism of the reaction. b. Catalysts are not consumed during the reaction. - Be familiar with the effect of order of a reaction on the units on the rate constant, k Collision theory: - Two factors that determine if a collision will result in a reaction: - The particles must be in the right orientation upon impact - The particles must collide with enough energy to meet the activation energy of the reaction

Third order reactions is 1/m^2*s - Determine when it is useful to use an integrated rate law When you need to determine the concentration of the reaction as a function of time. - Explain when the initial concentration is and is not important in half-life reactions Initial concentration is how concentrated the solution is at the beginning of the reaction. It’s not important in first order half-life reactions because it gets canceled out within the equation. For a second order half-life reaction, the half-life decreases as the initial amount increases. For zero order processes, half-life increases as initial amount increases.

- Explain the conditions that must be met molecularly for a reaction to occur (collisions, activation energy, and geometry) 1. Collisions: The rate of reaction is proportional to the rate of reactant collisions. The rate constant can be calculated with the Arrhenius equation. k = Ae−Ea /RT Where rate constant k depends on: a. A, the frequency factor b. Ea, the activation energy c. T, the absolute temperature 2. The reacting species must collide in an orientation that allows contact between the atoms that will become bonded together in the product. 3. The collision must occur with adequate energy so that atoms can rearrange and form new bonds (and new chemical species) - Explain why a termolecular reaction is very rare In a termolecular reaction, three different molecules must collide. This is a rare occurrence as it occurs at 1/1000th the chance of a bimolecular reaction (two molecules colliding). *If k is measured at several temperatures, Ea can be calculated from the slope of a plot of ln(k) vs. 1/T. Where -Ea/R is the slope. - Explain the way a catalyst speeds up a reaction, thinking about this in terms of a reaction coordinate diagram -A catalyst speeds up the rate of reaction by providing a pathway that has a lower activation energy than the uncatalyzed process.

- Draw reaction coordinate diagrams for endothermic, exothermic, catalyzed, and uncatalyzed reactions

- Determine which step is the rate limiting step and why

The rate limiting step occurs in multi-step reactions. The slowest step is the rate determining step. In general the slow step is the first step. If it’s not, then the calculations are more complex.

Equilibrium - Explain what is happening molecularly in a system at equilibrium In a chemical equilibrium, the forward and reverse reactions occur at equal rates, and the concentrations of products and reactants remain constant. 1. As the system approaches equilibrium, both the forward and reverse reactions are occurring. 2. At equilibrium, the forward and reverse reactions are proceeding at the same rate. 3. Once equilibrium is achieved, the amount of each reactant and product remains constant. 4. Equilibrium depends on temperature 5. The speed at which the system achieves equilibrium is proportional to T (kinetics) ***At equilibrium moles of product = moles of reactant but their concentrations are NOT equal - Determine whether an equilibrium is product or reactant favored given its equilibrium constant, K ● If K >> 1, the reaction is product-favored; product predominates at equilibrium. ● If K 0 and ΔH2 > 0) ● Energy is released when solute-solvent interactions are formed. (ΔH3 < 0) - Define a saturated solution, unsaturated solution, and supersaturated solution ● Saturated solution: solvent that has solvated as many solute particles as possible at a certain temperature. ● Unsaturated solution: The solvent holds a given amount of solute with ability to solvate more. The solution contains less than the maximum amount of solute at a certain temperature. ● Supersaturated solution: A case of nonequilibrium where the solvent holds more solute particles than is usually possible at a certain temperature. This is unstable and will precipitate at any opportunity. - How does the solubility of solids in solution change with temperature? Gases? Solubility of gas decreases as temperature increases. It is directly proportional to its pressure over the solution. The stronger the attractions between solute and solvent molecules, the greater the solubility. - Describe what is meant by “like dissolves like” Polar substances dissolve in polar solvents and non-polar molecules will favor non-polar solvents. - Define molality and the four colligative properties that it is used to define Molality: moles of solute over kilograms of solvent. Does not fluctuate with temperature like molarity does. It defines these four properties: 1. Vapor pressure lowering 2. Boiling point elevation 3. Freezing point depression 4. Osmotic pressure These properties depend on the quantity of the solute particles present, NOT the identity of the

particles. - Explain how molality effects vapor pressure, boiling point, freezing point, and osmotic pressure ● Vapor pressure lowering: A high molality lowers vapor pressure. ● Boiling point elevation: A high molality increases boiling point temperature. ● Freezing point depression: A high molality decreases freezing point temperature. ● Osmotic pressure: Net movement of a solvent from an area of low solute concentration to an area of high solute concentration. Solvent moves through semipermeable membrane but solutes do not. ● The osmotic pressure, p, is the pressure needed to stop the movement of solvent through the membrane. - Define isotonic, hypertonic, and hypotonic ● Isotonic: solution concentration on both sides of the membrane are equal ● Hypertonic: more concentrated ● Hypotonic: less concentrated. - Explain what the vant hoff factor is and what it would be for a non-electrolyte, weak acid, and a strong electrolyte Moles of particles particle in solution over moles of formula units dissolved. Prominent in ionic compounds. Ex.) NaCl → Na + Cl i=2 Ex.) FeCl3 → Fe + 3Cl i=4 Lewis acid-base - Define a lewis acid and base and provide one example ● Lewis acid: accepts electron pair ● Lewis base: donates electron pair Electrochemistry - Define oxidation and reduction ● Oxidation: loses electrons ● Reduction: gains electrons (charge is reduced) - Outline the steps in the half reaction method in an acidic solution and a basic solution Acidic solution: 1. Assign oxidation numbers to determine what is oxidized and what is reduced. 2. Write the oxidation and reduction half-reactions. 1. Balance each half-reaction. 1. Balance elements other than H and O. 2. Balance O by adding H2O. 3. Balance H by adding H+. 4. Balance charge by adding electrons. 1. Multiply the half-reactions by integers so that the electrons gained and lost are the same. 1. Add the half-reactions, subtracting things that appear on both sides. 2. Make sure the equation is balanced according to mass. 3. Make sure the equation is balanced according to charge.

4. Make sure there are no electrons in the final equation. 5. Basic solution: ● Do the same exact thing but then add OH− to each side to “neutralize” the H+ in the equation and create water in its place once the equation is balanced. If this produces water on both sides, you might have to subtract water from each side. - Outline the processes that occur at the anode and cathode - Which electrode do the electrons flow towards It flows from the anode to the cathode. - Which electrode do cations and anions flow towards Cations move toward the cathode and anions move toward the anode. - Explain how to determine which species will be reduced and which will be oxidized if given their reduction potentials...