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Chem 1220 midterm 1 study guide

by: ask0429

Chem 1220 midterm 1 study guide CHEM 1220

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These notes cover what's in our first exam as well as some additional resources and practice
General Chemistry 1220
Dr. Fus
Study Guide
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This 7 page Study Guide was uploaded by ask0429 on Friday January 29, 2016. The Study Guide belongs to CHEM 1220 at Ohio State University taught by Dr. Fus in Spring 2016. Since its upload, it has received 50 views. For similar materials see General Chemistry 1220 in Chemistry at Ohio State University.


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Date Created: 01/29/16
Chemistry 1220 Midterm One Review Sheet: Ch. 13­14 Ch. 13 Properties of Solutions Solutions: formed when a solute is dispersed evenly throughout solvent Solute: substance added to another. Dissolved Solvent: Substance that takes the other. Mechanism for dissolving Gas: Mix and spread into larger volumes when not restrained in some way. Spontaneous as  entropy increases Liquids and solids: Disperses only if the intermolecular forces are over come Formation of solution is favored by the increase in entropy that accompanies mixing IMF influence: Solute­solute attractions must be overcome for solute to form attractions with the  solvent. Breaking these attractions is endothermic [requires energy]. Solvent­solvent attractions  must also be overcome for the solvent to form attractions with the solute. Again, breaking these  attractions is endothermic.  *If a solution forms, these previously mentioned attractions are overcome and solute­solvent  attractions form, which is exothermic [releases energy]  In solution: Solvent surrounding the solute is called solvation. I.e. water and NaCl. Na and Cl ­  molecules are surrounded by water, called hydration, a specific form of solvation. Energy summary:  1. (Solute)  n Solute   ΔH = + n solute  2. (Solvent) mm Solvent  ΔH solvent+ 3. n Solute + m SolventSolution  ΔH mix ­  ΔH soln ΔH solute ΔH solventΔH mix  For a solution to form, the absolute value of heat of vaporization of the mixing must  be comparably similar to the absolute values of the heats of vaporization of solute  and solvent combined.  In the reaction Solute + Solvent Solution, the forwards reaction is dissolving and the reverse  reaction is crystallization *If it’s in equilibrium, it’s called saturation.  Solubility: amount of solute needed to form a saturated solution in a given quantity of solvent at  a given temperature.  1. Solute < solubility  unsaturated 2. Solute = solubility  saturated 3. Solute > solubility  supersaturated *Supersaturated solutions are highly unstable and can crystallize given seed crystal The stronger the attractions between solute and solvent molecules, the greater the solubility of  the solute is in that solvent.  Like dissolves like. Polar solutes tend to dissolve more readily in polar solvents because partial charges form dipoles, and dipoles attract each other [dipole­dipole IMF]. Non­polar solutes tend to dissolve more  readily in non­polar solvents because they can both overcome their own London Dispersion  Forces and form new ones between solute and solvent. Miscible: liquids that mix in all proportions Immiscible: liquids that don’t dissolve in one another Pressure: really only affects gases   Pressure increases so solubility increases Henry’s law: S =g k)*(P ) g  S = solubility [expressed in molarity] g   k = proportional constant  Pg = partial pressure Temperature: solubility of most solid solute in water increases as the solution’s temperature  increases. Solubility of gases in water decrease with increasing temperature. Dilute: relatively less solute in solution, compared to other proportions of the same reaction Concentrated: relatively more solute in solution, compared to other proportions of the same  reaction Ways to express concentration:  1. Mass percentage: (mass of component in solution)/(total mass of solution)*100% a. Can be expressed as parts per million or billion [ppm, ppb]  2. Mole fraction: X component moles of component)/(total moles of all components in  solution) 3. Molarity: (moles solute)/(liters solution) M 4. Molality: (moles solute)/(kg of solvent) m Colligative properties: properties of a solvent that depends on the total concentration of solute  particles present 1. Vapor pressure lowering a. Vapor pressure exerted by the vapor [gas] at equilibrium with the liquid b. Volatile: pressure exerted c. Nonvolatile: no vapor pressure [solid solutes]  d. Solutions form spontaneously with volatile liquid solvents and nonvolatile solutes i. More solutes lower the vapor pressure of the solvent e. Raolt’s Law: P =(Xsoln solvent(P˚ solvent f. Ideal solutions obey Raolt’s law. Gases are assumed to have no IMF’s and solutions  are assumed to have a uniformity of IMF’s 2. Boiling point elevation a. Since solutions have lower vapor pressure, it takes more energy [higher  temperature] to get to the boiling point [1 atm]  b. Van’t Hoff equation: ΔT = (i)b K )*(m) b c. i = van’t hoff factor [number of ions] (assume 1 for nonelectrolytes) d. K  b molal boiling point elevation constant e. m = molality 3. Freezing point depression a. Freezing point for solution is lower than the pure solvent’s freezing point b. ΔT = f(i)*(K)*(mf c. K =f olal freezing point depression constant 4. Osmotic pressure a. Net movement of solvent (low solute) is always towards the solution with lower  solvent (high solute) concentration b. Osmotic pressure: osmosis stops [isotonic] c. iMRT d. i = van’t hoff constant e. M = molarity f. R = ideal gas constant  g. T = absolute temp in K Ch. 14 Chemical Kinetics Chemical kinetics: rates of reaction. Speed at which a chemical reaction occurs is the reaction  rate. It’s the step by step molecular­level view of the pathway from reactants to products is the  reaction mechanism. 4 factors affect reaction rates: 1. Physical state of the reactants a. More surface area exposed means particles can collide more and easily, leading to a  quicker reaction 2. Reactant concentration a. More reactant itself means more collisions due to the sheer number of particles,  leading to a quicker reaction 3. Reaction temperature a. Molecules that have a higher temperature move faster [with more energy] meaning  more collisions that have enough energy to form products, leading to a quicker  reaction 4. Presence of a catalyst a. Catalysts serve as a mechanism to either lower the activation energy or allow more  molecules to collide with the appropriate energy *The greater the frequency of collisions, the higher the reaction rate Reactions only happen when collisions occur with enough energy to overcome the  activation energy. Or else if the collision happens with enough energy, the reactant simply  bounce off each other. Speed: change that occurs in a given time interval. The average rate of appearance or  disappearance AB Appearance: (Δ[B])/(ΔT) Disappearance: ­(Δ[A])/(ΔT) Instantaneous rate: slope of the curve of concentration vs. time at any given time. (Δreactants)/ (ΔT) *Stoichiometry doesn’t always dictate the rate law In a given reaction, the rates of appearance and disappearance are proportional to the reactant  and product coefficients aA + bB  cC + dD ­(1/a)(Δ[A]/ΔT) = ­(1/b)(Δ[B]/ΔT) = (1/c)(Δ[C]/ΔT) = (1/d)(Δ[D]/ΔT) m n Rate law = k[A] [B] m and n dictate the order. They are low positive integers. k is the rate constant, which changes  with temperature. The order must be found experimentally, as the coefficients don’t dictate it.  The overall reaction order is m+n  ~ 10  (fast)                                                                                                  ~ 10 (slow) Rate of the reaction depends on the concentration and physical state, and the rate contant  depends on temperature and catalysts st nd rd Reactions occur in orders, either 1 , 2 , 3 , or higher (though that’s rarely seen) 1  order reactions depend on the concentration of a single reactant raised to the 1  power. To find components of the reaction, use the equation ln[A] =t­kt + ln[A] {t0 at the given time, 0= initial} Plotting this equations gives a linear line.   2  order reactions mean overall order, so this can be two reactants raised to the 1  or one  st reactant squared. To find components of this reaction, use the equation 1/[A] =t t + 1/[A]  Pl0 ting the inverse of the concentration vs time gives a linear line,  nd indicating 2  order Zero order reactions means rate of disappearance is independent of concentration. To find  components of this reaction, use the equation [A] t ­kt + [A]  Th0s will give a linear line that is just the concentration is plotted against time Half­life reactions: time required for the concentration to decrease to half the initial. It is used in  most radioactive reactions to express reaction rates. Most reactions like these are first order, so  use the equation .693/k For a second order reaction, use  1/k*[A]   0 Collision model: molecules must collide to react, so increasing the number of collisions increases the rate, which can be increases by the 4 factors  Orientation factor: molecules must be oriented in a certain way to form a product Activation energy: minimum amount of energy required to form a product A transition molecule is formed during the collision [reactants bonded together, not yet the  product] called the activated complex Rate depends on the magnitude of activation energy lower the E  then the fastea the reaction The fraction of molecules in a reaction that have the energy to react can befound using this  equation F = e^(E /RaT) To find the relationship between k and E use the a, henius Equation  k = A*e^(E /R*a) {A= frequency factor, a constant} To find the activation energy of a reaction of two different reactions, use the equation ln(K /1 )2= (E /Ra*(1/T – 1/T2  1 Reactions occur in multiple steps, called elementary reactions.  Molecularity: number of molecules of the reactants. I.e. a single reactant molecule would be  unimolecular. Two reactant molecules = bimolecular, three reactants molecules = termolecular  (these are rare and not very probable)  Products that form in the elementary steps that are also reactants, are called intermediates. 1. A + B    X + Y 2. X + B  Y + Z Overall: A + B  Y + Z Rates laws for the elementary step DO depend of coefficients [molecularity] To figure out the rate of the overall reaction, look at the rates of the elementary steps. One step  always happens slower than the other(s), making it the rate determining step.  1  step slow: the overall rate law is equal to the 1  step rate law 2  step slow: you have to take into account both steps in this case. First determine the rate law  for the slow step and the fast step separately. Then, notice the product of the fast step is part of  the rate law for the slow step. Take the reverse of the fast step and substitute the reverse rate law  into the slow step rate law. Solve for the concentration of the intermediate by assuming  equilibrium is reached in the fast step A + B  C (fast) C + A  D (slow) Rate of slow = [C][A] Rate of fast = [A][B]  reverse this: Rate of reverse = [C]  Solve for [C]: [C] = K forwardsreverse][B] Put this into the slow step K slow( K forwardsreverse][B][A]  k*[A] [B] Catalyst: substance that changes the speed of the reaction without being consumed. Lowers the  overall activation energy to be the most effective [most catalysts]  Homogenous catalyst: same phase as reactant Heterogeneous catalyst: phase different than reactant. Adsorption of the reactants to be reactive  [binding to the surface] Enzymes: biological catalysts. Substrates bind to the active site using a lock and key model to  undergo reaction Additional resources and practice:  Khan academy (explanation videos and practice quizzes) Textbook; Useful to do the last few questions in the problem set of the section because they are  the hardest and will give you good practice Best of luck for the exam!


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