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UNT CHEM 1415 - Final Exam Study Guide
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CHEM 1415 1nd EditionFinal Exam Study Guide Lectures: 23 - 27Lecture 23 (November 13)Chemical Equilibrium Concentration  Chemical Equilibriumo Dynamic Equilibrium Rates of evaporation or condensation become equal and do not stop at equilibrium, and do not go to zero Individual molecules continue to move from the liquid to the vapor phaseand back againo The rate of the forward reaction equals the rate of the reverse reactiono Arrows indicate emphasize the dynamic character of the processR ←→Po Reverse reaction Raterev=krev[ P]o Two rates equal at equilibrium Ratefor=Raterevof ∨¿[R]=krev[P]∧kforkrev=[P]eq[R]eqk¿ The Equilibrium Expressiono Using a general chemical reaction aA+bB ←→cC +dD o The ratio of concentration is Q=[C ]c[D]d[ A ]a[B]bo The expression for the equilibrium constant is always based directly on the reaction stoichiometryo Homogeneous equilibria – reactants and products are in the same phase, either gaseous or aqueouso Heterogeneous equilibrium – multiphase system Equilibrium Concentrationso Strategies to solve equilibrium concentration problems Write a balanced chemical equation for the relevant equilibrium or equilibria Write the corresponding equilibrium expression Create a table of concentrations for all of the reacting specieso Example 1 Hydrogen reacts with Iodine gas at elevated temperatures, k = 59.3, T = 400℃, 0.050 M of hydrogen and 0.050 M of Iodine. What are the equilibrium concentrations?-H2(g )+I2 (g)←→2 H I(g)-K=[HI ]2[H2][I2]=59.3H2I2HIInitial Concentration0.050 M 0.050 M 0 MChange in Concentration-x -x +2xFinal Concentration0.050 - x 0.050 - x 2x-K=[HI]2[H2] [I2]=59.3 = (2 x)2(0.050−x)2=59.3-2 x0.050−x=7.70-9.70 x=0.39- X = 0.040- Therefore[H2]=[I2]=0.050−0.040=0.010 M ∧[HI]=2 x= 0.080 MLecture 24 (November 18)Equilibrium Concentrations Equilibrium Concentrationso Strategies to solve equilibrium concentration problems Write a balanced chemical equation for the relevant equilibrium or equilibria Write the corresponding equilibrium expression Create a table of concentrations for all of the reacting specieso Example 1 Phosphorus trichloride reacts with Chlorine gas to form phosphorus pentachloride at k = 33, T = 250℃, 0.050 M of PC l3 and 0.015 M of Cl2. What are the equilibrium concentrations ? -Cl2 (g)+PCl3 (g)←→PCl5 (g)-K=[ PCl5]2[Cl2][ PC l3]=33Cl2PCl3PCl5Initial Concentration0.015 M 0.050 M 0 MChange in Concentration-x -x +xFinal Concentration0.015 - x 1.50 - x x-PCl[¿¿5][Cl2][PC l3]=33K =¿ = x(0.050−x)(0.015−x)=59.3-33 x2−3.145 x +0.025=0-x=0.087∨0.0088- Therefore [PCl5]=x=0.0088 M ,-[C l2]=0.015−x=0.015−0.0088=.006 M ,-[PC I3]=0.05−x=0.050−0.0088=0.042 M LeChatelier’s Principleo When a system at equilibrium is stressed, it responds by reestablishing equilibrium to reduce the applied stresso Effects of a Change in Concentration by considering the reaction quotient, Q withthe equilibrium constant, KType of Concentration ChangeResulting Change in Q Response of SystemProducts increased Q > K More reactants formedProducts decreased Q < K More products formedReactants increased Q < K More products formedReactants decreased Q > K More reactants formedo Effects of a change in Pressure on Equilibrium If number of moles of gases differs between reactants and products, a shift in pressure will result in a change in the equilibrium positiono Effects of change in temperature on equilibriumo Effect of o Effects of a catalyst on Equilibrium Has no effectLecture 25 (November 20)Acids and Bases Bronsted – Lowry definition – an acid is a proton +¿H¿¿ donor and a base is a proton acceptoro water can act as both an acid or a base, depending on the properties of the othersubstanceo amphoteric – refers to any substance that can be either an acid or a base Conjugate acid-base pairs o Conjugate acid – of a base is the acid formed when the base receives a protono Conjugate base – of an acid is the base formed when the acid gives a protono pH = +¿H3O¿−log ⁡¿ Example Problem 1: The Kaof acetic acidC H3COOH ,is 1.8∗10−5. what is the pH of a 0.10 M solution.−¿(aq)+¿(aq)+C H3COO¿C H3COOH(aq)+H2O(l)← H3O¿C H3COOH(aq)+¿(aq)H3O¿−¿(aq)C H3CO O¿Initial 0.10 0 0Type of Reaction Type of Temperature Change Response of SystemExothermic T increase More reactants formedExothermic T decrease More products formedEndothermic T increase More products formedEndothermic T decrease More reactants formedconcentrationChange in Concentration-x +x +xFinal Concentration 0.10 – x X X+¿H3O¿¿−¿C H3CO O¿¿¿Ka=¿x2=1.8∗10−5(0.10)=1.8∗10−6x=1.3∗10−3MpH=−log(1.3∗10−3)=2.9 Free Energy and Chemical Equilibriumo Equilibrium – a state of minimum free energyo Free energy decreases as you move toward equilibrium from either directiono Chemical systems tend to move spontaneously toward equilibriumo∆ G=∆ G°+RTlnQLecture 26 (November 25)Redox Reactions and Galvanic Cells Oxidation-Reduction and Half-Reactionso Oxidation-reduction – reactions involving the transfer of electrons (redox)o Oxidation – the loss of electrons from some chemical specieso Reduction – gain of electronso Half reactions – neither one can occur by itself because oxidation and reduction must take place concurrently Galvanic Cello Salt bridge – contains a strong electrolyte that allows either cations or anions to migrate into the solution where they are needed to maintain neutral chargeo Galvanic cell – any electrochemical cell in which a spontaneous chemic reaction can be used to generate an electric currento Electrochemistry – observation of an electric current in a cell ledo Electrodes – electrically conducting sites at which either oxidation or reduction take placeo Anode – where oxidation occurso Cathode – where reduction occurso Cell notation – lists metals and ions involved in the reaction Anode | electrolyte of anode || electrolyte of cathode | cathodeo Cell potential/electromotive force (EMF) – buildup of charge on the electrodes allows for a potential of electrical workwmax=qEo Nernst equationE=E°−RTnFlnQ where F=96,485 J V−1mol−1 Relationship between cell potential and free energy∆ G°=−nF E°∧E°=RTnFlnKLecture 27 (December 02)Corrosion and Batteries Corrosiono Degradation of metals by chemical reactions with the environment, slow combination of oxygen with metals to


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