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CHM 11600 Exam 2 Study Guide

by: Gayatri

CHM 11600 Exam 2 Study Guide CHM 116

Marketplace > Purdue University > Chemistry > CHM 116 > CHM 11600 Exam 2 Study Guide
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CHM 116
Dr. Nash

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CHM 116
Dr. Nash
Study Guide
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This 5 page Study Guide was uploaded by Gayatri on Saturday March 7, 2015. The Study Guide belongs to CHM 116 at Purdue University taught by Dr. Nash in Winter2015. Since its upload, it has received 658 views. For similar materials see CHM 116 in Chemistry at Purdue University.


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Date Created: 03/07/15
CHM 11600 Exam 2 Study Guide Difference between completion for irreversible chemical reactions and for reversible chemical reactions 0 Irreversible reactions cannot go back to reactants after products have formed Completion complete consumption of reactants O Reversible reactions are constantly going back and forth between reactants and products Completion when there is a large amount of product we say the reaction almost goes to completion until limiting reactant gets used up A system at chemical equilibrium when the amount of reactants and products remain constant over time and the rates of the forward and reverse reactions are equal The position of equilibrium specific concentrations or pressures at a certain time 0 Shifts in position occur called net reactions and a new position is obtained that reduces the effect of the disturbance Law of chemical equilibrium law of mass action At a given temp a chemical system reaches a state where a particular ratio of reactant and product concentration is constant Equilibrium constant expression for a reversible reaction in terms of reactant and product concentrations or partial pressures gases For the reaction aA bB I cC dD the equilibrium constant Kc P P j 1074 Pf Calculating the value of the equilibrium constant for a reversible chemical reaction given the equilibrium concentrations of all reactants and products plug in equilibrium concentrations into the Kc Produc2L coe icient Product coe iciem coe icient For a reaction involving only gases K 1s wr1tten us1ng part1al pressures K p equation Kc Reactant coe zczent Reactant Why the concentrations of the reactants and products do not change when a chemical reaction has reached equilibrium concentrations of reactants and products do not change when a reaction has reached equilibrium because a change in one direction is balanced by a change in the other direction since the forward and reverse rates are equal Chemical equilibrium in terms of the rates of the forward and reverse reactions 0 In chemical equilibrium the rates of forward and reverse reactions are equal so the equilibrium constant is a ratio of the two rates Information provided by the value of the equilibrium constant for a chemical reaction 0 Small K mostly reactants present at equilibrium 0 Large K mostly products present at equilibrium Calculating the value of Kp given the value of Kc and vice versa 0 KP KC RTchangeEm0lesn Calculating the value of the reaction quotient Q given the concentrations or partial pressures gases of reactants and products at any moment in time 0 Q the reaction quotient changes value as concentrations of products and reactants change until equilibrium is reached 0 At equilibrium Q K O Follows the same formula as K Using Q to tell whether a reaction is in equilibrium and determine direction it must go in to achieve equilibrium 0 If Q K reaction is in equilibrium 0 If Q gt K reaction must proceed to the left toward reactants to achieve equilibrium 0 If Q lt K reaction must proceed to the right toward products to achieve equilibrium 0 LeChatelier s principle a change in one or more variables describing a system at equilibrium produces a shift in the system at equilibrium position that counteracts the effect of the change 0 Describe the effects on a system at chemical equilibrium when each of the following is changed 0 Concentration either reactants or products I No matter the disturbance the system will react to consume the added substance or produce removed substance I When reactant concentration increases the other reactant concentration will decrease because the reaction moves to the right to consume the reactants and make products therefore increasing product concentration 0 Pressure amounts of either reactants or products I An increase in pressure will cause the reaction to shift towards the side with fewer moles of gas I A decrease in pressure will cause the reaction to shift towards the side with more moles of gas 0 Addition of an inert gas volume of the container I Adding an inert gas will cause an increase in the total pressure of the system but will have no effect on equilibrium condition no effect on concentrations or partial pressures of reactants and products 0 Temperature reaction enthalpies I An increase in temperature will lead to an increase in K shift to the right for a system with a positive enthalpy an endothermic reaction I An increase in temperature will lead to a decrease in K shift to the left for a system with a negative enthalpy an exothermic reaction K 2 change enthalpy 1 1 K1 R08206 T2 T1 0 For endothermic reaction K2 gt K1 0 For exothermic reaction K1 gt K2 0 Interplay between the rate of reaction and the magnitude of the equilibrium constant for reversible chemical reactions 0 If K gtgt 1 the forward rate is faster products are favored 0 If K ltlt 1 the reverse rate is faster reactants are favored I HaberBosch Process and how the experimental conditions were chosen maximize the yield of ammonia o N2g3 H2g gt 2NH3g AH 918 k 0 Conditions to yield max ammonia I Decrease NH 3 I Decrease volume increase pressure I Decrease temperature I Effect of a catalyst on the equilibrium position for a reversible chemical reaction 0 A catalyst shortens the time it takes for a reaction to reach equilibrium but has no effect on equilibrium position 0 Arrhenius model for acids and bases 0 Acid substance with H in its formula that dissociates in water to yield H30 I Eg HCl HNO3 HCN 0 Base substance with OH in its formula that dissociates in water to yield OH I Eg NaOH KOH BaOH2 I van t Hoff equation In O Exceptions NH3 and K2CO3 also yield OH in water but don t have OH in their formula Strong Acids 1 HCl 2 HBr 3 HI 4 HNO3 5 HClO4 6 H2SO4 Strong Bases 1 LiOH 2 NaOH 3 KOH 4 RbOH 5 CsOH 6 CaOH2 7 SrOH2 8 BaOH2 Differences between strong acidsbases and weak acidsbases 0 Strong acids and bases are those that dissociate completely into ions in water and have K gtgt1 They can sometimes be expressed with I because their reaction is essentially complete 0 Weak acids and bases are those that only dissociate slightly into ions and have K ltlt 1 and low concentrations of H30 and OH Molecular total ionic and net ionic equations for an acidbase reaction 0 Molecular shows reactants and products as if they were intact undissociated compounds 0 Total ionic shows all soluble substances dissociated into ions liquid and aqueous 0 Net ionic eliminates spectator ions and shows the actual chemical change only 0 Spectator ions present only as part of reactants appear unchanged on both sides of the equation BronstedLowry model for acids and bases 0 Acid proton donor any species that donates H ions and has H in its formula ALL ARRHENIUS ACIDS ARE BRONSTED LOWRY ACIDS 0 Base proton acceptor any species that accepts H ion has lone e pair to bind to H ALL ARRHENIUS BASES CONTAIN THE BL BASE OH O Idea of conjugate acidbase pairs 0 Reaction occurs when an acid and base react to form a conjugate base and acid I Conjugate base has 1 less H and 1 more charge than acid I Conjugate acid has 1 more H and 1 less charge than the base I Acid loses H base gains H Describe an amphoteric substance 0 Substance that can either accept or donate a proton eg H2O Write the equilibrium reaction for the dissociation of pure water to produce H3O and OH ions 0 2H0z gtH0g0HgKwH0H0H 1xlOA14 25 C water doesn t appear in the equilibrium reaction because it is a liquid Molar concentrations of hydronium and hydroxide 0 H30 OH 1 x 10quot7 M Describe pH and pOH 0 pH logH30 O pOH logOH O pKw pH pOH 1400 Describe the strength of an acid or base in terms of the extent to which its molecules donate acids or accept bases protons 0 If an acid donates gives up its protons easily it has a high Ka low pKa so it is a Strong Acid 0 If a base accepts protons easily it has a high Kb low pr so it is a Strong Base Write the equilibrium constant expression for the reaction of an acid Ka or a base Kb with water H A 0 Acid HAH20H30AKazw HA BH 0H 0 Base BH20 gtOHBHKb Describe how the strength of either an acid or a base is indicated by the magnitude of its equilibrium constant ie Ka or Kb 0 High KaKb strong acidbase because dissociation occurs completely therefore products are favored which is why equilibrium constant is large 0 Ionic strength M1Z12 M2 23 O Ionic strength u 1 2 r M molar concentration of ions Z charges of ions 0 Relationship between both equilibrium constant expressions and pH and activitiesactivity coefficients 0 KwKaKb1x10quot14 O pKw pKa pr 1400 0 Activity A fa assume fa 1 0 Describe the main assumption that is used when an equilibrium constant or pH expression is written using molar concentrations 0 Describe the conditions where molar concentration and activity are not equal to one another 0 Determining if a substance will behave as an acid a base or neither 0 Acids 39 Group 7 with H or OH 39 Carboxylics or alcohols O Bases 39 Groups 1 or 2 with H or OH I Groups 1 or 2 with NH2 amine 0 Trends in acidbase properties for compounds having the structure HX 0 When neither H or X are more polar I nonpolar hydrides 0 When H is more electronegative X is anything under H I basic 0 When X is more electronegative I acidic strength trend is opposite of EN 0 Trends in acidbase properties for compounds having the structure HOX and how these properties are affected by bonding of electronegative atoms to atom X 0 If X is a metal I basic 0 If X is a nonmetal I acidic O Acidity can be increased by the addition of electronegative atoms to atom X 0 Zwitterions neutral molecules with no net charge but different charges on different sides 0 Conditions under which the contribution of H30 OH from the dissociation of water will be important in a solution containing a strong monoprotic acid base 0 If the concentration of the acid base is less than 45 x 10A7 then water contributes MORE than 5 H30 OH and must be taken into account and quadratic equation must be solved 0 If concentration of the acid base is more than or equal to 45 x 10quot7 then water dissociation contributes less than 5 and concentration of acid base H30total 0 Steps to calculate the pH pOH H3O tot OH tot H3O water and OH water in a solution containing a strong acid base given the initial concentration of the acid base 0 Check rules Write equation and expression and Q Evaluate the magnitude of K find out which is favored ICE table fill in with x for changes Write out K with x s Solve Check assumptions 0 Recheck 0 The leveling effect 0 Water exerts a leveling effect on any strong acid or base by reacting with it to form the product of water s autoionization and equalizes their strengths OOOOOO O The strongest acid that can exist in water is H30 strongest base is OH


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