Wright CHM 1220 - CHM 1220 SI_Exam 2 Review_Rev 2_ANSWERS (2 pages)

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CHM 1220 SI_Exam 2 Review_Rev 2_ANSWERS



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CHM 1220 SI_Exam 2 Review_Rev 2_ANSWERS

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Wright State University
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Chm 1220 - General Chemistry II
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Questions Comments Suggestions Email me at stevens 161 wright edu SI Activity Exam II Review ANSWERS CHM 1220 Multiple Choice 1 Consider the gas phase equilibrium system 2 H2O g 2 H2 g O2 g Given that the forward reaction is endothermic which of the following changes will decrease the amount of H2O adding more oxygen adding a solid phase catalyst decreasing the volume of the container increasing the temperature at constant pressure adding He gas 2 The combustion of ethane is represented by the equation 2 C2H6 g 7 O2 g 4 CO2 g 6 H2O l Which of the below statements is true The rate of consumption of ethane is seven times faster than the rate of consumption of oxygen The rate of formation of CO2 equals the rate of formation of water Water is formed at a rate equal to two thirds the rate of formation of CO2 The rate of consumption of oxygen equals the rate of consumption of water CO2 is formed twice as fast as ethane is consumed 3 The equilibrium constant for reaction 1 is K Which expression is the equilibrium constant for reaction 2 1 SO2 g O2 g SO3 g 2 2 SO3 g 2 SO2 g O2 g K 2 K 1 2 K 2 K K 2 4 The rate law of a reaction is rate k A 2 B What are the units of the rate constant M 1 s 1 M3 s 2 1 M 2 s 1 Ms M2 s 1 5 At 22 C Kp 0 070 for the equilibrium NH4HS s NH3 g H2S g A sample of solid NH4HS is placed in a closed vessel and allowed to equilibrate What is the equilibrium partial pressure atm of ammonia assuming that some solid NH4HS remains 0 26 0 26 0 0049 K p PNH3 PH2S 0 070 x 2 x 0 070 0 26 0 070 3 8 0 52 6 At elevated temperatures methylisonitrile CH3NC isomerizes to acetonitrile CH3CN according to the reaction CH3NC g CH3CN g The dependence of the rate constant on temperature is studied and the graph below is prepared from the results 6 6 The energy of activation of this reaction is kJ mol 166 1 66 10 5 E 1 ln k a ln A 4 42 10 7 R T 4 42 10 4 10 5 6 5 y2 y1 a 1 94 104 Slope E R x2 x1 0 00215 0 00195 Slope 20000 Ea 20000 R 166 kJ mol Problems 7 A proposed mechanism for the reaction Br2 g 2 NO g 2 NOBr g is found below Determine the rate law for the overall reaction Note that a proper rate law cannot depend on the concentration of an intermediate because intermediates are generally unstable and are present in low non measurable concentrations k1 NO g Br2 g NOBr2 g k 1 fast reversible NOBr2 g NO g k 2 NOBr g very slow 2 First the rate determining step is the very slow one The rate law for this step is Rate k2 NOBr2 NO This rate law contains NOBr2 so we must exploit the fact that the first step is reversible Since the reaction is fast we assume that it rapidly reaches equilibrium At equilibrium the forward and reverse reaction rates are equal so we may equate their rate laws as follows k1 Br2 NO k 1 NOBr2 We then solve the preceding expression for NOBr2 and substitute it into the RDS rate law k NOBr2 1 Br2 NO k The equation K 1 1 kf k 1 may be used to further simplify the rate expression This is optional kr k 1 k1k2 Br2 NO NO k 1 Rate k2 NOBr2 NO k1k2 2 2 NO K Br2 Br2 NO Rate 1k2 k 1 8 Consider the gas phase thermal E1 elimination of HCl from tert butyl chloride tBuCl to form isobutylene at 500 0 K below Suppose 2 000 moles of tBuCl are initially injected into an empty 15 00 L vessel maintained at 500 0 K After the reaction was allowed to equilibrate an analysis shows that the partial pressure of HCl in the vessel is 2 950 atm Using this information and assuming that all three gases behave ideally determine the equilibrium constant Kp at 500 0 K H3C H3C C Cl CH2 H3C tert butyl chloride C H3C H Cl CH3 isobutylene First we employ the ideal gas equation to calculate the INITIAL pressure of tBuCl pV nRT p nRT V L atm 2 000 mol 0 08206 mol K 500 0 K 15 00 L 5 47067 atm Next we set up an ICE table for this situation BuCl g 5 47067 atm x x 2 950 2 52067 atm t I C E isobutylene g 0 x 2 950 atm Because of the 1 1 stoichiometric relationship of isobutylene and HCl the equilibrium partial pressures of both are equal to the given HCl pressure of 2 950 atm This allows us to readily arrive at x because 0 x 2 950 atm so x 2 950 atm HCl g 0 x 2 950 atm Lastly we set up our law of mass action and plug in the ICE table values Kp Rev 1 Fall 2017 J Stevens 2 950 2 950 p p isobutylene pt BuCl HCl 2 52067 3 452


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