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1.-100 kJ2.100 kJcorrect3.550 kJ
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Version 173 – Exam 4 302 – sutcliffe – (51060)64.200 kJ5.450 kJExplanation:The minimum energy required by A to reactis represented by the height from A to thepeak.01910.0 pointsIn this question, n=8. This is because C startsas CH4and becomes CO2.If the standard free energy change for com-bustion of 1 mole of CH4(g) is-818 kJ·mol−1, calculate the standard voltage thatcould be obtained from a fuel cell using thisreaction.1.+ 4.24 V2.-1.06 V3.+ 8.48 V4.+ 0.53 V5.+ 1.06 VcorrectExplanation:02010.0 pointsCatalysts work by1.increasing the exothermicity of the reac-tion, which increases entropy in the surround-ings, making the reaction more favorable.2.increasing the diffusion rates (collisionrate) of molecules in solution.3.increasing the kinetic energy of the reac-tant molecules.4.creating a new reaction path with a loweractivation barrier.correctExplanation:Definition02110.0 pointsThis is a BONUS question: get it correct yourmax points are 250/240; get it wrong and yourmax points are 240/240.The overall reaction for the discharge of anickel/cadmium cell isCd + NiO2+ 2 H2O→Cd(OH)2+ Ni(OH)2Which of the following statements is true?1.Cadmium metal loses 2 electrons and isconverted to Cd2+at the cathode.2.Itisimpossibletorechargeanickel/cadmium cell.3.The nickel is oxidized during this reac-tion.4.Insoluble Cd(OH)2is deposited on theanode.correctExplanation:Looking at the reaction above,Cd→Cd2++ 2e−is an oxidation reaction that will occur at theanode.The nickel is not oxidized because we haveshown that Cd is being oxidized.Nickel/cadmium batteries are rechargable.Insoluble CD(OH)2is deposited on the an-ode since Cd is oxidized at the anode.02210.0 pointsA certain reaction is found to have a rateconstant of 1.50×10−8sec−1at 0◦C and anactivation energy of 45.0 kJ·mol−1.Whatwill be its rate constant at 100◦C?1.k= 68.4 s−12.None of the other answers is correct.3.k= 7.37×10−11s−14.k= 3.05×10−6s−1correct5.k= 3.10×10−3s−1Explanation:
Version 173 – Exam 4 302 – sutcliffe – (51060)7T1= 273 KT2= 373 KEA= 45.0 kJk1= 1.50×10−8sec−1lnk2k1= lnk2-lnk1=EARparenleftbigg1T1-1T2parenrightbigglnk2= lnk1+EARparenleftbigg1T1-1T2parenrightbigg= ln 1.50×10−8+450008.314parenleftbigg1273-1373parenrightbigg=-12.699876k2= 3.05×10−6s−102310.0 pointsNote: use a = 1 in this question.For the reactioncyclopropane→propenea plot of ln[cyclopropane]vstime in secondsgives a straight line with slope-4.1×10−3s−1at 550◦C. What is the rate constant for thisreaction?
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