Part A.Answer all questions(5 marks each). 1.Define all terms in the equation k=Ae- k is a rate constant.A is a pre-exponential factor that includes the steric factor,Ea is the activation energy. R is the gas constant and T is the absolute temperature. 2.During the corrosion of a piece of iron,the loss of iron metal may occur at a different place than the deposition of iron oxide.Why can this occur? The place where the oxidation of iron and the place where the reduction of oxygen occurs are fmetal acts as the external ci ircuit in the overall e mmected by the piece of metal.This pieceof 3.The ee constant for a particular reaction has the value11MWhat is the overall reaction Since the units ofk are M2s,we know immediately that the reaction is third order overall. 4.What will be the effect on the equilibrium position of the reaction Hag)+lxg)=2 HIg)if we increase the pressure? This reaction has two moles of gas on either side.Hence changing the pressure will have no effect on the position of the equilibrium. Part B.Ans swer a any y three e of the six questions.If you answer more than three,the best three will be used to calculate your mark (20marks each) 1 In the commercial production of aluminum metal,Al'3 is reduced to Als (a)If the reduction cell is operated at a current of 100,000 A,how long will it take to produce 1000 kg of aluminum metal The reaction is Al3+3 eAl.It therefore requires three electrons per atom of Al produced.or 3 moles ofelectrons per mole of Al produced. 00=3703 ol Al 037 mol electrons=111111 mol electrons The total charge q on this many electrons is q=nF=111111 mol ex 96487 C/mol e=1.07x 10C B,q=ior1=q=1.07x10c100,000C=1.07x103s(=1787min=29.8h (b)How much electrical power is used if the cell is operated at 5 V?Give your answer in kW hours.(1 W=1J/s) Power Volts x Amps 5 Vx 10'A=5 J/C x 10'C/s =5 x 10'J/s 5x 10'W=500 kW 500kWx29.8h=14900kWh Part A. Answer all questions (5 marks each). 1. Define all terms in the equation a RT/E Aek − = k is a rate constant. A is a pre-exponential factor that includes the steric factor, Ea is the activation energy, R is the gas constant and T is the absolute temperature. 2. During the corrosion of a piece of iron, the loss of iron metal may occur at a different place than the deposition of iron oxide. Why can this occur? The place where the oxidation of iron occurs and the place where the reduction of oxygen occurs are electrically connected by the piece of metal. This piece of metal acts as the external circuit in the overall galvanic cell. 3. The rate constant for a particular reaction has the value k = 1.23 x 10-4 M-2 s -1. What is the overall reaction order? Since the units of k are M-2 s -1, we know immediately that the reaction is third order overall. 4. What will be the effect on the equilibrium position of the reaction H2(g) + I2(g) = 2 HI(g) if we increase the pressure? This reaction has two moles of gas on either side. Hence changing the pressure will have no effect on the position of the equilibrium. Part B. Answer any three of the six questions. If you answer more than three, the best three will be used to calculate your mark (20 marks each). 1. In the commercial production of aluminum metal, Al+3 is reduced to Al(s). (a) If the reduction cell is operated at a current of 100,000 A, how long will it take to produce 1000 kg of aluminum metal? The reaction is Al+3 + 3 e- → Al. It therefore requires three electrons per atom of Al produced, or 3 moles of electrons per mole of Al produced. 1000 kg Al = 106 g/(27 g/mol) = 37037 mol Al which will require 3 x 37037 mol electrons = 111111 mol electrons The total charge q on this many electrons is q = nF = 111111 mol e- x 96487 C/mol e- = 1.07 x 1010 C But, q = it, or t = q/i = 1.07 x 1010C/(100,000 C/s) = 1.07 x 105 s (= 1787 min = 29.8 h) (b) How much electrical power is used if the cell is operated at 5 V? Give your answer in kW hours. (1 W = 1 J/s) Power = Volts x Amps = 5 V x 105 A = 5 J/C x 105 C/s = 5 x 105 J/s = 5 x 105 W = 500 kW 500 kW x 29.8 h = 14900 kWh