山东大学 Physica| Chemistry(2)课程试卷 20202021学年_Fa|学期 (4)[6 points] After electrolysis for 30 min, write the electrode reaction of the cathode and determine how many mole of NaoH can be produced in the cathodic compartment. How many electric energy is consumed? what is the energy efficiency of this electrolysis? 0.10Acm-2×100cm2×30×60%6500=1.865×103 mol NaOH 0.10Acmr2×100cm2×30min×60s×44V=0.792kJ 14-21[14 points] There are two first-order competing reactions A-ii+z (side reaction) (1)(4 points] If the side reaction can be neglected, at 800 K, half-life of A is 1386s, then hot long does it taken to achieve 99% conversion? 0691,=5:0×103s35hm5=k,1=9210s (2)[2 points] If the side reaction cannot be neglected, at 800 K for 99% conversion need 837 s, what is the apparent rate constant? k=(k+k2厘=5.50×103s (3)(4 points] At 800K, if kI=4.7x10-3sI, what is kz? What is the ratio Y/Z? k=0.8×103s1;Y=5.88 (4)[2 points) If the pre-exponential factor of these two reactions is just the same, Ea, I 80 k 线 mol-l, what is Eaz? (5)(2 points] If we want to increase yield of Y, should we perform this reaction at higher or At lower temperature.2020 -2021 学年 Fall 学期 ( 4) [6 points] After electrolysis for 30 min, write the electrode reaction of the cathod e and determine how many mole of NaOH can be produced in the cathodic compartment. How many electric energy is consumed? What is the energy efficiency of this electrolysis? 0.10 A cm - 2 1.00 cm 2 30 60/96500 = 1.865 10 - 3 mol NaOH 0.10 A cm - 2 1.00 cm 2 30 min 60 s 4.4 V = 0.792 kJ. 1.782/4.4 = 40.5% [ 4 -2] [14 points] There are two first -order competing reactions: (1) [4 points] If the side reaction can be neglected, at 800 K, half-life of A is 138.6 s, then how long does it taken to achieve 99% conversion? ， k=5.0×10 - 3 s - 1 ; , t =921.0 s (2) [2 points] If the side reaction cannot be neglected, at 800 K for 99% conversion need 837 s, what is the apparent rate constant? k=(k1+k2)=5.50×10-3 s-1 (3) [4 points] At 800K, if k 1=4.7×10 - 3 s - 1 , what is k2? What is the ratio Y/Z? k2=0.8 ×10 - 3 s - 1 ; Y/Z=5.88 (4) [2 points] If the pre -exponential factor of these two reactions is just the same, Ea,1=80 kJ mol - 1 , what is Ea, 2 ? Ea,1=91.78 kJ mol - 1 , (5) [2 points] If we want to increase yield of Y, should we perform this reaction at higher or lower temperature? At lower temperature. 山东大学 Physical Chemistry (2) 课程试卷 学院 专业 级 学号 姓名 ……………………………………密…………………………封…………………………线…………………………………