Homework-2020-10-15 【1】Yin,p.390,ex.14 【2】Yin,p.390,ex.16 【3】Yin,p.396,ex.19 【4】Yin,p.407,ex.29 【5】Yin,p.412,ex.32 【6】 Levine,p.596,17.89 Due time: Oct 20 2020
4-1 由题意Dk+== 光气的铁:1=14=p[ea1c 花的斛备:r=B=B2[cCC 由于反应们、16乖((均易到平续 k c 则:k==乙c6 (付=(k【 [cocl]=k2[co].K CchJ 贝小:光气的生戍速平公式 光乞的每岛速半公式 k[,()2=3LocE以 k-[cocl]Ccla 其中, ku ky k2 r=k/RE
4-1
P291.6×.16. 4-2根混两和调“[01,k2[eMCe 用稳壳面纵压:。p[RL-k[]R-2/LA]0 k-CRQJLRH]+k, [- [RH]-ksCR-]Co] k[AUm]RA了2=0 dIRo.] k[RDDHJ-kTRa: ]TRH) 配=A2【m-k2+1RH]= Autocatalysis ·()w A8代0,y=20:(卡kn RLROOH' CRH] 则Ea=E5+tLE-E 92 r k/nc 根据速公式可,反速与生成+的平机成正比,k应 初叫Ro的浓度發小反相较慢.反应的进,Dam 越来越大,所以应越来越快直到[的减小抵消3[n的 加速作用,速华降下来
4 - 2 Autocatalysis
习题1 4-3 ll第-没实验:由于B/4)3NO, 则=CPM“P写为P=P 时于kt困亲线,则为应 2 所以.区=1 Prota / 10Pa 10.0 11. 由次实验:设从平始到时应物湖耗 plp 1.15 1.09 N0y+M0→30 h B-x P 则隐=2B+x=12 员:x=25 +, RP:x=32 r=kP(N,OS] IP(NO) 即?:t2h时物反应了一半即5 toth. F-x th日,隐=2+X=15 t从h÷ih,502 则可近似着作反应速华与反应物起流度无 明此。t2h范星第实中反的半观句种的起,t k=2×3600 捉:P=.6 ,该反星一级反应,“(1=1∠
4 - 3 t/h 0 1 2 ptotal/10 3Pa 10.0 11.5 12.5 p /p* 1.15 1.09 N O NO 3NO 2 5 2 + ⎯⎯→ 2 5 r k p p [ (N O )] [ (NO)] =
a2速剑式r=-at=APoB 4-3 dCPA]:k+ [imo J-k. []cAwg))-k IPawo)JCRwo] r=AD(NO刀 稳态近似,at NOt>NO+ No b+LImOs)J No+no->2NO 则:F=2[Po] k+[(MOs) 上RwJ+P 当满海是D0×8P(上2CP忽略计 Physical meaning? Ry: r- k+ [P(Mon)] =kLpimi0s] 与安验果一 3).由得:1=ECPA2)=6xPA可 则:t= [M0)=1x(7反辰一曰丁 21:P=B-kx2÷2=15P
4-3 Physical meaning? 2 5 2 3 3 2 N O NO NO NO NO 2NO ⎯→ + + ⎯⎯→ 2 5 r k p = [ (N O )]
44,由题啊 图表标题 2500000 m Six 2000000 Lineweaver-Burk plot tool y=38941x+832448 900000 1500 2000 2500 贝,司大一 二58|1 LOFs 应:pA=468X10+m 当1时,长=、68m Ym=2r:. 20 x10 mol-L/
4-4 M m m 1 1 1 [S] k r r r = + Lineweaver-Burk plot
4-5 习题32.m),中 始反的物至一 h收岁量子的晕 且307=nL 九C h=663810 C kron m/s L 307入 L 2)p=2,说明以收一个光B量有5解 则2小:H1 H.+ +H1→h12+ 1+k→12 2HI+hy->H2+I
4-5 2 2 2HI H I + ⎯⎯→ + h
4-6 11D=42×0cm6=42×07myk 1=2x00m=2×0d kp=Z2 NAX 2YL X2D 2×引1+×602X03x4x42X0x2X10 127×C(ms 708×102(m15 The estimated value is close to the observed value This reaction is diffusion-controlled [Why do we take the recombination reaction as an example? The estimated value is a little larger than the observed one(by 58.75%) The I atom diffused into a solvent cage can not completely react. Some I atom diffuses way after several collisions. I
4-6 The estimated value is close to the observed value. This reaction is diffusion-controlled. [Why do we take the recombination reaction as an example?] The estimated value is a little larger than the observed one (by 58.75%). [The I atom diffused into a solvent cage can not completely react. Some I atom diffuses way after several collisions.]
