總復(fù)習(xí)提要2012 - 副本

1、2012 Physical Chemistry Xiamen University2021-11-16重點(diǎn)：重點(diǎn)：1. 具有簡(jiǎn)單反應(yīng)級(jí)數(shù)的反應(yīng)動(dòng)力學(xué)特點(diǎn)具有簡(jiǎn)單反應(yīng)級(jí)數(shù)的反應(yīng)動(dòng)力學(xué)特點(diǎn)2. 反應(yīng)級(jí)數(shù)、速率常數(shù)計(jì)算，速率方程的建立反應(yīng)級(jí)數(shù)、速率常數(shù)計(jì)算，速率方程的建立3. 半衰期、反應(yīng)時(shí)間的計(jì)算半衰期、反應(yīng)時(shí)間的計(jì)算4. 反應(yīng)速率常數(shù)與溫度的關(guān)系、活化能的計(jì)算反應(yīng)速率常數(shù)與溫度的關(guān)系、活化能的計(jì)算Chapter 8 Chemical Kinetics1Part I2012 Physical Chemistry Xiamen University2021-11-161. 具有簡(jiǎn)單反應(yīng)級(jí)數(shù)的反應(yīng)動(dòng)力

2、學(xué)特點(diǎn)動(dòng)力學(xué)特點(diǎn)：動(dòng)力學(xué)特點(diǎn)：lnA0A= ktt1/2 = ln2/k一級(jí)反應(yīng)二級(jí)反應(yīng)= n nkt1A1A0- -A0 = B0t1/2 = 1/(n nkA0)n n= 1 or 2ln A0BB0A1B0-A0= ktA0 B0平均壽命平均壽命t t = 1/kt1/2 : 2t1/2 : 3t1/2 = 1 : 2 : 32012 Physical Chemistry Xiamen University2021-11-162. 反應(yīng)級(jí)數(shù)、速率系數(shù)計(jì)算，速率方程的建立F積分法F微分法反應(yīng)級(jí)數(shù)反應(yīng)級(jí)數(shù)1級(jí)反應(yīng)k =1 t A0Aln2級(jí)反應(yīng)111tAA0- -k = A0= B0 初速法

3、lnr0 = lnk + nlnA0從從r0A0求求n2012 Physical Chemistry Xiamen University2021-11-16解t/h 4 8 12 16 /(mg/100cm3) 0.480 0.326 0.222 0.151 某抗菌素注入人體后，在血液中呈現(xiàn)簡(jiǎn)單的級(jí)數(shù)反應(yīng)，如果在人體中某抗菌素，然后在不同時(shí)間 t 測(cè)定它在血液中的濃度得到 下面的數(shù)據(jù)(1) 反應(yīng)級(jí)數(shù) (2) 反應(yīng)速率常數(shù) (3) 半衰期 (4) 若要使血液中抗菌素濃度不低于0.37mg/100cm3，問(wèn)幾小時(shí)后注射第二針？ k =1 t 0ln2012 Physical Chemistry X

4、iamen University2021-11-16例例2 對(duì)于對(duì)于A + B + C P的反應(yīng)，改變的反應(yīng)，改變A0、B 0和和C0 測(cè)得下列一組數(shù)據(jù)：測(cè)得下列一組數(shù)據(jù)：實(shí)驗(yàn)次數(shù)實(shí)驗(yàn)次數(shù)A0 /moldm-3B0 /moldm-3C0 /moldm-3P增加增加0.0020 moldm-3所需所需的時(shí)間的時(shí)間 /s10.100.0500.104020.100.0500.154008040.200.0500.1014試確定該反應(yīng)對(duì)各反應(yīng)物的級(jí)數(shù)，并計(jì)算速率常數(shù)。試確定該反應(yīng)對(duì)各反應(yīng)物的級(jí)數(shù)，并計(jì)算速率常數(shù)。5310,11P0.00205.0 10 mol dms40rt

5、-5310,22P0.00205.0 10 mol dms40rt-該題需要以平均速率代替反應(yīng)初期的瞬時(shí)速率（須注意這只有在反應(yīng)時(shí)該題需要以平均速率代替反應(yīng)初期的瞬時(shí)速率（須注意這只有在反應(yīng)時(shí)間較短或反應(yīng)速率較慢時(shí)才是可行的，否則誤差較大）。由此可求得上述間較短或反應(yīng)速率較慢時(shí)才是可行的，否則誤差較大）。由此可求得上述四組實(shí)驗(yàn)中四組實(shí)驗(yàn)中P的初始生成速率為的初始生成速率為解2012 Physical Chemistry Xiamen University2021-11-165310,33P0.00202.5 10 mol dms80rt-5310,44P0.002014.3 10 mol dm

6、s14rt-50,150,415 100.35214.3 10rr-lg0.351.5lg0.5設(shè)速率方程為設(shè)速率方程為r = kABC根據(jù)第根據(jù)第(1)和第和第(4)組數(shù)據(jù)，可得組數(shù)據(jù)，可得因此因此 同理，根據(jù)第同理，根據(jù)第(1)和第和第(3)組數(shù)據(jù)，可得組數(shù)據(jù)，可得 = 1同理，根據(jù)第同理，根據(jù)第(1)和第和第(2)組數(shù)據(jù)，可得組數(shù)據(jù)，可得 = 0因此該反應(yīng)對(duì)因此該反應(yīng)對(duì)A、B、C的級(jí)數(shù)分別為的級(jí)數(shù)分別為1.5、1、0。 將各組初始速率的實(shí)驗(yàn)數(shù)據(jù)代入速率方程即可得到速率常數(shù)將各組初始速率的實(shí)驗(yàn)數(shù)據(jù)代入速率方程即可得到速率常數(shù)k 7.910-5 (moldm-3)1/2s-1 2012 Ph

7、ysical Chemistry Xiamen University2021-11-16例例3：氣相反應(yīng)中的總壓與反應(yīng)物壓力的關(guān)系：氣相反應(yīng)中的總壓與反應(yīng)物壓力的關(guān)系 對(duì)分子數(shù)發(fā)生變化的氣相反應(yīng)常用總壓來(lái)描述反應(yīng)進(jìn)程，要對(duì)分子數(shù)發(fā)生變化的氣相反應(yīng)常用總壓來(lái)描述反應(yīng)進(jìn)程，要學(xué)會(huì)學(xué)會(huì)從總壓求反應(yīng)物壓力，或兩者間的關(guān)系，從而求速率系數(shù)從總壓求反應(yīng)物壓力，或兩者間的關(guān)系，從而求速率系數(shù)。t/min 6.5 13.0 26.5 52.6 pt/kPa 54.4 65.1 83.2 103.9 124.1例：對(duì)二甲醚的分解CH3OCH3(g) CH4 (g) + H2 (g) + CO(g)測(cè)得777 K

8、時(shí)總壓和時(shí)間呈如下的關(guān)系，求反應(yīng)級(jí)數(shù)和速率系數(shù)。CH3OCH3(g) CH4 (g) + H2 (g) + CO(g)t = 0 p0 0 0 0t = t pA p0- pA p0- pA p0- pAt = 0 p0 p0 p0 解析:總壓ptp03p0 -2pA3p0 2012 Physical Chemistry Xiamen University2021-11-16氣相反應(yīng)中的總壓與反應(yīng)物壓力的關(guān)系氣相反應(yīng)中的總壓與反應(yīng)物壓力的關(guān)系假設(shè)一級(jí)反應(yīng)，則t/min 0 6.5 13.0 26.5 52.6 pt/kPa 54.4 65.1 83.2 103.9 124.1pA/kPa 41

9、.4 34.9 29.5 20.5 10.1 0嘗試法：k = 1/t ln(pA0/pA)t/min 0 6.5 13.0 26.5 52.6 pA/kPa 41.4 34.9 29.5 20.5 10.1 0k/10-4 s-1 4.38 4.36 4.43 4.47 故：故：n = 1k = 4.41 10-4 s-1基本為常數(shù)2012 Physical Chemistry Xiamen University2021-11-16F半衰期法（分?jǐn)?shù)壽期法）1級(jí)反應(yīng)t1/2 =ln2k與起始濃度無(wú)關(guān)與起始濃度無(wú)關(guān)2級(jí)反應(yīng)t1/2 =1kA0與起始濃度成反比與起始濃度成反比t1/2 : t3/4

10、 : t7/8 = 1 : 2 : 350%:75%:87.5%t1/2 : t3/4 : t7/8 = 1 : 3 : 750%:75%:87.5%3. 半衰期、反應(yīng)時(shí)間的計(jì)算3級(jí)反應(yīng)t1/2 =32kA02t1/2 : t3/4 : t7/8 = 1 : 5 : 2150%:75%:87.5%與起始濃度平方成反比與起始濃度平方成反比2012 Physical Chemistry Xiamen University2021-11-163. 半衰期、反應(yīng)時(shí)間的計(jì)算半衰期、反應(yīng)時(shí)間的計(jì)算t1/2 =(n-1)kA0n-1n級(jí)反應(yīng)11/21/2ntata-1/21/2ln(/ )1ln( / )t

11、tna a 或 該式適用于任一分?jǐn)?shù)衰期，起始濃度可能為起始?jí)毫υ撌竭m用于任一分?jǐn)?shù)衰期，起始濃度可能為起始?jí)毫?n-1-12012 Physical Chemistry Xiamen University2021-11-161級(jí)反應(yīng)3. 半衰期、反應(yīng)時(shí)間的計(jì)算計(jì)算反應(yīng)時(shí)間計(jì)算反應(yīng)時(shí)間0A11111ln = ln= lnA1tkkykz-y反應(yīng)掉的份數(shù)反應(yīng)掉的份數(shù)z剩余的份數(shù)剩余的份數(shù)2012 Physical Chemistry Xiamen University2021-11-16從測(cè)量物理性質(zhì)確定反應(yīng)級(jí)數(shù)AA0=l l l0 l 對(duì)對(duì)1 1級(jí)反應(yīng)級(jí)反應(yīng)可使用的物理量包括：總壓總壓、膨脹值、旋

12、光度、吸光度、折射率、電導(dǎo)率、電動(dòng)勢(shì)、粘度等等對(duì)對(duì)2 2級(jí)反應(yīng)級(jí)反應(yīng)0011Aktllll-00AlnlnAktllll-2012 Physical Chemistry Xiamen University2021-11-16Consider a reaction0 = nBB設(shè)l0、l及l(fā)分別是時(shí)間為0, t及時(shí)體系中某物理量的值；B0, B為0及t時(shí)刻的某物種的濃度；A為某反應(yīng)物, t=時(shí)反應(yīng)完全，當(dāng)反應(yīng)進(jìn)度為x，體積為V時(shí)，l0 lM fBB0l = lM + fB(B0 + nB x/V)l = lM + fB(B0nB A0/ nA )l0 l = A0 fB nB / nAl l =

13、 A fB nB / nAAA0=l ll0 l2012 Physical Chemistry Xiamen University2021-11-164. 反應(yīng)速率常數(shù)與溫度的關(guān)系、活化能的計(jì)算（1）指數(shù)式： )exp(RTEAka-（2）對(duì)數(shù)式：BRTEk-alnArrhenius公式公式（4）定積分式)11(ln2112TTREkk-aa2dlndEkTRT（3）微分式k1dkdT=EaRT22012 Physical Chemistry Xiamen University2021-11-16Part II重點(diǎn)重點(diǎn)1. 三類典型的復(fù)合反應(yīng)的動(dòng)力學(xué)特點(diǎn)三類典型的復(fù)合反應(yīng)的動(dòng)力學(xué)特點(diǎn)2. 運(yùn)用

14、穩(wěn)態(tài)近似法和平衡假設(shè)從機(jī)理推導(dǎo)速率方程運(yùn)用穩(wěn)態(tài)近似法和平衡假設(shè)從機(jī)理推導(dǎo)速率方程3. 從鍵能估算簡(jiǎn)單基元反應(yīng)活化能從鍵能估算簡(jiǎn)單基元反應(yīng)活化能4. 從基元反應(yīng)活化能求復(fù)合反應(yīng)活化能從基元反應(yīng)活化能求復(fù)合反應(yīng)活化能5. 推測(cè)反應(yīng)歷程推測(cè)反應(yīng)歷程Chapter 8 Chemical Kinetics22012 Physical Chemistry Xiamen University2021-11-16對(duì)峙反應(yīng)的微分式1-1級(jí)對(duì)峙反應(yīng)A Bk1k-1dAdt= k1A-k-1BA0+B0=A+B=Ae+Bek1/k -1 =Be/Ae1 各物質(zhì)間的質(zhì)量關(guān)系2 平衡關(guān)系B = Ae+ Aek1/k -

15、1A dAdt= (k1+ k-1)(A- Ae)ln A0-Ae= (k1 + k-1) t A-Aet = 1/(k1 + k-1)K= Be/Ae= k1/k-12012 Physical Chemistry Xiamen University2021-11-16Relaxation Time for 1st-order Reversible Reactions1-1級(jí)對(duì)峙反應(yīng)級(jí)對(duì)峙反應(yīng)A BkfkrIntegrated rate equation:0efreAAln()A Akk t-A- -Ae = (A0- -Ae) exp- -(kf + kr) tAssume微擾微擾 0 0

16、AAe - - AA0， AAe - -AA 0 0exp- -(kf + kr) t = 0 0exp (- -t/t tR)t tR = 1/(kf + kr)以實(shí)驗(yàn)方法監(jiān)測(cè)以實(shí)驗(yàn)方法監(jiān)測(cè) t，求出求出t tR，再結(jié)合平衡常數(shù)分別求再結(jié)合平衡常數(shù)分別求kf 及及 kr2012 Physical Chemistry Xiamen University2021-11-16Example: Relaxation Time for 2nd-order Reversible Reactions2-2級(jí)對(duì)峙反應(yīng)級(jí)對(duì)峙反應(yīng)A + BkfkrC + DAssume: A Ae - ; B Be - ; C

17、 Ce ; D De ; kf AAe BBe= kr C e D e= - kf (Ae Be) + kr(Ce De) (kf - kr) 2對(duì)于很小的擾動(dòng)值對(duì)于很小的擾動(dòng)值, 20dAABCDdfrkkt-eeeed(A)(B)(C)(D)dfrkkt- -2012 Physical Chemistry Xiamen University2021-11-16Relaxation Time for 2nd-order Reversible Reactions可見(jiàn)，可見(jiàn)， 向平衡的趨近可近似為一級(jí)向平衡的趨近可近似為一級(jí)弛豫時(shí)間弛豫時(shí)間eeeed(AB )(CD )dfrkkt-Reeee1

18、(AB )(CD )frkkt2012 Physical Chemistry Xiamen University2021-11-16一些重要對(duì)峙反應(yīng)弛豫時(shí)間一些重要對(duì)峙反應(yīng)弛豫時(shí)間一些重要類型反應(yīng)的1/tR表達(dá)式反應(yīng)弛豫時(shí)間倒數(shù)表達(dá)式1/tRA Bkfkrkf + krA + BkfkrC + DA + BkfkrCAkfkrB + C2A A2kfkr4kf Ae+ krkf (Ae Be) + kr(Ce De)kf (Ae Be) + krkf + kr(Be Ce)2012 Physical Chemistry Xiamen University2021-11-16具有相同級(jí)數(shù)的平行反

19、應(yīng)ABCDk1k2k3dAdt= (k1+ k2 + k3) An = k表An（1）k表 = k1+ k2 + k3 = ki（2）當(dāng)產(chǎn)物的初始濃度為0時(shí)，B:C:D = k1:k2:k3（3）B/C = k1/k2 = exp( ) E1-E2RT如E1E2,可以通過(guò)調(diào)節(jié)溫度來(lái)調(diào)變產(chǎn)物組成；也可以通過(guò)改變活化能，如加催化劑的方式改變產(chǎn)物組成。平行反應(yīng)的速?zèng)Q步驟？：快步驟2012 Physical Chemistry Xiamen University2021-11-16例4：具有相同級(jí)數(shù)的平行反應(yīng)例題題目：具有相同級(jí)數(shù)的下列平行反應(yīng)對(duì)A, B均為一級(jí)若C0=D0=0，推出C/D用溫度和活化

20、能表示的表達(dá)式；討論提高主產(chǎn)物F選擇性的動(dòng)力學(xué)途徑。A + BG + CG + Dk1, E1k2, E2解析：C/D = k1/k2 = exp(E2-E1)/RT途徑有2條(1) 使用適當(dāng)?shù)拇呋瘎┦笶2-E1增加(2) 若E2E1，則應(yīng)降低溫度，但此時(shí)反應(yīng)速率會(huì)降低； 若E1E2，則應(yīng)升高溫度，此時(shí)反應(yīng)速率也會(huì)升高。G + F2012 Physical Chemistry Xiamen University2021-11-16連續(xù)反應(yīng)的微、積分式等A B Ck1k2設(shè)t=0時(shí)，A = A0, B0 = C0 =0dAdt= k1 AdBdt= k1 A k2 BdCdt= k2 B(1)(

21、2)(3)從(1)式得A = A0e-k1t代入(2)式得dBdt= k1A0e-k1tk2 B解一階線性微分方程得B =(e-k1te-k2t)k1A0k2-k1C = A0e-k1tk2k2-k1k1k2-k1e-k2t+(1-)2012 Physical Chemistry Xiamen University2021-11-16連續(xù)反應(yīng)的特點(diǎn): 中間產(chǎn)物極大值中間產(chǎn)物濃度B會(huì)出現(xiàn)極大值121021AB()k tk tkeekk-21102121AdB0dk tk tkk ek etkk-21212121lnlnln(/)mkkkktkkkk-2121k tk tk ek e-2012 P

22、hysical Chemistry Xiamen University2021-11-16連續(xù)反應(yīng)的特點(diǎn): 中間產(chǎn)物極大值221ln(/)021A1/1mk tkkeekk-1210max21AB()mmk tk tkeekk-221()1021A1mmk tkktkeekk-2012 Physical Chemistry Xiamen University2021-11-16從鍵能估算活化能的方法(1) 分子裂解為自由基的反應(yīng)Ea H = DCl-Cl(2) 自由基復(fù)合反應(yīng)， 如Ea 0對(duì)于放熱方向有：Ea = DBC 5.5%(3) 自由基和分子間的反應(yīng)A + BC AB C AB + C

23、對(duì)于吸熱方向有：Ea = Ea rHm (H0)Ea = (EA-B + EC-D) 30%(4) 分子間的基元反應(yīng)對(duì)于放熱方向有：A-B + C-D A-C + B-D對(duì)于吸熱方向有：Ea = Ea rHm (H0)2012 Physical Chemistry Xiamen University2021-11-16正反應(yīng)和逆反應(yīng)活化能間關(guān)系微觀可逆性原理：一個(gè)基元反應(yīng)的逆過(guò)程一定也是一個(gè)基元反應(yīng)，而它在正、反兩個(gè)方向上進(jìn)行時(shí)都一定經(jīng)過(guò)同一個(gè)活化體，此過(guò)程表示如下：A + B(AB)*C + D根據(jù)微觀可逆性原理，正、逆反應(yīng)的活化能Ea和Ea與反應(yīng)焓變間的關(guān)系為：EaEa = H2012 P

24、hysical Chemistry Xiamen University2021-11-16Activation Energy of a Composite Reaction-11. kobs = kiaiExample:Consider the following mechanism for A + B PA + B Dk1k-1D P rdsk2Assume a Pre-equilibriumdPdt=k2D = (k2k1/k-1)ABkobs = k2k1k-1-1 Ea = RT2dlnkdT= RT2dlnkiaidT RT2dlnki dT=ai()=aiEiEa= E2 + E1

25、E-12012 Physical Chemistry Xiamen University2021-11-16Activation Energy of a Composite Reaction-22. kobs = kiFor parallel reactionsABCDk1k2k3kobs = k1 + k2 + k3 iaiia332a21a1332322221121321222a)dd()dd()dd(1)dddddd(1)dd1(dlndkEkkEkEkEkTkkRTkTkkRTkTkkRTkkTkTkTkkRTTkkRTTkRTEnnkkkktAA)(ddAobs321-2012 Ph

26、ysical Chemistry Xiamen University2021-11-16例5：穩(wěn)態(tài)近似法:從機(jī)理推導(dǎo)速率方程O(píng)3 O2 + O (1)k1k-1O + O3 2O2 (2)k2Oss = k1 O3k-1 O2 + k2 O3r = k1k2 O32k-1 O2 + k2 O3dO3dt= 2k2O O3 =例例 2O3 3O2 dOdt= k1 O3 - k-1 O2O - k2OO3 = 0= (k1 O3 - k-1 O2O k2OO3)2012 Physical Chemistry Xiamen University2021-11-16Empirical Rules f

27、or Mechanism-1規(guī)則I 如果由實(shí)驗(yàn)確定的總反應(yīng)的速率方程為r = k RiniRi為計(jì)量方程中出現(xiàn)的穩(wěn)定組分，ni為Ri的反應(yīng)級(jí)數(shù)，則速控步的反應(yīng)物元素總組成為niRi。根據(jù)平衡假設(shè)，ni3, 由于4分子反應(yīng)不大可能，因而速控步前有若干快速平衡存在。II-2若總反應(yīng)方程中某反應(yīng)物計(jì)量數(shù)反應(yīng)級(jí)數(shù)，則速控步之后必有該反應(yīng)物參加。II-3 某組分在速率方程中存在,而計(jì)量方程中不存在，該組分為催化劑，其級(jí)數(shù)為正，在速控步前或速控步反應(yīng)物一方，且在隨后的快速反應(yīng)中再生。級(jí)數(shù)為負(fù)，則在速控步前平衡的產(chǎn)物一方，且參與速控步之后的反應(yīng)被消耗。某組分在速率方程中存在某組分在速率方程中存在,而計(jì)量方程

28、中不存在，該組而計(jì)量方程中不存在，該組分為水溶液中的分為水溶液中的OH-或或H+, 可在反應(yīng)中使用可在反應(yīng)中使用H2O或產(chǎn)生或產(chǎn)生H2O2012 Physical Chemistry Xiamen University2021-11-16Empirical Rules for Mechanism-3規(guī)則III 反應(yīng)A+BP, 如速率方程中, A出現(xiàn)分?jǐn)?shù)級(jí)數(shù)，則在速控步前有反應(yīng)物分子的離解平衡，一般有兩種: (1)A離解產(chǎn)生中間物直接參加速控步; (2) B解離的中間物與A反應(yīng)，產(chǎn)生另一個(gè)中間物，參加速控步。規(guī)則規(guī)則IV 若反應(yīng)無(wú)簡(jiǎn)單級(jí)數(shù)，速率方程的分母為幾項(xiàng)若反應(yīng)無(wú)簡(jiǎn)單級(jí)數(shù)，速率方程的分母為幾

29、項(xiàng)加和，這種反應(yīng)不同極限條件下有不同級(jí)數(shù)。此時(shí)可加和，這種反應(yīng)不同極限條件下有不同級(jí)數(shù)。此時(shí)可從極限情況入手，運(yùn)用前幾條規(guī)則推出各極限時(shí)的機(jī)從極限情況入手，運(yùn)用前幾條規(guī)則推出各極限時(shí)的機(jī)理，再由極限于推廣至一般。理，再由極限于推廣至一般。2012 Physical Chemistry Xiamen University2021-11-16Empirical Rules for Mechanism-4規(guī)則V 根據(jù)化學(xué)變化的微觀可逆性原理和精細(xì)平衡原理，任一基元反應(yīng)的逆反應(yīng)具有相同但逆向的反應(yīng)途徑，因此，總反應(yīng)無(wú)論正向與逆向進(jìn)行，構(gòu)成其反應(yīng)歷程的基元反應(yīng)序列相同，只是方向相反，且速控步驟也可能不同

30、。2012 Physical Chemistry Xiamen University2021-11-16例例6：推測(cè)反應(yīng)機(jī)理：推測(cè)反應(yīng)機(jī)理For the hydrolysis of alkyl halides: RCl + OH= ROH + Cl(R=t-butyl)The rate law obtained from experiments is as follows:r = kRCl/(1 + kCl-/OH-).Propose a reasonable mechanism for this reaction.Answer: With rule IV, for the reaction

31、without simple reaction order, we should consider the extreme situations2012 Physical Chemistry Xiamen University2021-11-16Exampler = kRCl/(1 + kCl-/OH-)(1) Assume Cl-OH-, so kCl-/OH-1, r = k”RClOH-/Cl-. From this rate law，we can propose the following mechanismKRCl R+ + Cl fast equilibriumR+ + OH RO

32、H rdsk2r = k2R+OH-Consider the Pre-equilibrium：K = R+ Cl-/RClr = k2KRClOH- Cl-1k” = k2K(I)2012 Physical Chemistry Xiamen University2021-11-16ExamplesRCl R+ + Cl rdsR+ + OH ROH fastk2k1(2) Assume Cl-OH- so kCl-/OH-OH-, R+ + OH- may be the rate determining step, while at Cl-OH-, R+ + OH- is so fast th

33、at the decomposition of RCl will be the rate-determining step. So, in a common situation, the decomposition of RCl and the reaction of R+ + OH- may proceed in similar ratesUsing stead-state approximationdR+/dt = k1RCl-k2R+Cl-k3R+OH- = 0R+ss = k1RCl/(k2Cl- + k3OH-)r = dROH/dt = k3R+OH- = k1k3RClOH/(k

34、2Cl- + k3OH-) = kRCl/(1 + kCl-/OH-) consistent with experimental results (k = k1, k= k2/k3)R+ + OH ROHk3k1R+ + Cl RClk2RCl R+ + Cl(III)2012 Physical Chemistry Xiamen University2021-11-16重點(diǎn)重點(diǎn)1. 兩大理論的速率常數(shù)表達(dá)式兩大理論的速率常數(shù)表達(dá)式2. 活化能和活化能和反應(yīng)閾能反應(yīng)閾能Ec、活化能和活化能和活化焓的關(guān)系活化焓的關(guān)系3. 指前因子項(xiàng)的物理意義，碰撞次數(shù)（頻率）計(jì)算、指前因子項(xiàng)的物理意義，碰撞次數(shù)（

35、頻率）計(jì)算、活化熵的計(jì)算活化熵的計(jì)算Chapter 9 Rate Theories for Elementary Reactions2012 Physical Chemistry Xiamen University2021-11-16碰撞理論反應(yīng)速率常數(shù)表達(dá)方式考慮A + B Pr = = ZABq/LdAdtdNALdt= ZAB = p dAB2 NA NB( )1/28 RTpm= L2p dAB2 A B( )1/28 RTpmZAA = s NA NA = 2p dAA2 NA2( )1/2RTpMA L22p dAA2 A 2( )1/2RTpMAk單位：?jiǎn)挝唬簃ol-1 m3 s

36、-11/22cAB8( )exp()ERTk TL dRTppm-2012 Physical Chemistry Xiamen University2021-11-16閾能與實(shí)驗(yàn)活化能的關(guān)系閾能與實(shí)驗(yàn)活化能的關(guān)系實(shí)驗(yàn)活化能的定義：實(shí)驗(yàn)活化能的定義：TkRTEdlnd2a將與T無(wú)關(guān)的物理量總稱為B：c1 lnlnln2EkTBRT -有TRTETk21dlnd2cRTEE21ca碰撞理論計(jì)算速率常數(shù)的公式：k(T) = Lp dAB2( )1/28 RTpmexp ( )EcRT將Ec= Ea1/2RT 代入并比較得指前因子:A = Lp dAB2( )1/28 RTepmA一般在一般在1010

37、1011 mol-1dm3s-1之間之間2012 Physical Chemistry Xiamen University2021-11-16過(guò)渡態(tài)理論反應(yīng)速率常數(shù)表達(dá)式過(guò)渡態(tài)理論反應(yīng)速率常數(shù)表達(dá)式0Bjexp()jEk TqkLhqRT-ABC0BABCexp()qEk TkLhq qRT-A + BC ABC AB + C對(duì)任一基元反應(yīng)對(duì)任一基元反應(yīng)2012 Physical Chemistry Xiamen University2021-11-16Structureless Particles (Diatomic) ReactionsFor reactions between struc

38、tureless particlesA + BAB PqA = qA,t/V = (2p pmAkBT/h2)3/2qB = qB,t/V = (2p pmBkBT/h2)3/2Being the same with that from SCT, E0 corresponds to Ec0BABABexp()Ek T qkLhq qRT-For A and B, only translational terms contributed to their partition functions3/22ABBABABABtr222(+)8/()Bmm k Tm mmmk Tqq qhhp21/20

39、BAB8(+ ) ()exp()Ek TkLrrRTm-概率因子概率因子P= 12012 Physical Chemistry Xiamen University2021-11-16過(guò)渡態(tài)理論預(yù)測(cè)概率因子過(guò)渡態(tài)理論預(yù)測(cè)概率因子1. 原子A + 原子B雙原子活化絡(luò)合物 PA L ft3 fr2ft3 ft3kBTh= Lfr2ft3kBTh= 1012 dm3 mol-1 s-1P 12. 原子A + 線性B(N)線性活化絡(luò)合物(N+1) PA Lft3 fr2 fv3(N+1)-6ft3 ft3 fr2 fv3N-5kBTh= Lfv2ft3kBTh= 1010 dm3 mol-1 s-1P

40、= fv2/fr2 10-23. 原子A + 線性B(N)非線性活化絡(luò)合物(N+1) PA Lft3 fr3 fv3(N+1)-7ft3 ft3 fr2 fv3N-5kBTh= Lfr fvft3kBTh= 1011 dm3 mol-1 s-1P = fv/fr 10-12012 Physical Chemistry Xiamen University2021-11-16過(guò)渡態(tài)理論的熱力學(xué)處理過(guò)渡態(tài)理論的熱力學(xué)處理1B1B/Brm()e p)x(ncnCck Tk TGkKchhkTcKRTh-$1Brmrm()expexpnk TSHkchRRT-$2012 Physical Chemist

41、ry Xiamen University2021-11-16標(biāo)準(zhǔn)活化內(nèi)能、活化焓的計(jì)算標(biāo)準(zhǔn)活化內(nèi)能、活化焓的計(jì)算2armdln dkERTURTT $1B/()nc ck TkcKh-$對(duì)凝聚相反應(yīng)：對(duì)凝聚相反應(yīng)：armEHRT $對(duì)氣相反應(yīng)：對(duì)氣相反應(yīng)：( (設(shè)設(shè)n為氣相反應(yīng)物分子數(shù)為氣相反應(yīng)物分子數(shù)) )armEHnRT $標(biāo)準(zhǔn)活化焓的計(jì)算標(biāo)準(zhǔn)活化焓的計(jì)算標(biāo)準(zhǔn)活化內(nèi)能的計(jì)算標(biāo)準(zhǔn)活化內(nèi)能的計(jì)算rma UERT-$2012 Physical Chemistry Xiamen University2021-11-16活化熵與指前因子間的關(guān)系活化熵與指前因子間的關(guān)系由上式可計(jì)算標(biāo)準(zhǔn)活化熵由上式可計(jì)

42、算標(biāo)準(zhǔn)活化熵1Brmrm()expexpnk TSHkchRRT-$2. For liquid-phase reactions:armEHRT $1. For gas-phase reactions:armEHnRT $1Brme ()exp()nnk TSAchR-$1Brme()exp()nk TSAchR-$2012 Physical Chemistry Xiamen University2021-11-16活化熵和活化自由能的計(jì)算活化熵和活化自由能的計(jì)算rmrmrmGHTS - $1Brme ()exp()nnk TSAchR-$對(duì)雙分子氣相反應(yīng)：對(duì)雙分子氣相反應(yīng)：RTpc2012 P

43、hysical Chemistry Xiamen University2021-11-16單分子反應(yīng)理論：?jiǎn)畏肿臃磻?yīng)理論：Lindemann MechanismThe first successful explanation of “unimolecular reactions” was given by Lindemann in 1921, and then elaborated by Hinshelwood.1) Reactant molecule A becomes energetically excited (A*) by collision with another molecule

44、 (A); 2) A* might lose its excess energy by collision with another molecule or might form product P “Time lag” hypothesispseudo-unimolecular reaction2012 Physical Chemistry Xiamen University2021-11-16Lindemann Mechanismk1A + MA* + Mk-1A*k2PMA or other inert moleculesr = k2A*dA*dt= k1AM k-1A*M k2A*=0

45、A*ss= k1 M A /(k-1M + k2)r = k2A* =k1 k2MAk-1M + k2Using steady-state approximationku=k1 k2Mk-1M + k2kuA2012 Physical Chemistry Xiamen University2021-11-16Reaction-order of “Unimolecular Reactions”(2) At low concentration, k-1M k2ku= k = k1 k2/ k-1 First-orderr =k-1k1 k2Aku=k1 k2Mk-1M + k2r = kuAk-1 k2k2 k12012 Physical Chemistry Xiamen University2021-11-16Reaction-or