【文章內(nèi)容簡介】 x+y) 反應(yīng)級數(shù)可以是整數(shù)、零、分數(shù)或小數(shù) n can be a integer, zero or fraction even negative . The order of a reaction It is the sum of the exponents on the concentration terms in the rate equation. aA+bB=cC+dD yBxA CkCv ?the order of reaction: n=(x+y) 若為元反應(yīng)，則化學(xué)反應(yīng)速率方程： V=kc a (A)c b (B) x=a； y=b 反應(yīng)級數(shù) = a+b=反應(yīng)分子數(shù) 如： NO2+CO=CO2+NO (元反應(yīng) ) V=kc (CO)c (NO2) 反應(yīng)級數(shù) =1+1=2反應(yīng)分子數(shù) =1+1=2 aA+bB=cC+dD If it is a elementary reaction: abABv kc c?The order of the reaction = a+b =molecularity of the reaction x=a； y=b ： NO2+CO=CO2+NO (elementary reaction) 2C O N Ov k c c?The order of the reaction = 1+1=2 =molecularity of the reaction Reaction mechanism of CO+Cl2=COCl2 (phosgene)： (1)Cl2?2Cl（ quick） 212[ C l][ C l ]K ?322[ Cl ][ Cl ] [ Cl]K ?(2)Cl2+ Cl ? Cl3（ quick） (3)Cl3+CO→COCl2+ Cl （ slow， ratelimiting step） 3C l C Ov k c c?322 2 23 C l C O 3 2 C l C l C O323 2 C l 1 C l C O C l C Ov k c c k K c c ck K c K c c k c c? ? ???32C l 2 C l C lc K c c?2Cl 1 Clc K c?The overall reaction order is 1+3/2= ?反應(yīng)級數(shù)與反應(yīng)分子數(shù)的比較 反應(yīng)級數(shù) 反應(yīng)分子數(shù) 使用范圍 任何反應(yīng) 元反應(yīng) 取值 整數(shù)、分數(shù) 或負數(shù) 1， 2， 3 與反應(yīng)物系數(shù)和的關(guān)系 不一定相等 相等 一級反應(yīng) 反應(yīng)速率與反應(yīng)物濃度的一次方成正比的反應(yīng) 定積分得： ——一級反應(yīng)方程式 d c d cv k c k d td t c? ? ? ? ? ?00l n ktc k t c c ec??? 或0l g l g2 .3 0 3ktcc? ? ?Firstorder reactions A firstorder reaction is one whose rate depends on the concentration of a single reactant raised to the first power d c d cv k c k d td t c? ? ? ? ? ?000l n r l gc l gcktck t o c c eckt?????Definite integral Y = mx + b Characteristic of firstorder reaction 一級反應(yīng)特征 lgc t 0 The curve of firstorder reaction sl op e k?(1) (2) k is independent on the concentration. units： [time]1 . s1 , min1,h1 (3) The halflife ( t1/2) of a reaction is the time required for the concentration of a reactant to drop to one half of its initial value, c1/2=1/2c0 . 012012l n 12l n 2 93 cktctkk?????、 k的數(shù)值與濃度無關(guān) 量綱： [時間 ]1 → s1 、 min h1 ??、半衰期：當(dāng)反應(yīng)物濃度由 CO → 時所需時間為半衰期，用 t1/2表示 一級反應(yīng) : c0 ln ??? = kt1/2 t1/2= ln2/k 1/2co t1/2=021 CSecondorder reactions 二級反應(yīng) is one whose rate depends on the reactant concentration raised to the second power or on the concentrations of two different reactants each raised to the first 22 d c d cv k c k d td t c? ? ? ? ? ?Definite integral 011 ktcc??Y = mx + b —— 二級反應(yīng)方程式 1/c t 0 Slope=k Characteristic of secondorder reaction 二級反應(yīng)特征 01I n te r c e p tc?The curve of secondorder reaction (1) (2) k Units: Lmol1s 1（ min h1） (3) Half life ( t1/2) ： 1201tkc?Zeroorder reactions反應(yīng)速率與反應(yīng)濃度無關(guān) The rate of reaction is independent on the concentration of the reactant. 0 dcv k c kdt? ? ? ?Definite integral c = kt +co Y = mx + b —— 零級反應(yīng)方程式 c t 0 Characteristic of zeroorder reaction (1) Slope= k The curve of zeroorder reaction (2) k Units: mol L 1s 1（ min h1） (3) Half life ( t1/2) : 012 2ctk?簡單反應(yīng)級數(shù)的特征 反應(yīng)級數(shù) 一級反應(yīng) 二級反應(yīng) 零級反應(yīng) 方程式 lgc = kt/ +lgc0 1/c = kt +1/c0 c = kt + c0 直線關(guān)系 lgc∽ t 1/c∽ t c∽ t 斜率 k/ k k k單位 s1 Lmol1s 1 molL1s 1 t1/2 Summary Order First Second Zero Equation lgc =kt/ + lgc0 1/c=kt +1/c0 c=kt+c0 Straight line lgc～ t 1/c～ t c～ t Slope k/ k k k (unit) s1 Lmol1s 1 molL1s 1 t1/2 例 1 ?反應(yīng) 2N2O5→NO2+O2 服 從 速 率 方 程 式V=kCN2O5， k= 102S1,， 如在一個 的容器中放入 N2O5， 在該溫度下反應(yīng)進行 1min， 問 N2O5的剩余量及 O2的生成量各為多少 ？ ?解：從速率方程式 V=kCN2O5知 ， 反應(yīng)符合 一級反應(yīng) ， 且 C0=L1 根據(jù)一級反應(yīng)方程式： lgC0/C=kt/ 得 102 60/ c= molL1 則知 N2O5剩余量 n(N2O5)=cV= 5= 參加反應(yīng) N2O5量 n(N2O5)== 設(shè) O2的生成量為 x，則由反應(yīng)方程式得： 2N2O5→NO2+O2 2 1 x 生成氧的量： x=Example k is 102s1. Add N2O5 in a container of L, please calculate how much N2O5 remained and how much O2 created after 1 min. The rate equation of the reaction:2N2O5→NO2+O2 is listed below, 25 NOv k c?according to 25 NOv k c?The reaction is a firstorder reaction: Answer: 02l g 2 .3 0 30 .5 1 .6 8 1 0 6 0l g 0 .1 8 2 m o l /L2 .3 0 3 c ktccc????? ? ?102 .5 0 .5 m o l L5c ? ? ?Assume the amount of O2 is x， 2N2O5→NO2+O2 2 1 x x=N2O5 remains n(N2O5)=cV= 5= The consumption of N2O5 n(N2O5)== 例題 2 在肺部血液中存在 Hb+O2?HbO2反應(yīng) ， 對Hb及 O2均為一級反應(yīng) ， 血中 Hb正常濃度 106 molL1， O2 106 molL1,在 37℃時 ， k= 106mol1L 1S 1 計算： （ 1） 在肺臟血液中 ， HbO2的生成速率 ? （ 2） 在某種疾病中 ， V(HbO2)= 104 molL1S1,為保持 cHb不變 ， 需輸氧 ， 此時 c（ O2） 應(yīng)高達多少才行 ？ ?解：由于 Hb及 O2均為一級反應(yīng) ， 所以 (1) V(HbO2)=k cHb c（ O2） = 106 106 106 = 105 molL1S1 (2) c（ O2） = V(HbO2)/k cHb = 104/ 106 106 = 106 molL1 Example There is a reaction : H b O 2 H b O 2+in the blood of lung, it is firstorder in Hb (hemoglobin) and also firstorder in O2, the normal concentration of Hb and O2 are 106 molL1 and 106 molL1 37℃ ， k= 106mol1L 1s 1 (1) please calculate how quickly the HbO2 was created in the blood of lung (2) if suffering some kind of disease, V(HbO2)= 104 molL1s1 in order to keep cHb unchanged,how much the c(O2) is in oxygen therapy. (1) V(HbO2)=k cHb c(O2) = 106 106 106 = 105 molL1S1 (2) c(O2) = V(HbO2)/k cHb = 104/ 106 106 = 106