公式也可以表示成k=Aexp(-Ea/RT) 一般化学反应的活化能在40~400kJmo-范围之内(exp: exponent) Reactions Solvent Ea/kJ·mol-l CH3 COOC2H5 NaOH 水 47.3 n-CsH1Cl+KI 丙酮 77.0 ChS ONa +CH3I 乙醇 CHs Br+ Naoh 乙醇 2HI一 气相 H+I- 2HI 气相 165.3 N2O4+1/20 气相 03.4 (CH2)3 CH3 CH=CH 气相 272.0 E 1 对 Arrhenius公式两边取自然对数,得:lnk=lnA-(),以nk对1T作图,是一 条直线,其斜率为-Ea/R,截距为ln4。因此活化能可以通过实验来测定:用在不同温度 下观察到的k值的自然对数对1/T作图,斜率为-Ea/R,从而求得Ea(见图3.5) 4.不同温度下,速率常数之间的关系 已知温度为T1时,速率常数为k1:温度为T2时,速率常数为k。由 Arrhenius uation得:lnk=lnA-Ea/Rn①,ink=lnA-Ea/R72② ②得 Ink,-Ink, =- RT-7)或者ln()=E1_1 Sample Exercise 1: The following table shows the rate constants for the rearrangement of methyl isonitrile at various temperature at various temperatures(these are the data that are graphed Temperature(℃) k(s-1) 189.7 2.52×10- 1989 5.25×10-5 230.3 10 6.30×10 2512 3.16×10-3 (a) From these data calculate the Fig 3. 6 The natural logarithm of the rate constant for the rearrangement of methyl isonitrile as a function activation energy for the reaction. (b)what is the value of the rate constant at 4300K? Solution:(a)We must first convert the temperatures from degrees Celsius to Kelvins. We then take the inverse of each temperature, 1/T, and the natural log of each rate constant, Ink. This gives us the following table 7(K) 462.9 472.1 1/7(K-1)2.160×10-32.118×103 1986×10 1.907×10-3 -10.589 9.855 7.370 5.757 The slope of the line is obtained by choosing two well-separated points55 公式也可以表示成 k = A·exp (− Ea / RT ) 一般化学反应的活化能在 40～400 kJ · mol−1 范围之内（exp.：exponent） Reactions Solvent Ea / kJ · mol−1 CH3COOC2H5 + NaOH 水 47.3 n − C5H11Cl + KI 丙酮 77.0 C2H5ONa + CH3I 乙醇 81.6 C2H5Br + NaOH 乙醇 89.5 2HI I2 + H2 气相 184.1 H + I 2HI 气相 165.3 N2O5 N2O4 + 1/2O2 气相 103.4 (CH2)3 CH3CH＝CH2 气相 272.0 对 Arrhenius 公式两边取自然对数，得： 1 ln ln ( ) E a k A R T = − ，以 lnk 对 1/T 作图，是一 条直线，其斜率为 − Ea / R，截距为 lnA。因此活化能可以通过实验来测定：用在不同温度 下观察到的 k 值的自然对数对 1/T 作图，斜率为 − Ea / R，从而求得 Ea（见图 3.5）。 4．不同温度下，速率常数之间的关系 已知温度为 T1 时，速率常数为 k1；温度为 T2 时，速率常数为 k2。由 Arrhenius equation 得： lnk1 = lnA − Ea / RT1 ① ， lnk2 = lnA − Ea / RT2 ② ① − ② 得： ) 1 1 ln ln ( 1 2 a 1 2 R T T E k − k = − − 或者 1 a 2 2 1 1 1 ln( ) ( ) k E k R T T = − Sample Exercise 1：The following table shows the rate constants for the rearrangement of methyl isonitrile at various temperature at various temperatures (these are the data that are graphed in right figure： Temperature (℃) k (s−1 ) 189.7 2.52  10−5 198.9 5.25  10−5 230.3 6.30  10−4 251.2 3.16  10−3 (a) From these data calculate the activation energy for the reaction. (b) What is the value of the rate constant at 430.0 K ? Solution：(a) We must first convert the temperatures from degrees Celsius to Kelvins. We then take the inverse of each temperature, 1 / T, and the natural log of each rate constant, lnk. This gives us the following table： T( K ) 462.9 472.1 503.5 524.4 1 / T ( K−1 ) 2.160  10−3 2.118  10−3 1.986  10−3 1.907  10−3 lnk −10.589 −9.855 −7.370 −5.757 The slope of the line is obtained by choosing two well-separated points： Fig 3.6 The natural logarithm of the rate constant for the rearrangement of methyl isonitrile as a function of 1/T