18.10 从实验数据确定费率法

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  从实验数据中确定费率法

  

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  How fast?
  ::有多快? 有多快? How fast?

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  Determining the amount of time a process requires calls for a timer. These devices can be simple kitchen timers (not very precise) or complex systems that can measure to a fraction of a second. Accurate time measurement is essential in kinetics studies for assessing rates of .
  ::确定一个过程需要一个计时器的时间。 这些装置可以是简单的厨房计时器(不太精确)或复杂的系统,可以测量到秒的一小部分。 精确的时间测量对于评估...的动能研究至关重要。

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  Determining the Rate Law from Experimental Data
  ::从实验数据中确定费率法

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  In order to experimentally determine a rate law , a series of experiments must be performed with various starting of reactants . The initial rate law is then measured for each of the reactions. Consider the reaction between nitrogen monoxide and hydrogen gas to form nitrogen gas and water vapor .
  ::为了实验性地确定比例法,必须对各种反应器进行一系列实验,然后对每一种反应进行初始比例法的测量。考虑一氧化氮和氢气之间的反应,形成氮气和水蒸气。

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  $$\begin{array}{r}2\text{NO}(g)+2{\text{H}}_2(g)\to {\text{N}}_2(g)+2{\text{H}}_2\text{O}(g)\end{array}$$

  
  ::2NO(g)+2H2(g)+N2(g)+2H2(g)+2H2O(g)

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  The following data were collected for this reaction at 1280°C ( Table ).
  ::以下数据是在1280°C时为这一反应收集的(表)。

  	[NO]	[H 2 ]	Initial Rate (M/s)
  1	0.0050	0.0020	$\begin{array}{r}1.25\times {10}^{-5}\end{array}$
  2	0.010	0.0020	$\begin{array}{r}5.00\times {10}^{-5}\end{array}$
  3	0.010	0.0040	$\begin{array}{r}1.00\times {10}^{-4}\end{array}$

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  Notice that the starting concentrations of NO and H ₂ were varied in a specific way. In order to compare the rates of reaction and determine the order with respect to each reactant, the initial concentration of each reactant must be changed while the other is held constant.
  ::注意NO和H2的起始浓度因具体方式而不同。为了比较反应率和确定每个反应器的顺序,必须改变每个反应器的初始浓度,而另一反应器保持恒定。

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  Comparing experiments 1 and 2 : the concentration of NO was doubled, while the concentration of H ₂ was held constant. The initial rate of the reaction quadrupled, since $\begin{array}{r}\frac{5.00\times {10}^{-5}}{1.25\times {10}^{-5}}=4\end{array}$ . Therefore, the order of the reaction with respect to NO is 2. In other words, rate α [NO] ² . Because $\begin{array}{r}{2}^2=4\end{array}$ , the doubling of [NO] results in a rate that is four times greater.
  ::对比实验1和2:NO的浓度翻了一番,而H2的浓度保持不变。自5.00x10-51.25x10-55=4以来,最初的反应率翻了四倍。因此,NO的反应顺序是2.,换句话说,ERA[NO]2.,因为22=4,[NO]的倍数增加了四倍。

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  Comparing experiments 2 and 3 : the concentration of H ₂ was doubled while the concentration of NO was held constant. The initial rate of the reaction doubled, since $\begin{array}{r}\frac{1.00\times {10}^{-4}}{5.00\times {10}^{-5}}=2\end{array}$ . Therefore, the order of the reaction with respect to H ₂ is 1, or rate α [H ₂ ] ¹ . Because $\begin{array}{r}{2}^1=2\end{array}$ , the doubling of H ₂ results in a rate that is twice as great.
  ::比较实验2和实验3:H2的浓度翻了一番,而NO的浓度保持不变。最初的反应率翻了一番,自1.00x10-45.00x10-5=2以来,H2的反应率为1,或者说,H2的反应率为α[H2]1。因为21=2,H2的两倍导致H2的两倍,其反应率是两倍。

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  The overall rate law then includes both of these results.
  ::然后,总体费率法包括了这两项结果。

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  $$\begin{array}{r}\text{rate}=k[\text{NO}{]}^2[{\text{H}}_2]\end{array}$$

  
  ::速率=k[NO]2[H2]

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  The sum of the exponents is $\begin{array}{r}2+1=3\end{array}$ , making the reaction third-order overall. Once the rate law for a reaction is determined, the specific rate constant can be found by substituting the data for any of the experiments into the rate law and solving for $\begin{array}{r}k\end{array}$ .
  ::指数的总和是 2+1=3, 使反应第三阶整体。 一旦确定反应的费率法, 具体比率常数可以通过将任何实验的数据替换为速度法和 k 的计算方法来找到 。

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  $$\begin{array}{r}k=\frac{\text{rate}}{[\text{NO}{]}^2[{\text{H}}_2]}=\frac{1.25\times {10}^{-5}\text{ M/s}}{(0.0050\text{ M}{)}^2(0.0020\text{ M})}=250{\text{ M}}^{-2}{\text{s}}^{-1}\end{array}$$

  
  ::k=rat[NO]2[H2]=1.25×10-5 M/s(0.0050 M)(0.0020 M)=250 M-2s-1

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  Notice that the rate law for the reaction does not relate to the balanced equation for the overall reaction. The coefficients of NO and H ₂ are both 2, while the order of the reaction with respect to the H ₂ is only one. The units for the specific rate constant vary with the order of the reaction. So far, we have seen reactions that are first or second order with respect to a given reactant. Occasionally, the rate of a reaction may not depend on the concentration of one of the reactants at all. In this case, the reaction is said to be zero-order with respect to that reactant.
  ::注意反应的速率法与总体反应的平衡等式无关。NO和H2的系数是2,而H2的系数是2,而H2的系数是1。具体率常数的单位随反应的顺序而不同。到目前为止,我们看到的反应是某一反应的先后顺序。有时,反应的速度可能不取决于反应者之一的集中程度。在这种情况下,该反应者的反应据说是零顺序。

   

   

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  Summary
  ::摘要

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  • The process of using experimental data to determine a rate law is described.
    ::介绍了使用实验数据确定费率法的过程。

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  Review
  ::回顾

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  1. How do you carry out experiments for determining rate constants?
     ::您如何进行确定利率常数的实验?
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  2. Why is the reaction order with regard to NO a value of 2?
     ::为什么对NO的反应顺序值为2?
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  3. Why is the reaction order with regard to hydrogen value of 1?
     ::为什么氢值为1的反应顺序是1?