19.10 查特利尔原则与平衡常数

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• 选择节勒查特利尔原则与平衡常数

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  勒查特利尔原则与平衡常数

  

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  How much is in savings this month?
  ::这个月储蓄多少?

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  With online banking, management of your personal finances can become less complicated in some ways. You can automatically deposit paychecks, pay bills, and designate how much goes into savings or other special accounts each month. If you want to maintain 10% of your bank account in savings, you can set up a program that moves money in and out of the account when you get a paycheck or pay bills. The amount of money in savings will change as the money comes in and out of the bank, but the ratio of savings to checking will always be constant.
  ::在网上银行业务中,个人财务管理在某些方面会变得不那么复杂。您可以自动存款支票、账单和指定每个月存入储蓄账户或其他特殊账户的金额。如果您想要将10%的银行账户存入储蓄账户,您可以在拿到支票或账单时建立一个将资金从账户中流出的方案。当钱进出银行时,储蓄金额将发生变化,但储蓄与支票的比例将保持不变。

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  Le Châtelier’s Principle and the Equilibrium Constant
  ::勒查特利尔原则与平衡常数

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  Occasionally, when students apply Le Chatelier’s principle to an equilibrium problem involving a change in , they assume that  $\begin{align*}K_{eq}\end{align*}$ must change. This seems logical since we talk about “shifting” the equilibrium in one direction or the other. However,  $\begin{align*}K_{eq}\end{align*}$ is a constant, for a given equilibrium at a given temperature , so it must not change. Here is an example of how this works. Consider the simplified equilibrium below:
  ::偶尔,当学生对涉及变化的平衡问题适用勒查特利尔的原则时,他们认为克克必须改变。 这似乎合乎逻辑,因为我们谈论“将平衡转移到”一个方向或另一个方向。 然而,克克是一个常数,对于特定温度的某一平衡来说,它是一个常数,因此它不能改变。这里是这一平衡如何运作的一个例子。 想想下面的简化平衡:

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  $$\begin{align*}A \rightleftarrows B\end{align*}$$
  ::AB

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  Let’s say we have a 1.0 liter container. At equilibrium the following amounts are measured.
  ::假设我们有一个1.0升的容器。 在平衡时,测量以下数量。

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  $$\begin{align*}A &=0.50 \text{ mol} \\ B &=1.0 \text{ mol}\end{align*}$$
  ::A=0.50 molB=1.0 mol

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  The value of  $\begin{align*}K_{eq}\end{align*}$ is given by:
  ::Keq 值的给定值为 :

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  $$\begin{align*}K_{eq}= \frac{\left [B \right ]}{\left [A \right ]}=\frac{1.0 \text{ M}}{0.50 \text{ M}}=2.0\end{align*}$$
  ::Keq=[B][A]=1.0 M0.50 M=2.0

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  Now we will disturb the equilibrium by adding 0.50 mole of  $\begin{align*}A\end{align*}$ to the . The equilibrium will shift towards the right, forming more $\begin{align*}B\end{align*}$ . Immediately after the addition of  $\begin{align*}A\end{align*}$ and before any response, we now have 1.0 mol of  $\begin{align*}A\end{align*}$ and 1.0 mol of $\begin{align*}B\end{align*}$ . The equilibrium then shifts in the forward direction. We will introduce a variable $\begin{align*}(x)\end{align*}$ , which will represent the change in concentrations as the reaction proceeds. Since the of  $\begin{align*}A:B\end{align*}$ is 1:1, as $\begin{align*}[A] \end{align*}$ decreases by the amount $\begin{align*}x\end{align*}$ , the  $\begin{align*}[B]\end{align*}$ increases by the amount $\begin{align*}x\end{align*}$ . We set up an analysis called ICE , which stands for Initial, Change, and Equilibrium. The values in the table represent molar concentrations.
  ::现在,我们将通过在 A 中增加 0.50 摩尔来扰乱平衡。 平衡将转向右边, 形成更多的 B 。 在添加 A 之后, 在任何响应之前, 我们现在有1.0 摩尔 A 和 1.0 毫 摩尔 B。 平衡然后向前方向转移。 我们将引入一个变量( x) , 它将代表随反应过程而变化的浓度。 由于 A: B 是 1: 1, 1, 因为 [A] 减少 x , [B] 增加 x 。 我们设置了一个分析, 称为ICE, 代表初始、 变化和 平衡。 表中的值代表摩尔浓度 。

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  $$\begin{align*}& \qquad \qquad \qquad \ \ \underline{\;\; A \qquad \qquad B \; \; \; \; \; \; \; \;\;\;} \\ & \text{Initial} \qquad \qquad \ \ 1.0 \qquad \quad \ \ 1.0 \\ & \text{Change}\qquad \quad \ \ -x \qquad \quad \ +x \\ & \text{Equilibrium} \qquad 1.0 -x \qquad 1.0 +x \end{align*}$$
  ::AB_初步 1.0 1.0 变化-x +x 平衡-x1.0-x1.0+x

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  At the new equilibrium position, the values for  $\begin{align*}A\end{align*}$ and  $\begin{align*}B\end{align*}$ as a function of  $\begin{align*}x\end{align*}$ can be set equal to the value of the $\begin{align*}K_{eq}\end{align*}$ . Then, one can solve for $\begin{align*}x\end{align*}$ .
  ::在新的平衡位置上,A和B的值作为 x 的函数可以设定为等于 Keq 的值。然后,可以解决 x 的值。

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  $$\begin{align*}K_{eq}=2.0=\frac{\left [ B \right ]}{\left [ A \right ]}=\frac{1.0 + x}{1.0 - x}\end{align*}$$
  ::Keq=2.0=[B][A]=1.0+x1.0-x

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  Solving for $\begin{align*}x\end{align*}$ :
  ::x 的解决方案 :

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  $$\begin{align*}2.0(1.0-x) &=1.0 + x \\ 2.0 - 2.0 x &=1.0+x \\ 3.0 x &=1.0 \\ x &=0.33\end{align*}$$
  ::2.0(1.0-x)=1.0+x2.0-2.0x=1.0+x3.0x=1.0x=0.33

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  This value for  $\begin{align*}x\end{align*}$ is now plugged back in to the Equilibrium line of the table and the final concentrations of  $\begin{align*}A\end{align*}$ and  $\begin{align*}B\end{align*}$ after the reaction is calculated.
  ::X的这个值现在被塞回表格的平衡线和计算反应后A和B的最后浓度。

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  $$\begin{align*}\left [ A \right ]&=1.0-x=0.67 \text{ M} \\ \left [ B \right ]&=1.0+x=1.33 \text{ M}\end{align*}$$
  ::[A]=1.0-x=0.67 m[B]=1.0+x=1.33 m

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  The value of  $\begin{align*}K_{eq}\end{align*}$ has been maintained since $\begin{align*}\frac{1.33}{0.67}=2.0\end{align*}$ . This shows that even though a change in concentration of one of the substances in equilibrium causes a shift in the equilibrium position, the value of the equilibrium constant does not change.
  ::Keq值自1.330.67=2.0以来一直保持不变,这表明,即使平衡物质之一的浓度发生变化导致平衡位置发生变化,但平衡常值的价值并没有改变。

   

   

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  Summary
  ::摘要

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  • Maintenance of the constant  $\begin{align*}K_{eq}\end{align*}$ for a reaction is described.
    ::说明维持常数Keq用于反应的情况。

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  Review
  ::回顾

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  1. Does  $\begin{align*}K_{eq}\end{align*}$ change for a given reaction at a given temperature?
     ::Keq 在特定温度下是否改变特定反应 ?
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  2. What does ICE stand for?
     ::ICE代表什么?
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  3. Will the equilibrium position change if materials are added to or removed from the reaction?
     ::如果将材料添加到反应中或从反应中去除,平衡位置是否会改变?
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  4. How does addition or removal of materials affect the $\begin{align*}K_{eq}\end{align*}$ ?
     ::材料的添加或移走对Keq有何影响?