19.4 用平衡常数计算

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  与平衡常数的计算

  

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  Iron-poor blood?
  ::缺铁血?

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  Iron is an important component of red cells. Patients who have low iron will usually be anemic and have a lower than normal number of red blood cells. One way to assess serum iron is with the use of Ferrozine, a complex organic molecule. Ferrozine forms a product with Fe ³⁺ , producing a pink color. In order to determine factors affecting the reaction, we need to measure the equilibrium constant . If the equilibrium does not lie far in the direction of products, precautions need to be taken when using this material to measure iron in serum.
  ::铁是红细胞的重要组成部分。 低铁患者通常贫血,红细胞数量低于正常数量。 评估铁血清的方法之一是使用复杂的有机分子Ferrozine。 Ferrozine 以Fe3+制成一种产品,产生粉色色。 为了确定影响反应的因素,我们需要测量平衡常数。 如果平衡不偏向于产品方向,在使用这种材料测量血清中的铁时需要谨慎。

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  Calculations with Equilibrium Constants
  ::与平衡常数的计算

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  The general value of the equilibrium constant gives us information about whether the reactants or the products are favored at equilibrium. Since the product concentrations are in the numerator of the equilibrium expression, a $\begin{align*}K_{eq} > 1\end{align*}$ means that the products are favored over the reactants. A  $\begin{align*}K_{eq}< 1\end{align*}$ means that the reactants are favored over the products.
  ::平衡常数的一般值为我们提供了关于反应器或产品是否在平衡时得到偏好的信息。由于产品浓度位于均衡表达式的分子中, Keq>1 意味着产品优于反应器。 A Keq < 1 意味着反应器优于产品。

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  Though it would often seem that the  $\begin{align*}K_{eq}\end{align*}$ value would have various units depending on the values of the exponents in the expression, the general rule is that any units are dropped. All  $\begin{align*}K_{eq}\end{align*}$ values will be reported as having no units.
  ::Keq 值似乎往往有不同的单位,取决于该表达式中的指数值,但一般规则是,任何单位都会被丢弃。所有Keq 值都将被报告为没有单位。

   

   

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  Sample Problem: Calculating an Equilibrium Constant
  ::样本问题: 计算平衡常数

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  Equilibrium occurs when nitrogen monoxide reacts with oxygen gas to form nitrogen dioxide gas.
  ::当一氧化氮与氧气发生反应形成二氧化氮气时,即出现平衡。

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  $$\begin{align*}2\text{NO}(g)+\text{O}_2(g) \rightleftarrows 2\text{NO}_2(g)\end{align*}$$
  ::2NO(g)+O2(g) @%%2NO2(g)

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  At equilibrium at 230°C, the concentrations are measured to be [NO] = 0.0542 M, [O ₂ ] = 0.127 M, and [NO ₂ ] = 15.5 M. Calculate the equilibrium constant at this temperature .
  ::在230°C的平衡时,测量的浓度为[NO]=0.0542 M,[O2]=0.127 M,[NO2]=15.5 M。 计算该温度的平衡常数。

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  Step 1: List the known values and plan the problem .
  ::第1步:列出已知值并规划问题。

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  Known
  ::已知已知

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  • [NO] = 0.0542 M
    ::[无]=0.0542米
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  • [O ₂ ] = 0.127 M
    ::[O2]=0.127米
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  • [NO ₂ ] = 15.5 M
    ::[注2] = 15.5米

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  Unknown
  ::未知

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  • $\begin{align*}K_{eq}\end{align*}$ value
    ::Keq 值

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  The equilibrium expression is first written according to the general form in the text. The equilibrium values are substituted into the expression and the value calculated.
  ::平衡表达式首先根据文本中的一般形式写成。均衡值被替换为表达式和计算值。

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  Step 2: Solve .
  ::步骤2:解决。

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  $$\begin{align*}K_{eq}=\frac{\left [\text{NO}_2 \right ]^2}{\left [\text{NO} \right ]^2 \left [\text{O}_2 \right ]}\end{align*}$$
  ::Keq=[NO2]2[NO2]2[O2]

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  Substituting in the concentrations at equilibrium:
  ::平衡时浓度的替代:

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  $$\begin{align*}K_{eq}=\frac{\left ( 15.5\right )^2}{\left (0.0542 \right )^2 \left (0.127\right )}=6.44 \times 10^5\end{align*}$$
  ::Keq=(15.5)2(0.005422(0.127)=6.44x105

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  Step 3: Think about your result .
  ::步骤3:想想你的结果。

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  The equilibrium concentration of the product NO ₂ is significantly higher than the concentrations of the reactants NO and O ₂ . As a result, the  $\begin{align*}K_{eq}\end{align*}$ value is much larger than 1, an indication that the product is favored at equilibrium.
  ::产品NO2的均衡浓度明显高于反应剂NO和O2的浓度。 因此,Keq值大大大于1,这表明该产品在平衡时得到偏好。

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  The equilibrium expression only shows those substances whose concentrations are variable during the reaction. A pure solid or a pure does not have a concentration that will vary during a reaction. Therefore, an equilibrium expression omits pure solids and liquids and only shows the concentrations of gases and aqueous solutions . The decomposition of mercury(II) oxide can be shown by the following equation, followed by its equilibrium expression. 
  ::平衡表达式只显示反应过程中浓度可变的物质。纯固体或纯固体不具有在反应期间变化的浓度。因此,平衡表达式省略了纯固体和液体,只显示气体和水溶液的浓度。汞氧化物的分解可以通过以下方程式显示,然后是平衡表达式。

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  $$\begin{align*}2\text{HgO}(s) \rightleftarrows 2\text{Hg}(l)+\text{O}_2(g) \qquad \quad K_{eq}=[\text{O}_2]\end{align*}$$
  ::2HGO 2Hg(l)+O2(g)Keq=[O2]

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  The stoichiometry of an equation can also be used in a calculation of an equilibrium constant. At 40°C, solid ammonium carbamate decomposes to ammonia and carbon dioxide gases.
  ::在40°C时,固态碳酸铵分解成氨和二氧化碳气体。

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  $$\begin{align*}\text{NH}_4\text{CO}_2\text{NH}_2(s) \rightleftarrows 2\text{NH}_3(g)+\text{CO}_2(g)\end{align*}$$
  ::NH4CO2NH2(s)2NH3(g)+CO2(g)

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  At equilibrium, the [CO ₂ ] is found to be 4.71 × 10 ^-3 M. Can the  $\begin{align*}K_{eq}\end{align*}$ value be calculated from just that information? Because the ammonium carbamate is a solid, it is not present in the equilibrium expression.
  ::在平衡时,[CO2]为4.71 × 10-3 M。 Keq值能否仅从这一信息中计算出来?由于氨氨氨氨是固体,因此在平衡表达式中不存在。

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  $$\begin{align*}K_{eq}=[\text{NH}_3]^2 [\text{CO}_2]\end{align*}$$
  ::Keq=[NH3]2 [CO2]

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  The stoichiometry of the chemical equation indicates that as the ammonium carbamate decomposes, 2 mol of ammonia gas is produced for every 1 mol of carbon dioxide. Therefore, at equilibrium, the concentration of the ammonia will be twice the concentration of carbon dioxide. So [NH ₃ ] = 2 × (4.71 × 10 ^-3 ) = 9.42 × 10 ^-3 M. Substituting these values into the  $\begin{align*}K_{eq}\end{align*}$ expression:
  ::化学方程式的声一测定法表明,由于氨氨氨氨氨基氨基甲酸分解,每1兆二氧化碳产生2兆瓦的氨气。因此,在平衡时,氨的浓度将是二氧化碳浓度的两倍。 因此,[NH3]=2×(4.71×10-3)=9.42×10-3M。

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  $$\begin{align*}K_{eq}=(9.42 \times 10^{-3})^2 (4.71 \times 10^{-3})=4.18 \times 10^{-7}\end{align*}$$
  ::Keq= (9. 42x10- 3) 2(4. 71x10-3) = 4. 18x10-7

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  Using Equilibrium Constants
  ::使用平衡常数

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  The equilibrium constants are known for a great many reactions. Hydrogen and bromine gases combine to form hydrogen bromide gas. At 730°C, the equation and  $\begin{align*}K_{eq}\end{align*}$ are given below.
  ::平衡常数以许多反应而著称,氢气和溴气结合形成氢溴气,在730°C时,公式和克克如下。

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  $$\begin{align*}\text{H}_2(g)+\text{Br}_2(g) \rightleftarrows 2\text{HBr}(g) \qquad \ \ K_{eq}=2.18 \times 10^6\end{align*}$$
  ::H2(g) +B(g) +B(g) +2HB(g) Keq= 2.18x106

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  A certain reaction is begun with only HBr. When the reaction reaches equilibrium at 730°C, the concentration of bromine gas is measured to be 0.00243 M. What is the concentration of the H ₂ and the HBr at equilibrium?
  ::当反应在730°C达到平衡时,溴气的浓度被测量为0.00243 M。 H2和HBR在平衡时的浓度是多少?

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  Since the reaction begins with only HBr and the of H ₂ to Br ₂ is 1:1, the concentration of H ₂ at equilibrium is also 0.00243 M. The equilibrium expression can be rearranged to solve for the concentration of HBr at equilibrium.
  ::由于反应只从HBR开始,而H2至BR2的浓度为1:1, 平衡时H2的浓度也是0.00243 M。 平衡表达式可以重新排列,以解决平衡时HBR的浓度问题。

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  $$\begin{align*}K_{eq} & = \frac{[\text{HBr}]^2}{[\text{H}_2][\text{Br}_2]}\\ \left [ \text{HBr} \right ] & = \sqrt{K_{eq}[\text{H}_2][\text{Br}_2]} \\ & = \sqrt{2.18 \times 10^6 (0.00243)(0.00243)} = 3.59 \text{ M}\end{align*}$$
  ::Keq=[HBR]2[H2][Br2][HBR]}Keq[H2][Br2]2.18×106(0.00243)(0.00243)=3.59M

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  Since the value of the equilibrium constant is very high, the concentration of HBr is much greater than that of H ₂ and Br ₂ at equilibrium.
  ::由于平衡常值值非常高,赫伯尔的浓度大大高于平衡时的H2和Br2的浓度。

   

   

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  Summary
  ::摘要

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  • Calculation of an equilibrium constant is described.
    ::对平衡常数的计算作了说明。

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  Review
  ::回顾

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  1. What are the units for $\begin{align*}K_{eq}\end{align*}$ ?
     ::Keq的单位是什么?
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  2. Why is the temperature specified in equilibrium problems?
     ::为什么在平衡问题中指定温度?
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  3. Why don’t we include solids or liquids in equilibrium calculations?
     ::为何我们不将固体或液体纳入均衡计算?