20.5 计算自由能源变化(G°)

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  计算自由能源变化 (G°)

  

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  Time for dessert!
  ::甜点时间到了!

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  When you are baking something, you the oven to the temperature indicated in the recipe. Then you mix all the ingredients, put them in the proper baking dish, and place them in the oven for a specified amount of time. If you had mixed the ingredients and left them out at room temperature, not much would change. The materials need to be heated to a given temperature for a set time in order for the ingredients to react with one another and produce a delicious final product.
  ::当你正在烤东西时,你将烤箱放在食谱中标明的温度上。然后你将所有成分混合起来,放在适当的烘烤盘中,并在一定的时间里把它们放在烤箱中。如果你把成分混合在一起,在室温下把它们留在外面,就不会有什么变化。材料需要加热到一定的温度,以便各成分相互反应,并生产一份美味的最后产品。

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  Calculating Free Energy  $\begin{align*}(\Delta G^\circ)\end{align*}$
  ::计算自由能源(G)

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  The change of a reaction can be calculated using the following expression:
  ::反应的改变可以用以下表述来计算:

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  $$\begin{align*}\Delta G^\circ=\Delta H^\circ - T\Delta S^\circ\end{align*}$$
  ::~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

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  where  $\begin{align*}\Delta G = \text{free energy change (kJ/mol)}\end{align*}$
  ::G=无能源变化(kJ/mol)

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  $\begin{align*}\Delta H = \text{change in enthalpy (kJ/mol)}\end{align*}$
  ::H = enthalpy (kJ/mol) 变化

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  $\begin{align*}\Delta S = \text{change in entropy (J/K} \cdot \text{mol)}\end{align*}$
  ::S = 酶变化 (J/Kmol)

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  $\begin{align*}T = \text{temperature (Kelvin)}\end{align*}$
  ::T=温度 (Kelvin)

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  Note that all values are for substances in their standard state. In performing calculations, it is necessary to change the units for $\begin{align*}\Delta S\end{align*}$  to kJ/K • mol, so that the calculation of $\begin{align*}\Delta G\end{align*}$  is in kJ/mol.
  ::请注意,所有值都是按其标准状态的物质值。在进行计算时,必须把 S 的单位改为 kJ/K / mol , 以便计算 G 的单位为 kJ/mol 。

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  Sample Problem: Gibbs Free Energy
  ::抽样问题:Gibbs免费能源

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  Methane reacts with water vapor to produce a of carbon monoxide and hydrogen according to the balanced equation below.
  ::甲烷与水蒸气发生反应,根据以下平衡方程式生产一氧化碳和氢。

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  $$\begin{align*}\text{CH}_4(g)+\text{H}_2\text{O}(g) \rightarrow \text{CO}(g)+3\text{H}_2(g)\end{align*}$$
  ::CH4(g)+H2O(g)+H2O(g)+CO(g)+3H2(g)

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  The  $\begin{align*}\Delta H^\circ\end{align*}$ for the reaction is +206.1 kJ/mol, while the  $\begin{align*}\Delta S^\circ\end{align*}$ is +215 J/K • mol . Calculate the  $\begin{align*}\Delta G^\circ\end{align*}$ at 25°C and determine if the reaction is spontaneous at that temperature.
  ::反应为 +206.1 kJ/mol, 反应为 +215 J/K / mol。 计算 25 °C 时的 +G , 确定反应是否自发 。

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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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  • $\begin{align*}\Delta H^\circ =206.1 \ \text{kJ/mol}\end{align*}$
    ::H206.1千焦/摩尔
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  • $\begin{align*}\Delta S^\circ = 215 \ \text{J/K}\cdot \text{mol}=0.215 \ \text{kJ/K}\cdot \text{mol}\end{align*}$
    ::215 J/Kmol=0.215 kJ/Kmol
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  • $\begin{align*}T =25^\circ \text{C}=298 \ \text{K}\end{align*}$
    ::T=25°C=298K

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  Unknown
  ::未知

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  • $\begin{align*}\Delta G^\circ =? \ \text{kJ/mol}\end{align*}$
    ::{\fn黑体\fs22\bord1\shad0\3aHBE\4aH00\fscx67\fscy66\2cHFFFFFF\3cH808080}

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  Prior to substitution into the Gibbs free energy equation, the entropy change is converted to  kJ/K • mol and the temperature to Kelvins.
  ::在替换为Gibbs免费能源方程式之前,将天体变化转换为kJ/K/Mol,温度转换为Kelvins。

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  Step 2: Solve.
  ::步骤2:解决。

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  $$\begin{align*}\Delta G^\circ =\Delta H^\circ - T\Delta S^\circ = 206.1 \ \text{kJ/mol} - 298 \ \text{K}(0.215 \ \text{kJ/K}\cdot \text{mol})=+142.0 \ \text{kJ/mol}\end{align*}$$
  ::GHTS206.1 kJ/mol-298 K(0.215 kJ/Kmol)142.0 kJ/mol

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  The resulting positive value of  $\begin{align*}\Delta G\end{align*}$ indicates that the reaction is not spontaneous at 25°C.
  ::由此得出的+G的正值表明,该反应在25°C时不是自发的。

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

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  The unfavorable driving force of increasing outweighed the favorable increase in entropy. The reaction will be spontaneous only at some elevated temperature.
  ::增长的不利驱动力比引温的有利增长要强。 只有在温度升高时反应才会是自发的。

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  Available values for enthalpy and entropy changes are generally measured at the standard conditions of 25°C and 1 atm pressure . The values are slightly temperature dependent and so we must use caution when calculating specific  $\begin{align*}\Delta G\end{align*}$ values at temperatures other than 25°C. However, since the values for  $\begin{align*}\Delta H\end{align*}$ and  $\begin{align*}\Delta S\end{align*}$ do not change a great deal, the tabulated values can safely be used when making general predictions about the spontaneity of a reaction at various temperatures.
  ::enthalpy 和 entropy 变化和 entropy 变化的可用值一般在25°C 和 1 atm 压力的标准条件下测量。 这些值与温度略有不同,因此在计算25°C 以外的温度的具体 G 值时必须谨慎。 然而,由于 H 和 S 的值没有发生很大变化,在对不同温度的反应的自发性作出一般预测时,可以安全地使用列表值。

   

   

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  Summary
  ::摘要

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  • Calculations of free energy changes are described.
    ::介绍了自由能源变化的计算方法。

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  Review
  ::回顾

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  1. What would happen to  $\begin{align*}\Delta H\end{align*}$ if you forgot to change the units for  $\begin{align*}\Delta S\end{align*}$ to kJ/K • mol ?
     ::若你忘了将QZS换成kJ/K-mol,
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  2. What are standard conditions for enthalpy and entropy changes?
     ::enthalpy 和 entropy 变化的标准条件是什么?
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  3. At what temperature would the reaction become spontaneous?
     ::在温度下反应会变成什么自发的?

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  Explore More
  ::探索更多

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  Use the resource below to answer the questions that follow.
  ::利用以下资源回答以下问题。

   

   

   

   

   

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  1. Why is  $\begin{align*}\Delta H\end{align*}$ negative in this example?
     ::这个例子为什么是否定的?
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  2. What would happen if you forgot to change the sign of the  $\begin{align*}T\Delta S\end{align*}$ value in the first calculation?
     ::如果你忘记在第一次计算中更改 TS 值的符号,会发生什么?
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  3. What indicates that the reaction is spontaneous?
     ::是什么表明反应是自发的?