20.7 国家和自由能源的改变

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  国家和自由能源的变化

  

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  How are energy and changes of state related?
  ::能源和国家变化如何相关?

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  Energy in a body of water can be gained or lost depending on conditions.  When water is heated above a certain temperature steam is generated.  The increase in energy creates a higher level of disorder in the water molecules as they boil off and leave the .
  ::水体中的能量可以随着条件而获取或丧失,当水在一定温度蒸汽之上加热时,产生的能量会增加,导致水分子在沸腾和离开时出现更高程度的紊乱。

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  Changes of State and Free Energy
  ::国家和自由能源的变化

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  At the temperature at which a occurs, the two states are in equilibrium with one another. For an ice-water system , equilibrium takes place at 0°C, so  $\begin{align*}\Delta G°^\circ\end{align*}$ is equal to 0 at that temperature. The heat of fusion of water is known to be equal to 6.01 kJ/mol, and so the Gibbs equation can be solved for the entropy change that occurs during the of ice. The symbol  $\begin{align*}\Delta S_{\text{fus}}\end{align*}$ represents the entropy change during the melting process, while  $\begin{align*}T_{\text{f}}\end{align*}$ is the point of water.
  ::当温度发生时,两个州之间是平衡的。对于冰水系统来说,平衡是在0°C,因此当温度为0时,G等于0。已知水聚变的热等于6.01kJ/mol,因此Gibbs等式可以解决冰下发生的星团变化。符号 Sfus 代表熔化过程中的星团变化,Tf是水的点。

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  $$\begin{align*}\Delta G & = 0 = \Delta H - T \Delta S\\ \Delta S_{\text{fus}} & = \frac{\Delta H_{\text{fus}}}{T_{\text{f}}} = \frac{6.01 \text{ kJ/mol}}{273 \text{ K}} = 0.0220 \text{ kJ/K} \cdot \text{mol} = 22.0 \text{ J/K} \cdot \text{mol}\end{align*}$$
  ::@G=0H-TSSSFus*HfusTf=6.01 kJ/mol273 K=0.0220 kJ/Kmol=22.0 J/Kmol

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  The entropy change is positive as the solid state changes into the liquid state. If the transition went from the liquid to the solid state, the numerical value for  $\begin{align*}\Delta S\end{align*}$ would be the same, but the sign would be reversed since we are going from a less ordered to a more ordered situation.
  ::随着固态向液态的转变,酶的变化是正的。 如果从液体向固态的转变发生,则-S的数值会相同,但信号会倒转,因为我们从一个更不固定的状态向一个更有序的局面转变。

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  A similar calculation can be performed for the vaporization of liquid to . In this case we would use the molar heat of vaporization . This value would be 40.79 kJ/mol. The  $\begin{align*}\Delta S_{\text{vap}}\end{align*}$ would then be as follows:
  ::对液体蒸发可作类似的计算。在这种情况下,我们将使用蒸发的摩尔热。这一数值为40.79kJ/mol。

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  $$\begin{align*}\Delta S = \frac{40.79 \text{ kJ/mol}}{373 \text{ K}} = 0.1094 \text{ kJ/K} \cdot \text{mol} = 109.4 \text{ J/K} \cdot \text{mol}\end{align*}$$
  ::S=40.79 kJ/mol373 K=0.1094 kJ/Kmol=109.4 J/Kmol

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  The value is positive, again reflecting the increase in disorder going from liquid to vapor . from vapor to liquid would give a negative value for $\begin{align*}\Delta S\end{align*}$ .
  ::其值为正值,再次反映从液体到蒸气的紊乱增加。 从蒸气到液体,其负值将为 ++S 带来负值。

   

   

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  Summary
  ::摘要

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  • Calculations are shown for determining entropy changes at transition temperatures (ice → water or water → vapor and reverse).
    ::为确定过渡温度(水或水蒸气和逆向)下的酶变化,列出了计算方法。

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  Review
  ::回顾

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  1. What precautions need to be taken in selecting a value for $\begin{align*}\Delta H\end{align*}$ ?
     ::选择 QH 值时需要采取哪些预防措施?
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  2. Why is temperature selection important?
     ::为什么选择温度很重要?
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  3. Why would the entropy of vaporization be so much larger than the entropy of fusion?
     ::为什么蒸发的环球比聚变的环球还要大?