节: 自由能源和克克的计算 | 20.8 自由能源和克克的计算

章节大纲

Cave formations of stalactites and stalagmites showcasing mineral deposition processes.

What are these formations called when they point down?
::当他们指向下时,这些阵形叫什么来着?

Formation of stalactites (pointing down) and stalagmites (pointing up) is a complex process. Solutions of minerals drip down and absorb carbon dioxide as water flows through the cave. Calcium carbonate dissolves in this and redeposits on the rock as the carbon dioxide is dissipated into .
::石化物的形成(指向下)和渣化物的形成(指向上)是一个复杂的过程。矿物质溶液滴落并吸收二氧化碳作为洞穴的水流。碳酸钙在此溶解,随着二氧化碳的消散,碳酸钙在岩石上重新沉淀。

Equilibrium Constant and $\begin{align}\Delta G\end{align}$
::平衡常数和+++G

At equilibrium the $\begin{align*}\Delta G\end{align*}$ for a is equal to zero. $\begin{align*}K_{eq}\end{align*}$ relates the of all substances in the reaction at equilibrium. Therefore we can write (through a more advanced treatment of thermodynamics) the following equation:
::平衡时 ++G 等于零。 Keq 与均衡反应中的所有物质有关。因此,我们可以(通过更先进的热力学处理)写下如下方程:

$\begin{align*}\Delta G^\circ=-RT \ln K_{eq}\end{align*}$
::G RTlnKeq

The variable $\begin{align*}R\end{align*}$ is the ideal gas constant (8.314 J/K • mol), $\begin{align*}T\end{align*}$ is the Kelvin temperature , and $\begin{align*}\ln K_{eq}\end{align*}$ is the natural logarithm of the equilibrium constant .
::变量R是理想气体常数(8.314 J/K / / mol),T是凯尔文温度,InKeq是平衡常数的自然对数。

When $\begin{align*}K_{eq}\end{align*}$ is large, the products of the reaction are favored and the negative sign in the equation means that the $\begin{align*}\Delta G^\circ\end{align*}$ is negative. When $\begin{align*}K_{eq}\end{align*}$ is small, the reactants of the reaction are favored. The natural logarithm of a number less than one is negative and so the sign of $\begin{align*}\Delta G^\circ\end{align*}$ is positive. Table summarizes the relationship of $\begin{align*}\Delta G^\circ\end{align*}$ to $\begin{align*}K_{eq}\end{align*}$ :
::当 Keq 很大时, 反应的产物得到偏好, 等式中的负符号表示 +G = 负。当 Keq 小时, 反应的反应者得到偏好。数字小于一个的自然对数是负的, 因此, QG = 表示是正的。表格总结了 +G +Q 与 Keq 的关系 :

Relationship of $\begin{align*}\Delta G^\circ\end{align*}$ and $\begin{align*}K_{eq}\end{align*}$
$\begin{align}K_{eq}\end{align}$	$\begin{align}\ln K_{eq}\end{align}$	$\begin{align}\Delta G^\circ\end{align}$	Description
>1	positive	negative	Products are favored at equilibrium.
1	0	0	are equally favored.
<1	negative	positive	Reactants are favored at equilibrium.

Knowledge of either the standard change or the equilibrium constant for a reaction allows for the calculation of the other. The following two sample problems illustrate each case.
::了解标准变化或反应的平衡常数,可以计算另一种情况。以下两个抽样问题说明每个情况。

Sample Problem: Gibbs Free Energy and the Equilibrium Constant
::抽样问题: Gibbs 自由能源和平衡常数

The formation of nitrogen monoxide from nitrogen and oxygen is a reaction that strongly favors the reactants at 25°C.
::来自氮和氧的一氧化氮形成是一种强烈有利于25°C反应剂的反应。

$\begin{align*}\text{N}_2(g)+\text{O}_2(g) \rightleftarrows 2\text{NO}(g)\end{align*}$
::N2(g) +O2(g) 2NO(g)

The actual concentrations of each gas would be difficult to measure, and so the _{$\begin{align*}K_{eq}\end{align*}$} for the reaction can more easily calculated from the $\begin{align*}\Delta G^\circ\end{align*}$ , which is equal to 173.4 kJ/mol.
::每种气体的实际浓度难以测量,因此,反应的Keq可以更容易地从等于173.4kJ/mol的QG计算出来。

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

Known
::已知已知

$\begin{align*}\Delta G^\circ=+173.4 \ \text{kJ} / \text{mol}\end{align*}$
::G173.4 kJ/mol

$\begin{align*}R=8.314 \ \text{J} / \text{K} \cdot \text{mol}\end{align*}$
::R=8.314 J/Kmol

$\begin{align*}T=25^\circ \text{C}=298 \ \text{K}\end{align*}$
::T=25°C=298K

Unknown
::未知

$\begin{align*}K_{eq}=?\end{align*}$
::凯克?

In order to make the units agree, the value of $\begin{align*}\Delta G^\circ\end{align*}$ will need to be converted to J/mol (173,400 J/mol). To solve for $\begin{align*}K_{eq}\end{align*}$ , the inverse of the natural logarithm, $\begin{align*}e^x\end{align*}$ , will be used.
::为了让单位同意, QG的值需要转换为 J/ mol (173, 400 J/ mol) 。要解决 Keq 的问题, 将使用自然对数的反义, 如。

Step 2: Solve .
::步骤2:解决。

$\begin{align*}\Delta G^\circ &=-RT \ln K_{eq} \\ \ln K_{eq}&=\frac{-\Delta G^\circ}{RT} \\ K_{eq} &=e^{\frac{-\Delta G^\circ}{RT}}=e^{\frac{-173, 400 \ \text{J} / \text{mol}}{8.314 \ \text{J} / \text{K} \cdot \text{mol}(298 \ \text{K})}}=4.0 \times 10^{-31} \end{align*}$
::GRTlnKeqlnKeqGRTKeq=eGRT=e-173 400 J/mol8.314 J/Kmol(298K)=4.0x10-31

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

The large positive free energy change leads to a _{$\begin{align*}K_{eq}\end{align*}$} value that is extremely small. Both lead to the conclusion that the reactants are highly favored and very few product molecules are present at equilibrium.
::巨大的正自由能源变化导致Keq值极小。两者都得出了反应器高度偏好和产品分子在平衡时很少存在的结论。

Sample Problem: Free Energy from $\begin{align}K_{sp}\end{align}$
::样本问题:来自 Ksp 的免费能源

The product constant $\begin{align*}(K_{sp}) \end{align*}$ of lead(II) iodide is 1.4 × 10 ^-8 at 25°C. Calculate $\begin{align*}\Delta G^\circ\end{align*}$ for the of lead(II) iodide in water.
::铅(II)碘化物的产品常量(Ksp)在25°C时为1.4 × 10-8。计算水中的铅(II)碘化物的产物常量(Ksp)。

$\begin{align*}\text{PbI}_2(s) \rightleftarrows \text{Pb}^{2+}(aq) + 2\text{I}^-(aq)\end{align*}$
::PbI2(s) Pb2+(aq)+2I-(aq)

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

Known
::已知已知

$\begin{align*}K_{eq}=K_{sp}=1.4 \times 10^{-8}\end{align*}$
::Keq=Ksp=1.4x10-8

$\begin{align*}R=8.314 \ \text{J} / \text{K} \cdot \text{mol}\end{align*}$
::R=8.314 J/Kmol

$\begin{align*}T=25^\circ \text{C}=298 \ \text{K}\end{align*}$
::T=25°C=298K

Unknown
::未知

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

The equation relating $\begin{align*}\Delta G^\circ\end{align*}$ to _{$\begin{align*}K_{eq}\end{align*}$} can be solved directly.
::和Keq有关的等式可以直接解决

Step 2: Solve.
::步骤2:解决。

$\begin{align*}\Delta G^\circ&=-RT \ln K_{eq} \\ &=-8.314 \ \text{J} / \text{K} \cdot \text{mol}(298 \ \text{K}) \ln (1.4 \times 10^{-8}) \\ &=45,000 \ \text{J} / \text{mol} \\ &=45 \ \text{kJ} / \text{mol}\end{align*}$
::GRTlnKeq8.314 J/Kmol(298 K)ln(1.4×10-8)=45 000 J/mol=45 kJ/mol

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

The large, positive $\begin{align*}\Delta G^\circ\end{align*}$ indicates that the solid lead(II) iodide is nearly insoluble and so very little of the solid is dissociated at equilibrium.
::大量呈阳性的G表示,固态铅(II)碘化物几乎无法溶解,在平衡时分解的固体极少。

Summary
::摘要

The relationship between $\begin{align*}\Delta G \end{align*}$ and _{$\begin{align*}K_{eq}\end{align*}$} is described.
::说明G和Keq之间的关系。

Calculations involving these two parameters are shown.
::列出了涉及这两个参数的计算结果。

Review
::回顾

When _{$\begin{align*}K_{eq}\end{align*}$} is large, what will be the sign of $\begin{align*}\Delta G \end{align*}$ ?
::当Keq是大,什么是G的迹象?

When _{$\begin{align*}K_{eq}\end{align*}$} is small, are reactants or products favored?
::当Keq面积小时,反应者或产品是否偏好?

What does $\begin{align*}R\end{align*}$ stand for?
::R代表什么?

章节大纲

What are these formations called when they point down? ::当他们指向下时,这些阵形叫什么来着?

Equilibrium Constant and Δ G \begin{align*}\Delta G\end{align*} ::平衡常数和+++G

Sample Problem: Gibbs Free Energy and the Equilibrium Constant ::抽样问题: Gibbs 自由能源和平衡常数

Sample Problem: Free Energy from K s p \begin{align*}K_{sp}\end{align*} ::样本问题:来自 Ksp 的免费能源

Summary ::摘要