第三章：化学热力学

Thus, for the Haber process,
The equilibrium-constant expression depends only on the stoichiometry of the reaction, not on its mechanism.平衡常
数表达式只与反应的计量数有关
Example: Converting between Kc and Kp
For the Haber process,
Kc = 9.60 at 300 °C. Calculate Kp for this reaction at this temperature.
Solve With 2 mol of gaseous products (2 NH3) and 4 mol of
The Magnitude of Equilibrium Constants
平衡常数没有单位。平衡常数的数值表源自了，反应达到平衡 时，各组分所占的量的多少。
In general, If K >> 1 (large K): Equilibrium lies to right, products predominate If K << 1 (small K): Equilibrium lies to left, reactants predominate
Relationship between magnitude of K and composition of an equilibrium mixture.
Example :Interpreting the Magnitude of an Equilibrium Constant
下列图表示三个处于平衡的体系, 容器的容积相同。(a) 不计 算, 将下列体系的Kc按升序排列. (b) 如果容器容积为1.0 L， 每个小球0.10 mol, 计算下列各体系的 Kc 值.
HETEROGENEOUS EQUILIBRIA异相平衡
In some cases, the substances in equilibrium are in different phases, giving rise to heterogeneous equilibria.
Whenever a pure solid or a pure liquid is involved in a heterogeneous equilibrium, its concentration is not included in the equilibrium-constant expression. Another example:
第三章 化学热力学初步 和化学平衡（III）
——化学平衡
一、化学平衡的概念：
一个化学反应达到平衡—反应物和产物的浓度不再随时间 变化而变化.
上述反应的反应速率可表示为:
达到平衡时:
上式重排 “the equilibrium constant” 平衡常数
• 平衡时, 反应物与产物浓度不随时间而变化. • 达到平衡, 体系中没有物质损失. • 平衡时, 各物质浓度之间的关系可表达为一个常数.
A general expression relating Kp and Kc :
Kp = Kc(RT)Δn The quantity Δn is the change in the number of moles of gas in the balanced chemical equation.
Δn = (moles of gaseous product) - (moles of gaseous reactant)
= 0.2122 = 0.0449
平衡体系中，各物质的浓度和反应方程式的写法无关, 但 Kc 值与平衡式的写法密切相关.
与Hess’s law相近，我们也可以利用分步反应的Kc计算出总反 应的Kc.
For example (at 100 °C )
The net sum of these two equations is
The same equilibrium mixture is produced regardless of the initial NO2 concentration.
Equilibrium Constants in Terms of Pressure, Kp
当反应物和产物为气体时, 我们可以分压的形式来表达 平衡常数. For the general reaction
Law of mass action(质量作用定律）, 表达对于任意一个反应， 达到平衡时反应物与产物浓度的关系. A general equilibrium equation
根据质量作用定律, 平衡时，各物质浓度可表达为：
“equilibrium-constant expression” 平衡常数表达式
(the subscript p stands for pressure) For example:
It is possible to calculate one from the other using the ideal-gas equation:
For substance A in our generic reaction, we therefore see that
Summarization: 1. 一反应的平衡常数与其逆反应的平衡常数为倒数关系:
2. 反应计量数发生改变时，平衡常数做相应的幂次方改变.
3. 总反应的Kc可由分步反应的Kc相乘得到:
Example: Combining Equilibrium Expressions
Given the reactions
Kc: the equilibrium constant 平衡常数
(The subscript c on the K indicates that concentrations expressed in molarity（摩尔浓度） are used to evaluate the constant. )
gaseous reactants (1 N2 + 3 H2), Δn = 2 - 4 = -2. (Remember that Δ functions are always based on products minus reactants.) The temperature is 273 + 300 = 573 K. The value for the ideal-gas constant, R, is 0.08206 L-atm/mol-K. Using Kc = 9.60, we therefore have
Reversing the second equation and again making the corresponding change to its equilibrium constant (taking the reciprocal) gives
Now we have two equations that sum to give the net equation, and we can multiply the individual Kc values to get the desired equilibrium constant.
Solve (a) The equilibrium-constant expression is Because H2O appears in the reaction as a liquid, its concentration does not appear in the equilibrium-constant expression. (b) The equilibrium-constant expression is Because SnO2 and Sn are pure solids, their concentrations do not appear in the equilibrium-constant expression.
determine the value of Kc for the reaction
Solve If we multiply the first equation by 2 and make the corresponding change to its equilibrium constant (raising to the power 2), we get
Solve
(a) Each box contains 10 spheres. The amount of product in each varies as follows: (i)6, (ii)1, (iii) 8. Therefore, the equilibrium constant varies in the order (ii) < (i) < (iii), from smallest (most reactant) to largest (most products). (b) In (i) we have 0.60 mol/L product and 0.40 mol/L reactant, giving Kc = 0.60/0.40 = 1.5. (You will get the same result by merely dividing the number of spheres of each kind: 6 spheres/4 spheres = 1.5.) In (ii) we have 0.10 mol/L product and 0.90 mol/L reactant, giving Kc = 0.10/0.90 = 0.11 (or 1 sphere/9 spheres = 0.11). In (iii) we have 0.80 mol/L product and 0.20 mol/L reactant, giving Kc = 0.80/0.20 = 4.0 (or 8 spheres/2 spheres = 4.0). These calculations verify the order in (a).