📚 Equilibria in Gas Reactions: the Equilibrium Constant, Kp | 气体反应平衡：平衡常数 Kp
For reversible reactions that take place entirely in the gas phase, the equilibrium position can be described quantitatively using partial pressures rather than concentrations. The resulting constant, Kp, is central to Cambridge A-Level Chemistry and appears frequently in Paper 2 and Paper 4 calculations.
对于完全在气相中进行的可逆反应，平衡位置可以使用分压而不是浓度来定量描述。由此得到的平衡常数 Kp 是剑桥 A-Level 化学的核心内容，经常出现在 Paper 2 和 Paper 4 的计算题中。
1. What Kp Measures | Kp 衡量的内容
Kp is an equilibrium constant expressed in terms of the partial pressures of gaseous reactants and products. It gives the ratio of product pressures to reactant pressures at equilibrium, with each pressure raised to the power of its stoichiometric coefficient.
Kp 是以气体反应物和产物的分压表示的平衡常数。它表示平衡时产物分压与反应物分压的比值，并且每种分压都要以其化学计量系数为指数。
A large Kp value, much greater than 1, means the equilibrium position lies well to the right and products are favoured. A very small Kp value means the equilibrium lies to the left and reactants are favoured.
Kp 值远大于 1 表示平衡位置明显偏向右侧，产物占优势；Kp 值非常小则表示平衡偏向左侧，反应物占优势。
2. Partial Pressure and Mole Fraction | 分压与摩尔分数
In a gas mixture, each gas exerts its own partial pressure. Dalton’s Law states that the total pressure is the sum of all partial pressures.
在气体混合物中，每种气体都产生自己的分压。道尔顿定律指出，总压等于所有分压之和。
The partial pressure of gas A is found by multiplying its mole fraction by the total pressure. The mole fraction is the number of moles of A divided by the total number of moles of gas present.
气体 A 的分压等于其摩尔分数乘以总压。摩尔分数是 A 的物质的量除以气体总物质的量。
pA = xA × Ptotal
xA = nA ÷ ntotal
These relationships are essential because Kp calculations usually start from initial moles, equilibrium moles and a quoted total pressure.
这些关系非常重要，因为 Kp 的计算通常从初始物质的量、平衡物质的量和给出的总压开始。
3. Defining the Equilibrium Constant Kp | 平衡常数 Kp 的定义
For a general gaseous equilibrium of the form aA(g) + bB(g) ⇌ cC(g) + dD(g), the expression for Kp is written as follows.
对于形如 aA(g) + bB(g) ⇌ cC(g) + dD(g) 的一般气体平衡，Kp 的表达式如下。
Kp = (pCc × pDd) ÷ (pAa × pBb)
Only gaseous species appear in the Kp expression. Solids and pure liquids are omitted because their concentrations or vapour pressures remain effectively constant during the reaction.
只有气体物种会出现在 Kp 表达式中。固体和纯液体被省略，因为它们在反应过程中浓度或蒸气压实际上保持不变。
4. Writing Kp Expressions for Common Reactions | 书写常见反应的 Kp 表达式
For the Haber process, N₂(g) + 3H₂(g) ⇌ 2NH₃(g), the equilibrium constant is given by the following expression.
对于哈伯法合成氨反应 N₂(g) + 3H₂(g) ⇌ 2NH₃(g)，平衡常数表达式如下。
Kp = p(NH₃)² ÷ [p(N₂) × p(H₂)³]
For the dissociation of dinitrogen tetroxide, N₂O₄(g) ⇌ 2NO₂(g), Kp is written as follows.
对于四氧化二氮的解离反应 N₂O₄(g) ⇌ 2NO₂(g)，Kp 的写法如下。
Kp = p(NO₂)² ÷ p(N₂O₄)
For the contact process equilibrium, 2SO₂(g) + O₂(g) ⇌ 2SO₃(g), the expression is shown below.
对于接触法平衡 2SO₂(g) + O₂(g) ⇌ 2SO₃(g)，表达式如下。
Kp = p(SO₃)² ÷ [p(SO₂)² × p(O₂)]
Notice that each partial pressure is raised to the power that matches the coefficient in the balanced chemical equation.
请注意，每种分压的指数必须与配平后的化学方程式中对应的化学计量数一致。
5. Homogeneous and Heterogeneous Gas Equilibria | 均相与异相气体平衡
A homogeneous equilibrium has all reactants and products in the same phase, most commonly the gas phase. Kp then includes every gaseous species.
均相平衡中所有反应物和产物都处于同一相，最常见的是气相。此时 Kp 包含每一种气体物种。
A heterogeneous equilibrium has species in more than one phase. For Kp, only the partial pressures of gases are included. Solids and pure liquids are left out of the expression.
异相平衡中的物种处于多个相。对于 Kp，只包括气体的分压，固体和纯液体不写入表达式。
For example, in CaCO₃(s) ⇌ CaO(s) + CO₂(g), the Kp expression is simply Kp = p(CO₂). The two solids do not appear.
例如，在 CaCO₃(s) ⇌ CaO(s) + CO₂(g) 中，Kp 表达式仅为 Kp = p(CO₂)，两种固体都不出现。
6. Units of Kp | Kp 的单位
The units of Kp depend on the change in the total number of moles of gas during the reaction, represented by Δn. If partial pressures are measured in atm, the units are atm raised to the power of Δn.
Kp 的单位取决于反应过程中气体总物质的量的变化，用 Δn 表示。如果分压以 atm 为单位，则 Kp 的单位是 atm 的 Δn 次方。
For N₂(g) + 3H₂(g) ⇌ 2NH₃(g), Δn = 2 − 4 = −2, so the units are atm⁻². If pressures are measured in Pa or kPa, the units become Pa⁻² or kPa⁻².
对于 N₂(g) + 3H₂(g) ⇌ 2NH₃(g)，Δn = 2 − 4 = −2，因此单位是 atm⁻²。如果压强以 Pa 或 kPa 为单位，则单位变为 Pa⁻² 或 kPa⁻²。
| Reaction | 反应 | Δn | Units of Kp if p in atm | Kp 单位（p 以 atm 计） |
|---|---|---|
| N₂O₄(g) ⇌ 2NO₂(g) | +1 | atm |
| N₂(g) + 3H₂(g) ⇌ 2NH₃(g) | −2 | atm⁻² |
| H₂(g) + I₂(g) ⇌ 2HI(g) | 0 | no units | 无单位 |
7. Relationship Between Kp and Kc | Kp 与 Kc 的关系
Kp can be related to the concentration-based constant Kc through the ideal gas behaviour of the species. The relationship is given by the equation below.
Kp 可以通过物种的理想气体行为与基于浓度的常数 Kc 联系起来，关系式如下。
Kp = Kc (RT)Δn
Here R is the gas constant, T is the absolute temperature in kelvin, and Δn is the change in moles of gas from reactants to products.
其中 R 是气体常数，T 是以开尔文为单位的绝对温度，Δn 是从反应物到产物的气体物质的量变化。
When Δn = 0, the two constants are numerically equal: Kp = Kc. This occurs for reactions such as H₂(g) + I₂(g) ⇌ 2HI(g).
当 Δn = 0 时，两个常数数值相等：Kp = Kc。例如 H₂(g) + I₂(g) ⇌ 2HI(g) 就属于这种情况。
8. Effect of Temperature on Kp | 温度对 Kp 的影响
For a given reaction, Kp changes only when the temperature changes. Pressure, concentration and catalysts do not alter the value of Kp at a fixed temperature.
对于给定的反应，只有温度变化时 Kp 才会改变。在温度固定时，压强、浓度和催化剂都不改变 Kp 的值。
If the forward reaction is exothermic, increasing the temperature shifts the equilibrium position to the left, so Kp decreases. If the forward reaction is endothermic, increasing the temperature shifts the equilibrium to the right and Kp increases.
如果正反应放热，升高温度会使平衡位置向左移动，因此 Kp 减小。如果正反应吸热，升高温度会使平衡位置向右移动，Kp 增大。
This behaviour is consistent with Le Chatelier’s
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