化学平衡

A reversible reaction 可逆反应 runs both ways. Chemical equilibrium 化学平衡 is reached when the forward reaction 正反应 and reverse reaction 逆反应 rates become equal, so concentrations 浓度 stop changing. Equilibrium is dynamic 动态的 – both reactions continue, but at matching rates, so nothing appears to change.

The rates graph above explains why concentrations settle; this next graph shows what the concentrations do – reactants 反应物 fall and products 生成物 rise until, once the rates match, both hold constant (they need not be equal).

At equilibrium the amounts of reactants and products are fixed but usually not equal. Whether the mixture favors products or reactants depends on the reaction; you compare the current state to the equilibrium condition to predict which way it shifts.

The reaction quotient 反应商 $Q$ has the same form as the equilibrium expression but uses current concentrations:

At equilibrium $Q$ equals the equilibrium constant 平衡常数 $K$. Comparing them predicts direction: $Q<K$ shifts forward (toward products); $Q>K$ shifts reverse; $Q=K$ means already at equilibrium.

Write $K$ from the balanced equation 配平方程式 (pure solids and liquids are left out). From equilibrium concentrations (or partial pressures 分压 for $K_p$), plug in and compute. $K_c$ uses molarities 摩尔浓度; $K_p$ uses pressures.

Worked example. For $\text{N}_2\text{O}_4\rightleftharpoons2\text{NO}_2$, the equilibrium concentrations are $[\text{N}_2\text{O}_4]=0.20\ \text{M}$ and $[\text{NO}_2]=0.10\ \text{M}$. Then

The NO$_2$ coefficient 系数 of $2$ becomes the power; N$_2$O$_4$ (coefficient $1$) is just to the first power.

• $K\gg 1$: products strongly favored (reaction nearly complete).
• $K\ll 1$: reactants favored (little reaction).
• $K\approx 1$: significant amounts of both.

The size of $K$ tells you where the equilibrium "sits."

$K$ changes in predictable ways: reversing a reaction inverts $K$ ($1/K$); multiplying coefficients by $n$ raises $K$ to the $n$th power; adding reactions multiplies their $K$ values. Only temperature changes the value of $K$ itself.

Use an ICE table 三段式 (Initial, Change, Equilibrium): fill in starting amounts, express the change with $x$ using the mole ratios 摩尔比, then substitute into the $K$ expression and solve for $x$. When $K$ is small, the approximation "$x$ is negligible 可忽略的" often simplifies the algebra.

Worked example. For $\text{H}_2+\text{I}_2\rightleftharpoons2\text{HI}$ with $K_c=50$, start with $0.100\ \text{M}$ each of H$_2$ and I$_2$. The ICE table gives equilibrium $[\text{H}_2]=[\text{I}_2]=0.100-x$ and $[\text{HI}]=2x$. Substitute:

so $[\text{HI}]=2x=0.156\ \text{M}$. Taking the square root of both sides works here because the expression is a perfect square.

A particulate 微粒 picture at equilibrium shows a fixed mix of reactant and product particles. Graphs of concentration vs time level off (never reaching zero) once equilibrium is reached – both curves flatten at the same moment.

Le Chatelier's principle 勒沙特列原理: if you disturb a system at equilibrium, it shifts to partly counteract the change. Adding a reactant shifts forward; removing product shifts forward; increasing pressure (by reducing volume) shifts toward the side with fewer gas moles; raising temperature shifts in the endothermic direction.

You can justify every Le Chatelier shift with $Q$ versus $K$: a disturbance changes $Q$, and the system reacts to bring $Q$ back to $K$. This quantitative view backs up the qualitative rule and is the fuller answer on the exam.

Worked example. For $\text{N}_2\text{O}_4(g) \rightleftharpoons 2\text{NO}_2(g)$ (forward endothermic), an equilibrium mixture has $[\text{N}_2\text{O}_4] = 0.40\ M$ and $[\text{NO}_2] = 0.20\ M$, so

• Add $\text{N}_2\text{O}_4$ until $[\text{N}_2\text{O}_4] = 0.70\ M$: $Q = \dfrac{0.20^2}{0.70} = 0.057 < K$, so the reaction runs forward until $Q = 0.10$ again.
• Halve the volume: every concentration doubles, so $Q = \dfrac{0.40^2}{0.80} = 0.20 > K$. The reaction runs in reverse, towards the side with fewer moles of gas. This works because the top of $Q$ is squared and the bottom is not.
• Add argon at constant volume: no concentration in $Q$ changes, so $Q = K$ and there is no shift, even though the total pressure rises.
• Raise the temperature: this time $Q$ does not change, but $K$ does. Heat favours the endothermic (forward) direction, so $K$ increases and now $Q < K$.

Common mistakes.

• Saying that adding a reactant "changes $K$". Only temperature changes $K$; other disturbances change $Q$.
• Saying that any pressure increase causes a shift. Adding an inert gas at constant volume does not change $Q$.
• Drawing the graph with every line jumping at a temperature change, or with the system returning to the original concentrations.
• Forgetting that a catalyst changes neither $Q$ nor $K$, so it causes no shift.

For a slightly soluble 微溶的 salt, the solubility product 溶度积 $K_{sp}$ is the equilibrium constant for its dissolving:

Compare $K_{sp}$ values directly only for salts with the same dissolution stoichiometry 化学计量 under the same conditions. For different stoichiometries, calculate molar solubility 摩尔溶解度 from each balanced equilibrium expression. The common-ion effect 同离子效应 – adding an ion already in the salt – pushes the equilibrium back and lowers solubility.

Worked example. Silver chloride has $K_{sp}=1.8\times10^{-10}$. If its molar solubility is $s$, then $[\text{Ag}^+]=[\text{Cl}^-]=s$, so $K_{sp}=s^2$ and

Adding NaCl (a common ion) would raise $[\text{Cl}^-]$, forcing $s$ down – far less AgCl would dissolve.

• At equilibrium the forward and reverse rates are equal, but the amounts are usually not.
• Leave pure solids and liquids out of the $K$ expression; a large $K$ favours products, a small $K$ favours reactants.
• Apply Le Chatelier: add reactant → shift forward; raise pressure → shift toward fewer gas moles; raise temperature → shift in the endothermic 吸热 direction.
• A catalyst 催化剂 does not shift the position of equilibrium — it only speeds the approach.
• Use an ICE table for equilibrium concentrations, and the small-$x$ approximation when $K$ is small.