化学动力学

Kinetics 动力学 studies reaction rate 反应速率 – how fast reactants become products. Rate is the change in concentration 浓度 per unit time, and it typically decreases as reactants are used up. Higher temperature 温度 and a suitable catalyst usually increase rate. Concentration effects depend on the rate law; greater exposed surface area 表面积 matters for reactions involving a solid.

The rate law 速率方程 relates rate to reactant concentrations:

where $k$ is the rate constant 速率常数 and $m,n$ are the orders 级数. Orders are found experimentally (not from the balanced coefficients) by seeing how the rate changes when you change one concentration at a time.

Worked example. In experiments, doubling $[\text{A}]$ makes the rate four times larger, while doubling $[\text{B}]$ leaves the rate unchanged. So the reaction is second order 二级 in A ($2^2=4$) and zero order 零级 in B, giving $\text{rate}=k[\text{A}]^2$. The overall order 总级数 is $2+0=2$. Never read the orders off the balanced coefficients – only experiment gives them.

The integrated rate laws 积分速率方程 describe how concentration falls with time and give straight-line tests:

• Zero order: $[\text{A}]$ vs $t$ is linear.
• First order 一级: $\ln[\text{A}]$ vs $t$ is linear; constant half-life 半衰期.
• Second order: $\dfrac{1}{[\text{A}]}$ vs $t$ is linear.

Whichever plot is straight tells you the order and gives $k$ from its slope.

Worked example. A first-order reaction has rate constant $k=0.030\ \text{s}^{-1}$. Its half-life is

and, being first order, that half-life stays the same no matter how much reactant remains – so after $46\ \text{s}$ ($2$ half-lives) one quarter is left.

A reaction mechanism 反应机理 is the sequence of elementary steps 基元反应 that actually occur. Their molecularity 分子数 (how many particles collide in a step) sets that step's rate law directly. Species made in one step and used up in a later one are intermediates 中间体.

Collision theory 碰撞理论: molecules must collide with enough energy (the activation energy 活化能, $E_a$) and the correct orientation 取向 to react. Higher temperature means more molecules exceed $E_a$, so the reaction speeds up sharply.

An energy profile 能量图 plots energy along the reaction path. The peak is the transition state 过渡态; the climb from reactants to the peak is $E_a$; the difference between reactant and product energies is the enthalpy change 焓变 $\Delta H$ (down for exothermic 放热).

In a multi-step mechanism, the steps must add up to the overall balanced equation (intermediates cancel). Each step has its own energy barrier, measured from the preceding minimum. A diagram helps suggest a controlling step, but the full mechanism and concentrations determine the rate.

The rate-determining step 决速步骤 is the slow step that controls rate in a suitable mechanism model. Its elementary rate law follows its molecularity. If this law contains an intermediate, use a justified relation, such as a fast pre-equilibrium, to express it in observable species before predicting the overall rate law. A valid mechanism must (1) add to the overall reaction and (2) predict the experimentally observed rate law.

Worked example. Suppose the slow (rate-determining) step is the bimolecular 双分子 collision $\text{NO}_2+\text{NO}_2\rightarrow\text{NO}_3+\text{NO}$. Its molecularity gives the rate law directly: $\text{rate}=k[\text{NO}_2]^2$. If experiment shows exactly this, the proposed mechanism is consistent; if experiment gave $\text{rate}=k[\text{NO}_2]$, the mechanism would be wrong.

If a fast step precedes the slow one, an intermediate appears in the slow step's rate law. Use the fast pre-equilibrium 快速预平衡 to rewrite that intermediate in terms of reactants, so the final rate law contains only measurable species.

Worked example (a fast first step). For $2\text{NO} + \text{Br}_2 \rightarrow 2\text{NOBr}$, this mechanism is proposed:

• Step 1 (fast, reversible): $\text{NO} + \text{Br}_2 \rightleftharpoons \text{NOBr}_2$
• Step 2 (slow): $\text{NOBr}_2 + \text{NO} \rightarrow 2\text{NOBr}$

1. Check the sum: $\text{NOBr}_2$ appears on both sides and cancels, leaving $2\text{NO} + \text{Br}_2 \rightarrow 2\text{NOBr}$. ✓
2. Write the rate law of the slow step from its molecularity: $\text{rate} = k_2[\text{NOBr}_2][\text{NO}]$.
3. $\text{NOBr}_2$ is an intermediate, so it cannot stay in the answer. Step 1 is at equilibrium, so $K_1 = \dfrac{[\text{NOBr}_2]}{[\text{NO}][\text{Br}_2]}$, which gives $[\text{NOBr}_2] = K_1[\text{NO}][\text{Br}_2]$.
4. Substitute: $\text{rate} = k_2K_1[\text{NO}]^2[\text{Br}_2] = k[\text{NO}]^2[\text{Br}_2]$.

The predicted law is second order in NO and first order in $\text{Br}_2$, which matches experiment. Notice that it looks as if three particles collide at once, but no single step has three particles.

A quick check that always works: the rate law of a mechanism with a fast first step contains every reactant species that appears up to and including the slow step, counted as many times as it appears there.

A multi-step reaction's profile shows several peaks (one per step) with valleys (intermediates) between them. Measure each barrier from its preceding minimum, and test any proposed controlling step against the observed rate law. The highest absolute peak alone does not establish the rate-determining step.

A catalyst 催化剂 speeds a reaction by providing a new pathway with a lower activation energy, without being consumed. It does not change $\Delta H$ or the equilibrium position 平衡位置 – only how fast equilibrium is reached. On an energy profile, a catalyst lowers the peak(s).

Catalyst or intermediate? Both cancel from the overall equation, so look at where each one first appears:

• an intermediate is made in an early step and used up in a later step.
• a catalyst is used up in an early step and made again in a later step, so it is there at the start.

Worked example. Iodide ions speed up the decomposition of hydrogen peroxide:

• Step 1 (slow): $\text{H}_2\text{O}_2 + \text{I}^- \rightarrow \text{H}_2\text{O} + \text{IO}^-$
• Step 2 (fast): $\text{H}_2\text{O}_2 + \text{IO}^- \rightarrow \text{H}_2\text{O} + \text{O}_2 + \text{I}^-$

Adding the steps gives $2\text{H}_2\text{O}_2 \rightarrow 2\text{H}_2\text{O} + \text{O}_2$. $\text{I}^-$ is used in step 1 and returned in step 2, so it is the catalyst. $\text{IO}^-$ is made in step 1 and used in step 2, so it is an intermediate. The slow step gives $\text{rate} = k[\text{H}_2\text{O}_2][\text{I}^-]$: a catalyst can appear in the rate law, even though it is not in the overall equation.

Common mistakes.

• Writing the rate law from the coefficients of the overall equation. Orders come from experiment, or from the slow step of a mechanism, never from the balanced equation.
• Leaving an intermediate in a final rate law. Replace it using the fast equilibrium.
• Using the orders of the slow step when a fast step comes before it, without substituting.
• Calling the highest point on the profile the "intermediate". Peaks are transition states; intermediates sit in the valleys between peaks.
• Saying a catalyst "lowers $\Delta H$" or "shifts the equilibrium". It only lowers $E_a$, for both directions equally.

• Use collision theory to explain temperature and suitable catalyst effects; check the rate law for concentration effects and whether a solid surface is involved.
• Temperature works mainly by getting more particles above the activation energy, not just more collisions.
• Reaction orders come from experiment, not the balanced coefficients — see how the rate changes when one concentration is varied.
• On an energy profile the hill height is the activation energy and the reactant–product gap is $\Delta H$.
• A catalyst lowers the activation energy (a new pathway) but leaves $\Delta H$ and the equilibrium position unchanged.