Kinetics

Snap a glow stick and drop it into hot water: it shines brightly but fades within an hour. Put an identical one in the freezer and it glows dimly, but for many hours. The same chemicals react in both, and only the speed differs. Food in a fridge, iron rusting and the engine of a car all depend on the same question: how fast? This unit measures reaction speed, explains it with colliding particles, and uses it to work out the steps a reaction really takes.

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 balanced equation links the rates of the different species. For $2\text{N}_2\text{O}_5\rightarrow4\text{NO}_2+\text{O}_2$, for every 2 mol of $\text{N}_2\text{O}_5$ used, 4 mol of $\text{NO}_2$ and 1 mol of $\text{O}_2$ form. If $\text{N}_2\text{O}_5$ is used up at $0.010$ M/s, then $\text{NO}_2$ forms at $0.020$ M/s and $\text{O}_2$ at $0.0050$ M/s. Always say which species a rate refers to.

To measure a rate in the lab, follow something that changes as the reaction goes: the volume of gas given off, the mass lost as gas escapes, a colour (with a colorimeter), the pressure, or the pH.

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.

Worked example (method of initial rates). For $2\text{NO}+\text{O}_2\rightarrow2\text{NO}_2$:

Experiment	$[\text{NO}]_0$ / M	$[\text{O}_2]_0$ / M	Initial rate / M s⁻¹
1	0.010	0.010	$2.5\times10^{-5}$
2	0.020	0.010	$1.0\times10^{-4}$
3	0.010	0.020	$5.0\times10^{-5}$

From 1 to 2, $[\text{NO}]$ doubles and the rate goes up 4 times: second order in NO. From 1 to 3, $[\text{O}_2]$ doubles and the rate doubles: first order in $\text{O}_2$. So $\text{rate}=k[\text{NO}]^2[\text{O}_2]$, third order overall. Using experiment 1:

The units of $k$ depend on the overall order: $\text{M s}^{-1}$ for zero order, $\text{s}^{-1}$ for first, $\text{M}^{-1}\text{s}^{-1}$ for second, $\text{M}^{-2}\text{s}^{-1}$ for third. The value of $k$ changes with temperature, but not with concentration.

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.

For times that are not whole half-lives, use the integrated law, $\ln[\text{A}]_t-\ln[\text{A}]_0=-kt$. For this reaction, starting at 0.50 M, after 60 s:

Radioactive decay 放射性衰变 is always first order. Carbon-14 has a half-life of 5730 years. A piece of wood with one-eighth of the carbon-14 of living wood has passed through 3 half-lives ($\tfrac12\times\tfrac12\times\tfrac12=\tfrac18$), so it is about $3\times5730\approx17{,}000$ years old. That is how carbon dating works.

Worked example (which plot is straight?). A reactant is measured at 0, 10, 20 and 30 s: 0.80, 0.40, 0.20 and 0.10 M. The concentration halves every 10 s, which is a constant half-life, so the reaction is first order and a plot of $\ln[\text{A}]$ against $t$ is a straight line. Its slope is $-k$, with $k=0.693/10=0.069\ \text{s}^{-1}$. For a second-order reaction, each half-life is twice as long as the one before.

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.

For an elementary step, and only for an elementary step, the rate law comes straight from the equation for that step:

Elementary step	Molecularity	Rate law
$\text{A}\rightarrow\text{products}$	unimolecular	$k[\text{A}]$
$\text{A}+\text{B}\rightarrow\text{products}$	bimolecular	$k[\text{A}][\text{B}]$
$2\text{A}\rightarrow\text{products}$	bimolecular	$k[\text{A}]^2$

A step in which three particles must collide at the same moment is very rare, because three particles almost never meet at one point at the same time.

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.

Most collisions do not lead to a reaction. In a gas at room temperature each molecule collides billions of times per second, so if every collision worked, every reaction would be over almost instantly. The Maxwell-Boltzmann distribution 麦克斯韦-玻尔兹曼分布 shows the spread of particle energies.

Raising the temperature makes particles collide a little more often, but the main effect is that a much larger fraction of collisions now have energy above $E_a$. That is why, for many reactions near room temperature, a rise of just 10 °C roughly doubles the rate. The Arrhenius equation describes this link between $k$, $E_a$ and temperature; AP asks you to explain it, not to calculate with it.

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).

Catalysts work in different ways:

Type	How it works	Example
Binding (enzyme)	holds reactants in the right orientation, lowering $E_a$	enzymes in living cells
Acid–base	a reactant gains or loses a proton, giving a new, easier intermediate	$\text{H}^+$ speeding up the making of esters
Surface	reactants bond to a metal surface, which weakens their bonds	platinum in a car's catalytic converter; iron in the Haber process

In each case the catalyst is used in one step, often the slow one, and made again in a later step, so its total amount does not change.

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.