反应动力学

A rate equation 速率方程 shows how the rate depends on the concentrations 浓度 of the reactants 反应物:

• $m$ is the order of reaction 反应级数 with respect to A, and $n$ the order with respect to B. Each is $0$, $1$ or $2$.
• the overall order of reaction 总反应级数 is $m + n$.
• $k$ is the rate constant 速率常数. The rate equation can only be found by experiment, not from the balanced equation 配平方程式.

Finding the order

• initial rates method 初始速率法: change one concentration at a time and see how the starting rate changes. If doubling $[\text{A}]$ doubles the rate, the order in A is 1; if it quadruples the rate, the order is 2; if the rate is unchanged, the order is 0.

Worked example. In experiments on $\text{A} + \text{B} \rightarrow$ products, doubling $[\text{A}]$ (with $[\text{B}]$ fixed) doubles the rate, and doubling $[\text{B}]$ (with $[\text{A}]$ fixed) quadruples the rate. Write the rate equation and give the overall order.

Doubling $[\text{A}]$ doubles the rate, so first order 一级 in A. Doubling $[\text{B}]$ quadruples ($2^2$) the rate, so second order 二级 in B. Hence

Units of $k$. Rearrange the rate equation so that $k$ is alone, then put in the units: rate in $\text{mol dm}^{-3}\,\text{s}^{-1}$, each concentration in $\text{mol dm}^{-3}$. For the example above:

Each extra order adds one $\text{mol}^{-1}\,\text{dm}^{3}$. Work the units out each time rather than learning the table: a mistake in the algebra shows up at once.

• graphs: a concentration–time graph 浓度–时间图 for a first-order reaction has a constant half-life 半衰期 (the time for the concentration to halve). A rate–concentration graph is a straight line through the origin 原点 for first order, and a curve for second order.

Half-life and the rate constant

For a first-order reaction the half-life is constant — it does not depend on the concentration. You can find the rate constant from it:

Worked example. A first-order reaction has a half-life of $120\ \text{s}$. Find its rate constant.

You can also find $k$ by putting initial-rate data into the rate equation.

Most reactions happen in several steps 步骤. The slowest step is the rate-determining step 决速步骤, and it controls the overall rate. Only the species 微粒 involved up to and including this step appear in the rate equation.

• an intermediate 中间体 is a species made in one step and then used up in a later step. It is not in the overall equation.
• you can suggest a mechanism 反应机理 that fits both the rate equation and the overall equation, predict the order from a given mechanism, or pick out the rate-determining step.

If you compare the initial rate of different mixtures, you can deduce the rate equation, and from that work out the mechanism.

Worked example (mechanism from a rate equation). For $2\text{NO}_2 + \text{F}_2 \rightarrow 2\text{NO}_2\text{F}$ the rate equation is $\text{rate} = k[\text{NO}_2][\text{F}_2]$. Suggest a two-step mechanism.

1. The rate equation has one $\text{NO}_2$ and one $\text{F}_2$, so the rate-determining step uses exactly one of each: $\text{NO}_2 + \text{F}_2 \rightarrow \text{NO}_2\text{F} + \text{F}$ (slow).
2. The second $\text{NO}_2$ must react in a later, fast step, with the F atom made in step 1: $\text{NO}_2 + \text{F} \rightarrow \text{NO}_2\text{F}$ (fast).
3. Check: adding the steps and cancelling F gives the overall equation. F is the intermediate.

The same reasoning explains the hydrolysis of 2-bromo-2-methylpropane, $\text{rate} = k[(\text{CH}_3)_3\text{CBr}]$. Hydroxide ions are not in the rate equation, so they react after the slow step: the C–Br bond breaks first to give a carbocation, which $\text{OH}^-$ then attacks. This is the $\text{S}_\text{N}1$ mechanism. A reactant that is zero order is still used up; it just reacts after the rate-determining step.

Effect of temperature

Raising the temperature increases the rate constant $k$ (more molecules pass the activation energy 活化能), so the rate goes up.

Catalysts 催化剂 can be homogeneous or heterogeneous.

Heterogeneous catalysts

A heterogeneous catalyst 多相催化剂 is in a different physical state 物态 from the reactants (usually a solid with gases). It works in three stages:

1. adsorption 吸附: reactant molecules stick to the catalyst surface.
2. the bonds in the reactants are weakened 削弱, so they react more easily.
3. desorption 脱附: the product molecules leave the surface.

Examples are iron in the Haber process 哈伯法, and platinum, palladium and rhodium in a catalytic converter 催化转化器.

Homogeneous catalysts

A homogeneous catalyst 均相催化剂 is in the same physical state as the reactants. It is used up in one step and then reformed 再生 in a later step, so it comes back unchanged. Examples are oxides of nitrogen helping to oxidise 氧化 atmospheric 大气中的 sulfur dioxide, and $\text{Fe}^{2+}$ or $\text{Fe}^{3+}$ speeding up the reaction between $\text{I}^-$ and $\text{S}_2\text{O}_8^{2-}$.

How the iron ions work. The reaction $\text{S}_2\text{O}_8^{2-} + 2\text{I}^- \rightarrow 2\text{SO}_4^{2-} + \text{I}_2$ is slow because both reactants are negative ions, which repel 互相排斥 each other, so the activation energy is high. With iron ions it goes in two steps:

Each step is between a positive and a negative ion, which attract, so each has a lower activation energy. The $\text{Fe}^{2+}$ used in the first step is made again in the second, and the steps can run in either order, which is why $\text{Fe}^{3+}$ works as well as $\text{Fe}^{2+}$. Iron can do this because it has two stable oxidation states, $+2$ and $+3$.

Nitrogen oxides do the same for sulfur dioxide in the air:

$\text{NO}_2$ is used in the first step and made again in the second; adding the steps gives $\text{SO}_2 + \tfrac{1}{2}\text{O}_2 \rightarrow \text{SO}_3$.

Common mistakes.

• Taking the orders from the balanced equation. In the $\text{NO}_2$ example above the equation has $2\text{NO}_2$, but the order in $\text{NO}_2$ is 1. Orders come only from experiment.
• Saying a zero-order reactant "is not needed" or "does not react". It is used up, but after the rate-determining step.
• Using the half-life to find $k$ for a reaction that is not first order. Only a first-order reaction has a constant half-life.
• Saying $k$ changes when a concentration changes. $k$ changes only with temperature (or a catalyst); a change of concentration changes the rate.
• Mixing up a catalyst and an intermediate. A catalyst is there at the start and made again at the end; an intermediate is made in one step and used up in a later one.

• Find orders from initial-rate data: doubling a concentration that doubles the rate is first order, quadruples it is second order, no change is zero order 零级.
• Write $\text{rate} = k[\text{A}]^m[\text{B}]^n$ and work out the units of $k$ from it.
• The rate-determining step contains the species (and orders) in the rate equation — use this to test a mechanism.
• A constant half-life means first order.