Dynamic equilibrium: when the colour stops changing, has the reaction stopped?

01预测Predict

02让烧瓶反应Run the flask

先让它跑到颜色稳定。然后给一部分 N₂O₄ 打上 ¹⁵N 标记：浓度不变，但你能追踪原子去了哪里。

Run it until the colour is steady. Then tag some N₂O₄ with ¹⁵N: the concentrations don’t change, but you can follow where the atoms go.

烧瓶里一开始装的是The flask starts with 只有 N₂O₄（40 mM，无色）Only N₂O₄ (40 mM, colourless) 只有 NO₂（80 mM，深棕色，氮原子总数相同）Only NO₂ (80 mM, dark brown, same number of N atoms)

把 25% 的 N₂O₄ 换成 ¹⁵N 标记的分子Swap 25% of the N₂O₄ for ¹⁵N-tagged molecules

03解释：停住的是颜色，不是反应Explain: the colour stops, the reaction doesn’t

颜色只取决于 [NO₂]。颜色稳定，说明 [NO₂] 不变，也就是每秒生成的 NO₂ 和每秒消耗的 NO₂ 一样多。两边都不是零。

标记实验就是证据：浓度一点没变，¹⁵N 却从 N₂O₄ 跑进了 NO₂。如果反应真的停了，标记会永远留在 N₂O₄ 里。

为什么两个速率会自己变得相等

分解速率 r→ = k→[N₂O₄]，结合速率 r← = k←[NO₂]²（两个 NO₂ 相撞才能结合，所以是平方）。一开始只有 N₂O₄，r← 为零；NO₂ 越积越多，r← 越来越大，r→ 越来越小，直到两者相等。令 r→ = r←，就得到

K = k→ / k← = [NO₂]² / [N₂O₄]

这就是平衡常数。它不是额外的定律，而是“两个速率相等”的另一种写法（古尔德贝格和瓦格在 1864 年提出的质量作用定律）。

试试改从“只有 NO₂”开始：两条路最后停在同一组浓度上。平衡位置只由 K 和原子总数决定，跟从哪边出发无关。

The colour depends only on [NO₂]. A steady colour means [NO₂] is constant: NO₂ is made exactly as fast as it is used up. Neither rate is zero.

The tags are the evidence. The concentrations never moved, yet ¹⁵N crossed from N₂O₄ into NO₂. If the reaction had stopped, every tag would have stayed in N₂O₄.

Why the two rates even out on their own

Splitting runs at r→ = k→[N₂O₄]; joining at r← = k←[NO₂]² (two NO₂ must collide, hence the square). At first there is no NO₂, so r← is zero. As NO₂ builds up, r← climbs and r→ falls until they meet. Set r→ = r← and you get

K = k→ / k← = [NO₂]² / [N₂O₄]

That is the equilibrium constant. It is not an extra law; it is “the two rates are equal” written another way (the law of mass action, Guldberg and Waage, 1864).

Now start from “only NO₂”. Both runs settle on the same concentrations: the equilibrium depends on K and the total number of atoms, not on which side you start from.

先看浓度：两条曲线走平了，颜色就不再变。Start with concentration: both curves flatten, so the colour stops changing.
再看速率：两条线在同一个高度汇合，都不是零。Now the rates: the two lines meet at the same height, and neither is zero.
看烧瓶：闪光一直在出现，每一个都是一次分解或结合。Look in the flask: flashes keep appearing, each one a split or a join.
看标记：浓度没动，¹⁵N 却进了 NO₂。Look at the tag: the concentrations hold still, yet ¹⁵N moves into NO₂.
最后看 Q：它停在 K 上，正是“两个速率相等”的另一种写法。Finally Q: it settles at K, which is “the two rates are equal” written another way.

用你自己的话写：In your own words:

04打破平衡Disturb it

温度Temperature

勒夏特列原理是速率的结果。加入 N₂O₄ 后 r→ 立刻变大，r← 要等 NO₂ 积累才跟上。于是净反应向右，直到 Q 回到 K。“平衡会抵消扰动”说的就是这件事（勒夏特列，1884）。

温度改的是 K 本身。N₂O₄ 分解要吸收约 57 kJ/mol，加热时 k→ 增长得远比 k← 快，所以 K = k→/k← 变大，热烧瓶更棕，冰水里几乎无色。

催化剂不改 K。它同时降低两个方向的能垒，k→ 和 k← 按同一倍数变大：到达平衡更快，平衡位置不变。

Le Chatelier’s principle falls out of the rates. Adding N₂O₄ raises r→ at once; r← only catches up as NO₂ builds. So the net reaction runs forward until Q is back at K. “Equilibrium counteracts a disturbance” describes exactly that (Le Chatelier, 1884).

Temperature changes K itself. Splitting N₂O₄ absorbs about 57 kJ/mol, so heating speeds up k→ far more than k←. K = k→/k← grows: the hot flask is browner, the iced one nearly colourless.

A catalyst leaves K alone. It lowers the barrier both ways, scaling k→ and k← by the same factor: equilibrium arrives sooner but in the same place.

05独立练习Independent practice

这一步收起辅助：图上的曲线、标签和读数都会隐藏，控件也会锁住。只看题目里的数字作答。

This step removes the supports: the plots, labels and readouts are hidden and the controls lock. Work from the numbers in each question.

06迁移：饱和食盐水Transfer: a saturated salt solution

一杯饱和食盐水，杯底有没溶完的盐晶体，放了一周，晶体质量一点没变。现在放进一粒用放射性钠标记的盐晶体。一天后，标记会在哪里？A saturated salt solution has undissolved crystals at the bottom; their mass hasn’t changed in a week. You drop in one crystal made with radioactive sodium. A day later, where is the label? 还在那粒晶体里：溶液饱和了，不会再溶解。Still in that crystal: a saturated solution can’t dissolve any more. 溶液里和其他晶体上都有了，但晶体总质量不变。In the solution and on other crystals too, while the total crystal mass stays the same. 进入了溶液，因为那粒晶体溶解变小了。In the solution, because that crystal dissolved and shrank.

这个模型的适用范围Where this model applies

时间被大大放慢了：真实的 N₂O₄ 在一眨眼间就到达平衡。速率常数的绝对值是为观看而选的，比值 K 才是真实的：25 °C 时 Kc ≈ 4.6 mM，ΔH ≈ +57.2 kJ/mol。温度依赖用范特霍夫方程计算，并把 ΔH 直接用于 Kc，这是近似。

模型把 k← 当成与温度无关（NO₂ 结合几乎没有能垒），把气体当作理想、均匀的。瓶里画出的每个分子代表 0.5 mM；粒子的位置和谁跟谁反应是随机示意，反应次数跟着速率模型走，所以图和读数不会对不上。

标记实验是示意：真实的同位素交换实验更常用于溶液和固体（比如迁移题里的盐），原理相同。

历史：古尔德贝格和瓦格（1864）提出质量作用定律；勒夏特列（1884）总结了平衡对扰动的响应；德·海韦西把放射性同位素用作示踪剂，获 1943 年诺贝尔化学奖。

Time is slowed down enormously: real N₂O₄ reaches equilibrium in the blink of an eye. The absolute rate constants were chosen for viewing; their ratio K is real: Kc ≈ 4.6 mM at 25 °C, ΔH ≈ +57.2 kJ/mol. Temperature dependence uses the van ’t Hoff equation with ΔH applied to Kc, an approximation.

The model treats k← as temperature-independent (NO₂ recombination has almost no barrier) and the gas as ideal and well mixed. Each drawn molecule stands for 0.5 mM. Positions and which molecules react are a random illustration; the number of reactions follows the rate model, so the flask never disagrees with the readouts.

The tagging run is illustrative. Real isotope-exchange experiments are more often done in solutions and solids (like the salt in the transfer question); the principle is the same.

History: Guldberg and Waage proposed the law of mass action (1864); Le Chatelier described how equilibria respond to disturbance (1884); George de Hevesy’s use of radioactive isotopes as tracers earned the 1943 Nobel Prize in Chemistry.

离线可用。用 Tab 切换控件，方向键调节滑块。没有快捷键会干扰你的写作。Works offline. Use Tab to reach controls and arrow keys on sliders. No keyboard shortcuts override your writing.

颜色不再变了，反应就停了吗？When the colour stops changing, has the reaction stopped?