化学反应

A chemical reaction 化学反应 rearranges atoms into new substances. Possible evidence includes a color change, a gas or precipitate 沉淀 forming, or a temperature change. These observations alone do not prove that new substances formed. Atoms are conserved, so an equation must be balanced – the same count of each element on both sides.

When writing aqueous ionic equations, split dissolved strong electrolytes into ions; keep insoluble solids and weak electrolytes in their appropriate forms. A net ionic equation 净离子方程式 shows only the species that actually change, leaving out the spectator ions 旁观离子 that appear unchanged on both sides. It captures the real chemistry (e.g. $\text{Ag}^+ + \text{Cl}^- \rightarrow \text{AgCl}(s)$).

The same reaction can be shown as a symbolic equation, a particulate drawing (atoms and molecules), and a macroscopic observation (what you see). Moving between these levels – connecting the equation to the particles to the beaker – is a core skill.

A physical change 物理变化 alters form but not identity (melting, dissolving); a chemical change 化学变化 makes new substances by breaking and forming bonds. Dissolving salt is physical; the salt is unchanged and recoverable.

Stoichiometry 化学计量 uses the balanced equation's mole ratios to relate amounts of reactants and products. The path is always grams → moles → (mole ratio) → moles → grams. The limiting reactant 限量反应物 runs out first and sets the maximum product (the theoretical yield 理论产量); the percent yield compares actual to theoretical.

Worked example. How much water forms when $4.0\ \text{g}$ of hydrogen burns in excess oxygen? $2\text{H}_2+\text{O}_2\rightarrow2\text{H}_2\text{O}$. Convert to moles, cross the mole ratio ($2:2=1:1$ here), convert back:

Worked example (limiting reactant). $10.0\ \text{g}$ of N$_2$ reacts with $5.0\ \text{g}$ of H$_2$ ($\text{N}_2+3\text{H}_2\rightarrow2\text{NH}_3$). Which runs out? Moles: $n(\text{N}_2)=10.0/28.0=0.357$, $n(\text{H}_2)=5.0/2=2.5$. The reaction needs $3$ H$_2$ per N$_2$; you have $2.5/0.357=7.0$, far more than $3$, so N$_2$ is limiting. It makes $2\times0.357=0.714\ \text{mol}$ of ammonia ($12.1\ \text{g}$), with hydrogen left over.

A titration 滴定 finds an unknown concentration by reacting it with a solution of known concentration until they reach the equivalence point 等当点 (stoichiometrically equal). From the known volume and concentration, use the mole ratio to find the unknown. An indicator or a pH curve signals the endpoint.

Worked example. $25.0\ \text{mL}$ of hydrochloric acid is exactly neutralized by $30.0\ \text{mL}$ of $0.100\ \text{M}$ NaOH. Find the acid's concentration. The reaction is $1:1$, so the moles match:

Common patterns include synthesis (combining), decomposition (breaking apart), combustion (with oxygen, releasing energy), precipitation (forming an insoluble solid), acid–base (proton transfer), and redox (electron transfer). Recognizing the type helps predict the products.

An acid 酸 donates a proton ($\text{H}^+$); a base 碱 accepts one (Brønsted–Lowry). For an aqueous acid reacting with a hydroxide base, the net neutralisation is: $\text{H}^+ + \text{OH}^- \rightarrow \text{H}_2\text{O}$. Strong acids ionise essentially completely in dilute water; soluble strong hydroxide bases dissociate into ions. Weak acids and bases establish equilibria.

In a redox 氧化还原 reaction, electrons transfer between species. Oxidation 氧化 is loss of electrons (oxidation number rises); reduction 还原 is gain (oxidation number falls). Track changes with oxidation numbers, and remember every oxidation is paired with a reduction – the electrons lost equal the electrons gained.

Worked example. Find the oxidation number of manganese in permanganate, $\text{KMnO}_4$. Potassium is $+1$ and each oxygen is $-2$ (four of them, $-8$). The whole formula is neutral, so $(+1)+\text{Mn}+(-8)=0$, giving $\text{Mn}=+7$ – its maximum, which is why permanganate is a powerful oxidizing agent (it can only gain electrons).

Common mistakes.

• Splitting a precipitate or a weak acid into ions. In a net ionic equation only dissolved strong electrolytes are written as ions; $\text{AgCl}(s)$, water and $\text{CH}_3\text{COOH}$ stay whole.
• Choosing the limiting reactant by the smaller mass. Compare moles, divided by the coefficients: in the example above, 10.0 g of N$_2$ is the larger mass but it runs out first.
• Rounding moles early. Carry an extra figure through the working ($0.357$, not $0.36$) and round only the final answer.
• Using mL in $n=cV$. Change to litres first: $30.0\ \text{mL}=0.0300\ \text{L}$.
• Calling dissolving a chemical change because the solid "disappears". Dissolving salt is physical: the ions are unchanged and the salt comes back when the water evaporates.
• Mixing up oxidation and reduction. Oxidation is loss of electrons (the oxidation number rises); the species oxidised is the reducing agent.

• Balance every equation and work in moles — convert grams→moles, cross the mole ratio, then convert back.
• Find the limiting reactant by comparing mole ratios; it sets the maximum product (theoretical yield).
• In redox, remember OIL RIG: oxidation is loss of electrons, reduction is gain; every oxidation is paired with a reduction.
• For titrations use the balanced ratio to link the known and unknown at the equivalence point.
• A net ionic equation shows only the species that change; leave out spectator ions.