U4.4 · Physical and Chemical ChangesPractice · 物理变化与化学变化

The test is whether a new substance forms. $\ce{O2}$ and $\ce{O3}$ both contain only O atoms, but their molecules differ, so they are different substances (allotropes): O=O bonds break and new bonds form, a chemical change. Boiling, dissolving, and compressing leave the $\ce{O2}$ molecules unchanged.

Electrolysis breaks O–H bonds and regroups the atoms into $\ce{H2}$ and $\ce{O2}$, new substances. Boiling only overcomes the hydrogen bonds between water molecules; the bubbles are water vapor, still $\ce{H2O}$. Needing energy, being reversible, or forming bubbles doesn’t decide whether a change is chemical.

The water in $\ce{CuSO4.5H2O}$ is water of crystallization, present in a fixed ratio as part of the compound. Heating produces a different substance, anhydrous $\ce{CuSO4}$, with a different color and composition, so it is a chemical change. Reversibility doesn’t decide it; water on wet sand just clings to the surface in no fixed ratio.

Liquid bromine turning to vapor leaves the $\ce{Br2}$ molecules intact; only the distance between them grows, so it is a physical change. In the other three the molecules change: $\ce{N2O4}$ splits into $\ce{NO2}$, $\ce{H2}$ and $\ce{O2}$ form water, and $\ce{O3}$ becomes $\ce{O2}$. Bonds break and form, so they are chemical changes.

Sucrose is a molecular solid. Dissolving overcomes only the forces between sucrose molecules; the molecules spread into the water and hydrogen-bond with it, and no covalent bonds inside them break. Sucrose is a nonelectrolyte, so it doesn’t ionize and the solution doesn’t conduct. Evaporating the water gives back sucrose.