IB Chemistry — Structure 1 · 鼎睿学苑

Models of the Particulate Nature of Matter物质粒子性模型

From states of matter to electron configurations, from the mole concept to ideal gases. This unit builds the foundational models that underpin all of chemistry.从物质三态到电子构型(electron configuration),从摩尔(mole)概念到理想气体(ideal gas)。本单元搭建起支撑整门化学的基础模型。

SL: 17 hrs · HL: 21 hrsSL: 17 学时 · HL: 21 学时 5 Sub-topics5 个子主题 HL adds 1.2.3, 1.3.6, 1.3.7HL 加考 1.2.3、1.3.6、1.3.7

Introduction to the Particulate Nature of Matter物质粒子性导论

All matter is composed of particles. The three classifications of matter — elements, compounds, and mixtures — describe how those particles are organized and bonded. Understanding this hierarchy is the starting point for everything else in chemistry.一切物质都由粒子构成。物质的三种分类——元素(element)、化合物(compound)、混合物(mixture)——描述了这些粒子的组合方式与成键关系。理清这一层次结构是后续所有化学内容的起点。

Elements, Compounds, and Mixtures Elements cannot be chemically broken down into simpler substances. They consist of one type of atom. Compounds consist of atoms of different elements chemically bonded in a fixed ratio. Their properties differ from their constituent elements. Mixtures contain more than one element or compound in no fixed ratio, not chemically bonded — they can be separated by physical methods like filtration, distillation, evaporation, recrystallization, and chromatography.
元素、化合物与混合物 元素(element无法通过化学方法分解为更简单的物质,由同一种原子组成。 化合物(compound由不同元素的原子按固定比例化学键合而成,其性质与组成它的元素不同。 混合物(mixture含有一种以上的元素或化合物,比例不固定,组分之间没有化学键合——可通过过滤、蒸馏、蒸发、重结晶、色谱等物理方法(filtration / distillation / evaporation / recrystallization / chromatography)分离。
Homogeneous vs. Heterogeneous Mixtures A homogeneous mixture has uniform composition throughout (e.g., salt water, air). A heterogeneous mixture has visibly different components or phases (e.g., sand and water, oil and vinegar). The distinction matters when choosing separation techniques.
均相混合物与非均相混合物 均相(homogeneous混合物各处组成均匀(如盐水、空气);非均相(heterogeneous混合物可见地由不同组分或相组成(如沙与水、油与醋)。选择分离方法时这一区分至关重要。

Kinetic Molecular Theory and States of Matter分子运动论与物质三态

The kinetic molecular theory models matter as particles in constant motion. The three states — solid, liquid, and gas — differ in the arrangement, spacing, and motion of their particles. Solids have fixed positions with vibrational motion. Liquids have particles close together but able to slide past each other. Gases have widely spaced particles moving rapidly in all directions.分子运动论(kinetic molecular theory把物质看作不停运动的粒子集合。固、液、气三态在粒子排列、间距和运动方式上各不相同:固体粒子位置固定,仅作振动;液体粒子彼此靠近,但可相互滑动;气体粒子间距很大,向各方向高速运动。

Changes of State Melting (solid to liquid), freezing (liquid to solid), vaporization (liquid to gas — includes evaporation and boiling), condensation (gas to liquid), sublimation (solid to gas), and deposition (gas to solid). Each involves energy transfer: breaking intermolecular forces requires energy (endothermic), forming them releases energy (exothermic).
物态变化 熔化(melting,固→液)、凝固(freezing,液→固)、汽化(vaporization,液→气,包括蒸发与沸腾)、凝结(condensation,气→液)、升华(sublimation,固→气)、凝华(deposition,气→固)。每一种都涉及能量传递:克服分子间作用力(intermolecular force)需要吸热(endothermic),形成分子间作用力则放热(exothermic)。
State Symbols in Equations Always include state symbols: (s) solid, (l) liquid, (g) gas, (aq) aqueous (dissolved in water). These are required in IB Chemistry equations and carry marks in exams.
化学方程中的状态符号 化学方程中必须写出状态符号(state symbols):(s) 固态、(l) 液态、(g) 气态、(aq) 水溶液(溶于水)。IB Chemistry 方程式题强制要求,缺失会被扣分。

Temperature and Kinetic Energy温度与动能

Temperature in Kelvin (K) is a measure of the average kinetic energy ($E_k$) of particles. The kelvin has the same increment size as the Celsius degree. Converting between them:开尔文(K)温标度量的是粒子的平均动能($E_k$)。开尔文与摄氏度的刻度大小相同。两者换算:

Celsius ↔ Kelvin Conversion摄氏度与开尔文的换算
$$T(K) = T(°C) + 273$$

At 0 K (absolute zero), particles have minimum kinetic energy. During a change of state, temperature remains constant even though heat is being added or removed — the energy goes into breaking or forming intermolecular forces.

在 0 K(绝对零度,absolute zero)时粒子的动能最小。物态变化(phase change)过程中,即使持续吸热或放热,温度也保持不变——能量全部用于克服或形成分子间作用力。

During the melting of ice at 0 degrees C, which statement is correct?冰在 0 °C 熔化过程中,下列说法正确的是?
The kinetic energy of particles increases粒子的动能增加
The temperature rises steadily温度持续升高
Energy is used to overcome intermolecular forces能量用于克服分子间作用力
The particles stop moving粒子停止运动
Correct! During a phase change, added energy goes into overcoming intermolecular forces rather than increasing kinetic energy. The temperature stays constant until the phase change is complete.正确!物态变化期间,吸收的能量用于克服分子间作用力,而非增加动能。直到相变完成,温度始终保持不变。
During melting, the temperature remains constant. The added energy overcomes intermolecular forces rather than increasing particle speed. Answer: (C).熔化过程中温度保持不变。吸收的能量克服分子间作用力,而不是加快粒子运动。答案:(C)。

The Nuclear Atom核式原子模型

Atoms consist of a positively charged, dense nucleus containing protons and neutrons (collectively called nucleons), surrounded by negatively charged electrons. The nuclear symbol $_Z^A X$ encodes the identity and composition of any atom or ion.原子由带正电的致密原子核(含质子与中子,二者统称核子)和围绕其外、带负电的电子组成。核符号(nuclear symbol)$_Z^A X$ 编码了任意原子或离子的身份与组成。

Nuclear Symbol核符号
$$_Z^A X$$

$Z$ = atomic number (protons), $A$ = mass number (protons + neutrons). Neutrons = $A - Z$. For neutral atoms, electrons = protons = $Z$.

$Z$ = 原子序数(atomic number,质子数),$A$ = 质量数(mass number,质子数+中子数)。中子数 = $A - Z$。中性原子的电子数等于质子数,即 $Z$。

Subatomic Particles亚原子粒子
ParticleRelative MassRelative ChargeLocation
Proton1+1Nucleus
Neutron10Nucleus
Electron~1/1840 (negligible)-1Electron shells
粒子相对质量相对电荷位置
质子 (proton)1+1原子核
中子 (neutron)10原子核
电子 (electron)约 1/1840(可忽略)-1电子壳层

Isotopes同位素

Isotopes are atoms of the same element (same number of protons) with different numbers of neutrons. They have identical chemical properties but different physical properties (mass, density, rate of diffusion). The relative atomic mass ($A_r$) of an element is a weighted average of its isotopic masses based on their natural abundances.同位素(isotope是同一元素(质子数相同)但中子数不同的原子。它们化学性质相同,但物理性质(质量、密度、扩散速率)有差异。元素的相对原子质量(relative atomic mass,$A_r$)是按自然丰度对各同位素质量取的加权平均值。

Worked Example — Relative Atomic Mass from Isotopic Data例题 — 由同位素数据求相对原子质量

Chlorine has two isotopes: $^{35}$Cl (75.77%) and $^{37}$Cl (24.23%). Calculate the relative atomic mass.

氯有两种同位素:$^{35}$Cl(75.77%)与 $^{37}$Cl(24.23%)。求其相对原子质量。

Weighted average
$$A_r = \frac{(35 \times 75.77) + (37 \times 24.23)}{100}$$
$$A_r = \frac{2651.95 + 896.51}{100} = \frac{3548.46}{100} = 35.48$$
This is why chlorine's relative atomic mass is approximately 35.5, not a whole number.
加权平均
$$A_r = \frac{(35 \times 75.77) + (37 \times 24.23)}{100}$$
$$A_r = \frac{2651.95 + 896.51}{100} = \frac{3548.46}{100} = 35.48$$
这就是氯的相对原子质量约为 35.5(而非整数)的原因。
HL Only — Mass Spectrometry (1.2.3) Mass spectra show peaks corresponding to each isotope. The $x$-axis shows mass-to-charge ratio ($m/z$), and the $y$-axis shows relative abundance. You can read off isotopic masses and their abundances directly from the spectrum and calculate $A_r$ from the data.
仅 HL — 质谱法(1.2.3) 质谱图(mass spectrum)上每个同位素对应一个峰。横轴是质荷比(mass-to-charge ratio,$m/z$),纵轴是相对丰度(relative abundance)。可直接从谱图读出各同位素的质量与丰度,并据此计算 $A_r$。
An atom of $_{26}^{56}$Fe$^{3+}$ contains:$_{26}^{56}$Fe$^{3+}$ 含有:
26 protons, 30 neutrons, 26 electrons26 个质子、30 个中子、26 个电子
26 protons, 30 neutrons, 23 electrons26 个质子、30 个中子、23 个电子
26 protons, 56 neutrons, 23 electrons26 个质子、56 个中子、23 个电子
56 protons, 26 neutrons, 53 electrons56 个质子、26 个中子、53 个电子
Correct! Protons = $Z$ = 26. Neutrons = $A - Z$ = 56 - 26 = 30. The 3+ charge means 3 electrons were lost: 26 - 3 = 23 electrons.正确!质子数 = $Z$ = 26;中子数 = $A - Z$ = 56 - 26 = 30;3+ 电荷意味着失去 3 个电子,故电子数 = 26 - 3 = 23。
$Z$ = 26 protons, neutrons = 56 - 26 = 30, and the 3+ charge means 3 fewer electrons than protons: 26 - 3 = 23. Answer: (B).$Z$ = 26 个质子;中子 = 56 - 26 = 30;3+ 表示电子比质子少 3 个:26 - 3 = 23。答案:(B)。

Electron Configurations电子构型

Emission Spectra and Energy Levels发射光谱与能级

When atoms absorb energy, electrons are excited to higher energy levels. When they fall back down, they emit photons of specific energies, producing a line emission spectrum. The hydrogen emission spectrum provides direct evidence that electrons exist in discrete (quantized) energy levels that converge at higher energies.当原子吸收能量时,电子被激发到更高的能级;当它们跃迁回较低能级时,发射出特定能量的光子,形成线状发射光谱(line emission spectrum。氢原子发射光谱直接证明:电子处于离散(量子化)能级,且能级在高能端逐渐汇聚。

Continuous vs. Line Spectra A continuous spectrum contains all wavelengths (like white light through a prism). A line spectrum contains only specific wavelengths, corresponding to specific energy transitions within atoms. Each element has a unique line spectrum — a chemical fingerprint.
连续光谱与线状光谱 连续光谱(continuous spectrum包含所有波长(如白光经棱镜色散);线状光谱(line spectrum只含特定波长,对应原子内特定的能级跃迁。每种元素的线状光谱独一无二——堪称化学指纹。

Main Energy Levels and Sublevels主能级与亚层

The main energy level is given by the integer $n$ (1, 2, 3...) and can hold a maximum of $2n^2$ electrons. Within each main level, electrons occupy sublevels: s, p, d, and f, each with successively higher energy.主能级由整数 $n$(1、2、3……)标记,最多容纳 $2n^2$ 个电子。在每个主能级内,电子分布于亚层(subshell):s、p、d、f,能量依次升高。

Sublevel Summary亚层总览
SublevelOrbitalsMax ElectronsShape
s12Spherical
p36Dumbbell (3 orientations)
d510Complex (various)
f714Complex (various)
亚层轨道数最大电子数形状
s12球形
p36哑铃形(3 种取向)
d510复杂(多种形状)
f714复杂(多种形状)

Electron Configuration Rules电子排布规则

Three principles govern how electrons fill orbitals:三条原则规定了电子如何填充轨道(orbital):

Filling Rules Aufbau principle: Electrons fill the lowest energy orbitals first (1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p...). Hund's rule: Electrons occupy degenerate orbitals singly before pairing up, all with the same spin. Pauli exclusion principle: Each orbital holds a maximum of 2 electrons, which must have opposite spins.
填充规则 构造原理(Aufbau principle):电子优先填入能量最低的轨道(1s、2s、2p、3s、3p、4s、3d、4p……)。 洪德规则(Hund's rule):电子先单占简并轨道、自旋方向相同,最后才两两配对。 泡利不相容原理(Pauli exclusion principle):每个轨道最多容纳 2 个电子,且二者自旋必须相反。
Exceptions: Chromium and Copper Cr (Z=24): expected [Ar] 3d$^4$ 4s$^2$, actual [Ar] 3d$^5$ 4s$^1$. Cu (Z=29): expected [Ar] 3d$^9$ 4s$^2$, actual [Ar] 3d$^{10}$ 4s$^1$. A half-filled or fully filled d-sublevel provides extra stability. You must know these two exceptions.
两个例外:铬与铜 Cr(Z=24):按规则应为 [Ar] 3d$^4$ 4s$^2$,实际为 [Ar] 3d$^5$ 4s$^1$。Cu(Z=29):按规则应为 [Ar] 3d$^9$ 4s$^2$,实际为 [Ar] 3d$^{10}$ 4s$^1$。半充满或全充满的 d 亚层更稳定。这两个例外必须记住。
Worked Example — Electron Configuration例题 — 电子构型

Write the full and condensed electron configurations for Fe (Z=26) and Fe$^{2+}$.

写出 Fe(Z=26)与 Fe$^{2+}$ 的完整电子构型与缩写电子构型。

Fe (26 electrons)
Full: 1s$^2$ 2s$^2$ 2p$^6$ 3s$^2$ 3p$^6$ 4s$^2$ 3d$^6$
Condensed: [Ar] 4s$^2$ 3d$^6$
Fe(26 个电子)
完整式:1s$^2$ 2s$^2$ 2p$^6$ 3s$^2$ 3p$^6$ 4s$^2$ 3d$^6$
缩写式:[Ar] 4s$^2$ 3d$^6$
Fe$^{2+}$ (24 electrons)
When transition metals form ions, electrons are removed from the 4s sublevel first (not 3d).
Fe$^{2+}$: [Ar] 3d$^6$
The 4s electrons are lost before the 3d electrons because in an ion, 3d is lower in energy than 4s.
Fe$^{2+}$(24 个电子)
过渡金属形成阳离子时,电子先从 4s 亚层失去,而不是 3d。
Fe$^{2+}$:[Ar] 3d$^6$
先失 4s 后失 3d 的原因是:在离子中 3d 的能量低于 4s。
HL Only — Ionization Energy from Spectra (1.3.6–1.3.7) The convergence limit of spectral lines corresponds to ionization. The first ionization energy can be calculated from the frequency of the convergence limit: $IE = hf$ or $IE = hc/\lambda$. Successive IE data reveal the electron configuration — large jumps between successive IEs indicate electrons being removed from a different shell. IE trends and discontinuities across a period provide evidence for the existence of sublevels.
仅 HL — 从光谱求电离能(1.3.6—1.3.7) 光谱线的汇聚极限对应电离过程。第一电离能(ionization energy)可由汇聚极限的频率求得:$IE = hf$ 或 $IE = hc/\lambda$。逐级电离能数据可揭示电子构型——相邻两级电离能突然变大,说明电子是从不同的电子壳层被移除的。同一周期内电离能的趋势与突变也为亚层的存在提供证据。
What is the electron configuration of Cu (Z=29)?Cu(Z=29)的电子构型是?
[Ar] 3d$^9$ 4s$^2$
[Ar] 3d$^8$ 4s$^2$ 4p$^1$
[Ar] 3d$^{10}$ 4s$^2$
[Ar] 3d$^{10}$ 4s$^1$
Correct! Copper is one of two required exceptions. A fully filled 3d$^{10}$ sublevel is more stable, so one 4s electron is promoted: [Ar] 3d$^{10}$ 4s$^1$.正确!铜是必考的两个例外之一。3d$^{10}$ 全充满更稳定,因此一个 4s 电子被"提"到 3d:[Ar] 3d$^{10}$ 4s$^1$。
Cu is an exception: [Ar] 3d$^{10}$ 4s$^1$, not the expected [Ar] 3d$^9$ 4s$^2$. Answer: (D).Cu 是例外:实际为 [Ar] 3d$^{10}$ 4s$^1$,不是 [Ar] 3d$^9$ 4s$^2$。答案:(D)。

Counting Particles by Mass: The Mole用质量数粒子:摩尔

The mole (mol) is the SI unit of amount of substance. One mole contains exactly the number of elementary entities given by the Avogadro constant ($N_A = 6.022 \times 10^{23}$ mol$^{-1}$). An elementary entity can be an atom, molecule, ion, electron, or any specified particle.摩尔(mole,mol)是 SI 单位制中物质的量的基本单位。1 mol 含有的基本粒子数由阿伏伽德罗常数(Avogadro's constant)给出:$N_A = 6.022 \times 10^{23}$ mol$^{-1}$。基本粒子可以是原子、分子、离子、电子或任何指明的粒子。

The Mole Triangle摩尔三角
$$n = \frac{m}{M} \qquad n = \frac{N}{N_A} \qquad n = C \times V$$

$n$ = amount (mol), $m$ = mass (g), $M$ = molar mass (g mol$^{-1}$), $N$ = number of particles, $N_A$ = Avogadro constant, $C$ = concentration (mol dm$^{-3}$), $V$ = volume (dm$^3$).

$n$ = 物质的量(mol);$m$ = 质量(g);$M$ = 摩尔质量(molar mass,g mol$^{-1}$);$N$ = 粒子数;$N_A$ = 阿伏伽德罗常数;$C$ = 浓度(concentration,mol dm$^{-3}$);$V$ = 体积(dm$^3$)。

Relative Masses相对质量

Masses of atoms are compared on a scale relative to $^{12}$C. The relative atomic mass ($A_r$) and relative formula mass ($M_r$) are dimensionless (no units). The molar mass ($M$) has units g mol$^{-1}$ and is numerically equal to $M_r$.原子质量以 $^{12}$C 为基准进行比较。相对原子质量(relative atomic mass,$A_r$)相对式量(relative formula mass,$M_r$)是无量纲量。摩尔质量(molar mass,$M$)的单位是 g mol$^{-1}$,数值上等于 $M_r$。

Empirical and Molecular Formulas实验式与分子式

The empirical formula gives the simplest whole-number ratio of atoms. The molecular formula gives the actual number of atoms in a molecule. To find the molecular formula, divide the molar mass by the empirical formula mass to get a multiplier.实验式(empirical formula给出原子数的最简整数比;分子式(molecular formula给出分子中各原子的真实个数。用摩尔质量除以实验式式量,即可得到倍数,进而得到分子式。

Worked Example — Empirical Formula from Percentage Composition例题 — 由质量百分比求实验式

A compound contains 40.0% C, 6.7% H, and 53.3% O by mass. Its molar mass is 180 g mol$^{-1}$. Find the empirical and molecular formulas.

某化合物按质量计含 40.0% C、6.7% H、53.3% O,其摩尔质量为 180 g mol$^{-1}$。求其实验式与分子式。

Step 1 — Assume 100 g, convert to moles
$$n_C = \frac{40.0}{12.01} = 3.33 \quad n_H = \frac{6.7}{1.01} = 6.63 \quad n_O = \frac{53.3}{16.00} = 3.33$$
第 1 步 — 设总质量为 100 g,换算为摩尔数
$$n_C = \frac{40.0}{12.01} = 3.33 \quad n_H = \frac{6.7}{1.01} = 6.63 \quad n_O = \frac{53.3}{16.00} = 3.33$$
Step 2 — Divide by smallest
$$C: \frac{3.33}{3.33} = 1 \quad H: \frac{6.63}{3.33} = 2 \quad O: \frac{3.33}{3.33} = 1$$
Empirical formula: CH$_2$O. Empirical formula mass = 12 + 2 + 16 = 30.
第 2 步 — 全部除以最小者
$$C: \frac{3.33}{3.33} = 1 \quad H: \frac{6.63}{3.33} = 2 \quad O: \frac{3.33}{3.33} = 1$$
实验式:CH$_2$O。实验式式量 = 12 + 2 + 16 = 30。
Step 3 — Find molecular formula
$$\text{Multiplier} = \frac{180}{30} = 6$$
Molecular formula: C$_6$H$_{12}$O$_6$ (glucose).
第 3 步 — 求分子式
$$\text{倍数} = \frac{180}{30} = 6$$
分子式:C$_6$H$_{12}$O$_6$(葡萄糖)。

Concentration and Dilution浓度与稀释

Molar concentration is determined by the amount of solute and the volume of solution. Square brackets denote concentration: [NaOH] means the molar concentration of NaOH in mol dm$^{-3}$. You should also be able to convert between g dm$^{-3}$ and mol dm$^{-3}$.摩尔浓度由溶质的物质的量与溶液体积决定。方括号代表浓度:[NaOH] 即 NaOH 的摩尔浓度(mol dm$^{-3}$)。你还应能在 g dm$^{-3}$ 与 mol dm$^{-3}$ 之间互换。

Avogadro's Law阿伏伽德罗定律

Equal volumes of all gases at the same temperature and pressure contain equal numbers of molecules. This means the mole ratio in a balanced equation directly gives the volume ratio for gaseous reactants and products.同温同压下,等体积的任何气体含有相同数目的分子(Avogadro's law)。因此,配平方程中的摩尔比就是气态反应物与产物的体积比。

How many molecules are in 0.50 mol of H$_2$O?0.50 mol H$_2$O 中有多少个分子?
$6.022 \times 10^{23}$
$3.011 \times 10^{23}$
$1.204 \times 10^{24}$
$9.033 \times 10^{23}$
Correct! $N = n \times N_A = 0.50 \times 6.022 \times 10^{23} = 3.011 \times 10^{23}$ molecules.正确!$N = n \times N_A = 0.50 \times 6.022 \times 10^{23} = 3.011 \times 10^{23}$ 个分子。
$N = n \times N_A = 0.50 \times 6.022 \times 10^{23} = 3.011 \times 10^{23}$. Answer: (B).$N = n \times N_A = 0.50 \times 6.022 \times 10^{23} = 3.011 \times 10^{23}$。答案:(B)。

Ideal Gases理想气体

An ideal gas is a theoretical model where particles have negligible volume and no intermolecular forces, and all collisions are elastic. Real gases approximate ideal behavior at high temperatures and low pressures (where particles are far apart and moving fast).理想气体(ideal gas是一个理论模型:粒子自身体积可忽略、粒子间无作用力,且所有碰撞均为弹性碰撞。真实气体在高温低压(粒子彼此距离远、运动快)时近似服从理想气体行为。

When Real Gases Deviate Real gases deviate from ideal behavior at low temperature (particles move slowly, intermolecular forces become significant) and high pressure (particles are close together, their volume is no longer negligible). Gases with stronger intermolecular forces (e.g., polar molecules, larger molecules) deviate more.
真实气体何时偏离理想行为 真实气体在低温(粒子运动慢,分子间作用力显著)与高压(粒子彼此靠近,自身体积不再可忽略)下偏离理想行为。分子间作用力越强(如极性分子、较大分子)偏离越显著。

Molar Volume and the Ideal Gas Equation摩尔体积与理想气体方程

At STP (273.15 K, 100 kPa), the molar volume of an ideal gas is 22.7 dm$^3$ mol$^{-1}$ (given in the data booklet). The ideal gas equation relates all four state variables:在 STP(STP,273.15 K、100 kPa)下,理想气体的摩尔体积(molar volume)为 22.7 dm$^3$ mol$^{-1}$(数据手册中给出)。理想气体方程把四个状态变量联系起来:

Ideal Gas Equation理想气体方程
$$PV = nRT$$

$P$ = pressure (Pa), $V$ = volume (m$^3$), $n$ = moles (mol), $R$ = 8.314 J K$^{-1}$ mol$^{-1}$, $T$ = temperature (K). Use SI units throughout.

$P$ = 压强(Pa),$V$ = 体积(m$^3$),$n$ = 物质的量(mol),$R$ = 8.314 J K$^{-1}$ mol$^{-1}$,$T$ = 温度(K)。全部使用 SI 单位。

Combined Gas Law气体联合定律
$$\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}$$

Used when comparing the same sample of gas under two different sets of conditions.

用于比较同一份气体在两种不同条件下的状态。

Worked Example — Ideal Gas Equation例题 — 理想气体方程

Calculate the volume occupied by 0.250 mol of gas at 300 K and 150 kPa.

求 0.250 mol 气体在 300 K、150 kPa 下所占的体积。

Convert units to SI
$P$ = 150 kPa = 150,000 Pa. $T$ = 300 K. $n$ = 0.250 mol.
单位换算为 SI
$P$ = 150 kPa = 150,000 Pa;$T$ = 300 K;$n$ = 0.250 mol。
Apply PV = nRT
$$V = \frac{nRT}{P} = \frac{0.250 \times 8.314 \times 300}{150\,000}$$
$$V = \frac{623.55}{150\,000} = 4.16 \times 10^{-3}\;\text{m}^3 = 4.16\;\text{dm}^3$$
代入 PV = nRT
$$V = \frac{nRT}{P} = \frac{0.250 \times 8.314 \times 300}{150\,000}$$
$$V = \frac{623.55}{150\,000} = 4.16 \times 10^{-3}\;\text{m}^3 = 4.16\;\text{dm}^3$$
Under which conditions does a real gas behave most like an ideal gas?真实气体在哪种条件下最接近理想气体行为?
High temperature and low pressure高温且低压
Low temperature and high pressure低温且高压
Low temperature and low pressure低温且低压
High temperature and high pressure高温且高压
Correct! At high temperature, particles move fast and intermolecular forces are insignificant. At low pressure, particles are far apart and their volume is negligible. Both conditions match the ideal gas assumptions.正确!高温下粒子运动快,分子间作用力可忽略;低压下粒子距离远,其自身体积可忽略——两者都恰好符合理想气体的假设。
Ideal gas behavior requires conditions where intermolecular forces and particle volume are negligible: high $T$, low $P$. Answer: (A).理想气体行为要求分子间作用力与粒子自身体积均可忽略,即高 $T$、低 $P$。答案:(A)。

Exam Strategy考试策略

Paper 1 (Multiple Choice)Paper 1(Multiple Choice)

Know the subatomic particle table cold. For electron configurations, watch for ions (remove 4s before 3d for transition metals) and the Cr/Cu exceptions. Mole calculations are frequent — practise converting between mass, moles, particles, and concentration quickly.

亚原子粒子表必须背得滚瓜烂熟。电子构型题留意离子(过渡金属先失 4s 再失 3d),以及 Cr / Cu 两个例外。摩尔计算高频出现——练习快速在质量、物质的量、粒子数与浓度间互换。

Paper 2 (Extended Response)Paper 2(Extended Response)

Show all working in mole calculations: write the formula, substitute, and solve with units. For gas calculations, always convert to SI units before applying $PV = nRT$. Empirical formula questions: set up a clear table showing element, mass, moles, and ratio.

摩尔计算题写清全过程:先写公式,再代入数值,最后带单位求解。气体题在使用 $PV = nRT$ 之前先把全部数值换成 SI 单位。求实验式时列出清晰的"元素—质量—摩尔数—比"表。

Data BookletData Booklet

Key items to locate quickly: relative atomic masses (periodic table), Avogadro constant, molar gas volume at STP, gas constant $R$, the equations $n = m/M$, $n = CV$, $PV = nRT$, and the combined gas law.

数据手册中必须能快速定位:相对原子质量(周期表)、阿伏伽德罗常数、STP 下的摩尔气体体积、气体常数 $R$,以及 $n = m/M$、$n = CV$、$PV = nRT$ 和联合气体定律。


Common Mistakes常见错误

Mistake 1 — Forgetting Unit Conversions for PV=nRT $P$ must be in Pa (not kPa), $V$ must be in m$^3$ (not dm$^3$ or cm$^3$). 1 kPa = 1000 Pa. 1 dm$^3$ = 10$^{-3}$ m$^3$. Most errors in gas calculations come from mismatched units.
错误 1 — 忘记 PV=nRT 的单位换算 $P$ 必须用 Pa(不是 kPa),$V$ 必须用 m$^3$(不是 dm$^3$ 或 cm$^3$)。1 kPa = 1000 Pa;1 dm$^3$ = 10$^{-3}$ m$^3$。气体计算题大部分错误都源于单位不一致。
Mistake 2 — Removing 3d Electrons First When Forming Ions When transition metals form cations, the 4s electrons are lost before 3d electrons. Fe loses its two 4s electrons first to become Fe$^{2+}$: [Ar] 3d$^6$, not [Ar] 4s$^2$ 3d$^4$.
错误 2 — 形成阳离子时先失去 3d 电子 过渡金属形成阳离子时,先失 4s 电子、再失 3d 电子。Fe 先失去 2 个 4s 电子,形成 Fe$^{2+}$:[Ar] 3d$^6$,而不是 [Ar] 4s$^2$ 3d$^4$。
Mistake 3 — Confusing $A_r$ and $M$ Relative atomic mass ($A_r$) and relative formula mass ($M_r$) have no units. Molar mass ($M$) has units of g mol$^{-1}$. Numerically they are the same, but examiners penalize missing or incorrect units.
错误 3 — 混淆 $A_r$ 与 $M$ 相对原子质量($A_r$)与相对式量($M_r$)无单位;摩尔质量($M$)单位为 g mol$^{-1}$。三者数值相同,但漏写或写错单位会被考官扣分。
Mistake 4 — Using Celsius in Gas Calculations Temperature must always be in Kelvin for $PV = nRT$ and the combined gas law. Add 273 to convert from Celsius. Using Celsius gives wildly wrong answers (and zero marks).
错误 4 — 气体计算中使用摄氏度 $PV = nRT$ 与气体联合定律中的温度必须用开尔文(K)。摄氏度加 273 即得。直接用摄氏度会算出离谱的答案,得 0 分。

Flashcards闪卡

Click a card to flip it.点击卡片翻面。

What are isotopes?什么是同位素(isotope)?
Atoms of the same element with different numbers of neutrons (same $Z$, different $A$).同种元素、中子数不同的原子($Z$ 相同、$A$ 不同)。
Maximum electrons in energy level $n$?第 $n$ 主能级最多容纳几个电子?
$2n^2$. So: $n$=1 holds 2, $n$=2 holds 8, $n$=3 holds 18, $n$=4 holds 32.$2n^2$。即:$n$=1 容 2 个、$n$=2 容 8 个、$n$=3 容 18 个、$n$=4 容 32 个。
State Avogadro's law叙述阿伏伽德罗定律
Equal volumes of all gases at the same $T$ and $P$ contain equal numbers of molecules.同 $T$ 同 $P$ 下,任何气体的等体积含有等数量的分子。
Electron config of Cr?Cr 的电子构型?
[Ar] 3d$^5$ 4s$^1$ (exception — not 3d$^4$ 4s$^2$). Half-filled d-sublevel is more stable.[Ar] 3d$^5$ 4s$^1$(例外,不是 3d$^4$ 4s$^2$)。d 亚层半充满更稳定。
When do real gases deviate most from ideal?真实气体在何条件下偏离理想气体最严重?
At low temperature and high pressure. Intermolecular forces and particle volume become significant.低温、高压。此时分子间作用力与粒子自身体积都不可忽略。
What is the molar volume at STP?STP 下的摩尔体积是?
22.7 dm$^3$ mol$^{-1}$ at 273.15 K and 100 kPa.在 273.15 K、100 kPa 下为 22.7 dm$^3$ mol$^{-1}$。
Empirical vs. molecular formula?实验式与分子式的区别?
Empirical = simplest whole-number ratio. Molecular = actual number of atoms. E.g., CH$_2$O vs. C$_6$H$_{12}$O$_6$.实验式 = 最简整数比;分子式 = 真实原子数。例:CH$_2$O 对应 C$_6$H$_{12}$O$_6$。
What happens to temperature during a phase change?物态变化过程中温度如何变化?
Temperature remains constant. Added energy overcomes intermolecular forces rather than increasing kinetic energy.温度保持不变。吸收的能量用于克服分子间作用力,而非增加动能。

Unit Quiz单元测验

1. Which separation technique is most appropriate for separating a dissolved solid from a solution?1. 从溶液中分离已溶解的固体,最合适的分离方法是?
Filtration过滤(filtration)
Chromatography色谱(chromatography)
Evaporation蒸发(evaporation)
Distillation蒸馏(distillation)
Correct! Evaporation removes the solvent (water), leaving the dissolved solid behind. Filtration separates insoluble solids, not dissolved ones.正确!蒸发可去除溶剂(水),留下溶质固体。过滤只能分离不溶性固体,对已溶解的固体无效。
A dissolved solid passes through filter paper. Evaporation removes the water, leaving the solid. Answer: (C).已溶解的固体会随溶液穿过滤纸。蒸发掉水后即可得到固体。答案:(C)。
2. What is the electron configuration of V$^{3+}$ (Z=23)?2. V$^{3+}$(Z=23)的电子构型是?
[Ar] 3d$^2$ 4s$^2$ minus 3 from 4s and 3d[Ar] 3d$^2$ 4s$^2$ 然后从 4s 与 3d 各失去若干
[Ar] 3d$^2$
[Ar] 3d$^3$ 4s$^2$ minus 3 from 3d[Ar] 3d$^3$ 4s$^2$ 仅从 3d 失去 3 个
[Ar] 4s$^2$ 3d$^0$
Correct! V is [Ar] 3d$^3$ 4s$^2$. Removing 3 electrons: 2 from 4s first, then 1 from 3d, giving [Ar] 3d$^2$.正确!V 的电子构型为 [Ar] 3d$^3$ 4s$^2$。失去 3 个电子时,先失 2 个 4s,再失 1 个 3d,剩 [Ar] 3d$^2$。
V = [Ar] 3d$^3$ 4s$^2$. Remove 4s electrons first (2), then 1 from 3d: V$^{3+}$ = [Ar] 3d$^2$. Answer: (B).V = [Ar] 3d$^3$ 4s$^2$。先失 4s(2 个),再失 3d(1 个):V$^{3+}$ = [Ar] 3d$^2$。答案:(B)。
3. What volume does 0.100 mol of an ideal gas occupy at STP (molar volume = 22.7 dm$^3$ mol$^{-1}$)?3. 在 STP(摩尔体积 = 22.7 dm$^3$ mol$^{-1}$)下,0.100 mol 理想气体占多大体积?
$0.227$ dm$^3$
$22.7$ dm$^3$
$227$ dm$^3$
$2.27$ dm$^3$
Correct! $V = n \times V_m = 0.100 \times 22.7 = 2.27$ dm$^3$.正确!$V = n \times V_m = 0.100 \times 22.7 = 2.27$ dm$^3$。
$V = n \times V_m = 0.100 \times 22.7 = 2.27$ dm$^3$. Answer: (D).$V = n \times V_m = 0.100 \times 22.7 = 2.27$ dm$^3$。答案:(D)。