- understand that chemical reactions are accompanied by enthalpy changes and these changes can be exothermic ($\Delta H$ is negative) or endothermic ($\Delta H$ is positive)
- construct and interpret a reaction pathway diagram, in terms of the enthalpy change of the reaction and of the activation energy
- define and use the terms: (a) standard conditions (this syllabus assumes that these are $298\text{ K}$ and $101\text{ kPa}$) shown by $^{\ominus}$. (b) enthalpy change with particular reference to: reaction, $\Delta H_r$, formation, $\Delta H_f$, combustion, $\Delta H_c$, neutralisation, $\Delta H_{\text{neut}}$
- understand that energy transfers occur during chemical reactions because of the breaking and making of chemical bonds
- use bond energies ($\Delta H$ positive, i.e. bond breaking) to calculate enthalpy change of reaction, $\Delta H_r$
- understand that some bond energies are exact and some bond energies are averages
- calculate enthalpy changes from appropriate experimental results, including the use of the relationships $q = mc\Delta T$ and $\Delta H = -mc\Delta T/n$
Chemical energetics
A-Level Chemistry · Topic 5
5.1
Enthalpy change, ΔH
Syllabus
Source: Cambridge International syllabus
Burning fuel is exothermic, releasing energy to the surroundings.
An instant cold pack uses an endothermic reaction that takes in heat.
Every chemical reaction takes in or gives out energy. This energy change, measured at constant pressure, is the enthalpy change 焓变, with symbol $\Delta H$.
- in an exothermic 放热 reaction the system gives out heat, so the products have less energy than the reactants and $\Delta H$ is negative.
- in an endothermic 吸热 reaction the system takes in heat, so the products have more energy than the reactants and $\Delta H$ is positive.
Reaction pathway diagrams
A reaction pathway diagram 反应路径图 shows the energy of the reactants and products, and the energy "hill" between them. The height of the hill is the activation energy 活化能 — the least energy the particles need before they can react.
- exothermic: products sit lower than reactants ($\Delta H < 0$).
- endothermic: products sit higher than reactants ($\Delta H > 0$).
Exothermic reactions end lower than they start ($\Delta H<0$); endothermic reactions end higher ($\Delta H>0$)
Standard conditions and types of enthalpy change
Energy values are compared under standard conditions 标准条件: $298\ \text{K}$ and $101\ \text{kPa}$, shown by the symbol $^{\ominus}$. Each substance is in its normal physical state at those conditions.
| Symbol | Name | Definition (per mole, under standard conditions) |
|---|---|---|
| $\Delta H_r^{\ominus}$ | enthalpy change of reaction 反应焓变 | for the amounts shown in the equation |
| $\Delta H_f^{\ominus}$ | enthalpy change of formation 生成焓变 | one mole of a compound forms from its elements |
| $\Delta H_c^{\ominus}$ | enthalpy change of combustion 燃烧焓变 | one mole of a substance burns completely in oxygen |
| $\Delta H_{\text{neut}}^{\ominus}$ | enthalpy change of neutralisation 中和焓变 | one mole of water forms from an acid and an alkali |
Don't confuse them: formation builds 1 mol of a compound from its elements; combustion burns 1 mol of a substance in oxygen
Energy from breaking and making bonds
During a reaction, old bonds break and new bonds form. Breaking a bond needs energy (endothermic); making a bond releases energy (exothermic). The enthalpy change of the reaction is the difference between the two:
$$\Delta H_r = \sum (\text{bond energies broken}) - \sum (\text{bond energies made})$$
The bond energy 键能 is the energy needed to break one mole of a particular bond in the gas state, so it is always positive. Some bond energies are exact (for one specific molecule); others are averages taken over many different molecules, so calculations using them are only approximate.
Worked example. Use bond energies to find $\Delta H$ for $\text{H}_2 + \text{Cl}_2 \rightarrow 2\text{HCl}$. Bond energies (kJ mol⁻¹): H–H $= 436$, Cl–Cl $= 242$, H–Cl $= 431$.
$$\Delta H = \sum(\text{broken}) - \sum(\text{made}) = (436 + 242) - (2 \times 431) = 678 - 862 = -184\ \text{kJ mol}^{-1}.$$
Breaking bonds takes energy in; making bonds gives energy out. $\Delta H$ is the difference between the two
Measuring enthalpy change in the lab
When a reaction heats up (or cools down) a known mass of water or solution, the heat transferred is:
$$q = mc\Delta T$$
where $m$ is the mass, $c$ is the specific heat capacity 比热容 (how much energy raises 1 g by 1 K), and $\Delta T$ is the temperature change. The enthalpy change per mole is then:
$$\Delta H = -\frac{mc\Delta T}{n}$$
The minus sign makes $\Delta H$ negative when the temperature rises (an exothermic reaction).
Worked example. Burning $0.50\ \text{g}$ of methanol ($M_r = 32$) raises the temperature of $100\ \text{g}$ of water by $18\ ^{\circ}\text{C}$. Find the enthalpy change of combustion per mole. ($c = 4.18\ \text{J g}^{-1}\,\text{K}^{-1}$.)
The heat released is $q = mc\Delta T = 100 \times 4.18 \times 18 = 7520\ \text{J}$. The amount burnt is $n = 0.50/32 = 0.0156\ \text{mol}$, so
$$\Delta H_c = -\frac{q}{n} = -\frac{7520}{0.0156} \approx -480\ \text{kJ mol}^{-1}.$$
An insulated cup and a thermometer measure the temperature change of a known mass of solution
Explore
Exothermic and endothermic reactions
Drag ΔH. An exothermic reaction releases energy (products lower); an endothermic one takes it in (products higher).
Vocabulary
Train
| English | Chinese | Pinyin |
|---|---|---|
| enthalpy change | 焓变 | hán biàn |
| exothermic | 放热 | fàng rè |
| endothermic | 吸热 | xī rè |
| reaction pathway diagram | 反应路径图 | fǎn yìng lù jìng tú |
| activation energy | 活化能 | huó huà néng |
| standard conditions | 标准条件 | biāo zhǔn tiáo jiàn |
| enthalpy change of reaction | 反应焓变 | fǎn yìng hán biàn |
| enthalpy change of formation | 生成焓变 | shēng chéng hán biàn |
| enthalpy change of combustion | 燃烧焓变 | rán shāo hán biàn |
| enthalpy change of neutralisation | 中和焓变 | zhōng hé hán biàn |
| bond energy | 键能 | jiàn néng |
| specific heat capacity | 比热容 | bǐ rè róng |
5.2
Hess's law
Syllabus
- apply Hess’s law to construct simple energy cycles
- carry out calculations using cycles and relevant energy terms, including: (a) determining enthalpy changes that cannot be found by direct experiment (b) use of bond energy data
Source: Cambridge International syllabus
Hess's law 盖斯定律 says that the total enthalpy change for a reaction is the same, no matter which route you take from reactants to products. This is because energy is conserved.
This lets you draw an energy cycle 能量循环: link the reactants and products by a direct step and by an indirect route, then add the steps so that both routes give the same total.
The direct route equals the indirect route, so $\Delta H_r = \Delta H_1 + \Delta H_2$
Hess's law is useful in two ways:
- it lets you find an enthalpy change that you cannot measure directly (for example, the formation of a compound that forms slowly or with side reactions).
- it lets you calculate $\Delta H_r$ from bond energy data, or from formation or combustion data given in the question.
Explore
Hess's law cycle
Enthalpy change is the same whichever route you take — so an unknown ΔH can be found by an alternative path.
Vocabulary
Train
| English | Chinese | Pinyin |
|---|---|---|
| Hess's law | 盖斯定律 | gài sī dìng lǜ |
| energy cycle | 能量循环 | néng liàng xún huán |
5.2
Exam tips
- Define each enthalpy change with its exact standard conditions (e.g. combustion = one mole burned completely in excess oxygen).
- Use $\Delta H = \sum(\text{bonds broken}) - \sum(\text{bonds made})$; breaking is endothermic ($+$), making is exothermic ($-$) — getting the sign the wrong way round is the classic error.
- In $q = mc\Delta T$ use the mass of the water/solution, then divide by moles and add the minus sign for an exothermic reaction.
- Draw Hess cycles with arrows the same way round, follow the alternative route, and always give $\Delta H$ a sign and units ($\text{kJ mol}^{-1}$).