Chemical energetics

5.1

Enthalpy change, ΔH

Syllabus

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 $\Delta H = -mc\Delta T/n$

Source: Cambridge International syllabus

Reaction profile: activation energy and ΔH

A blue Bunsen burner flame — 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 ($\Delta H > 0$).

Two reaction pathway diagrams side by side: an exothermic one with products below the reactants, and an endothermic one with products above — 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

Two processes: formation goes from elements to one mole of a compound; combustion goes from one mole of a substance plus oxygen to carbon dioxide and water — 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 $= 431$.

$$\Delta H = \sum(\text{broken}) - \sum(\text{made}) = (436 + 242) - (2 \times 431) = 678 - 862 = -184\ \text{kJ mol}^{-1}.$$

A stepped energy diagram: reactants rise to separate gaseous atoms as bonds break, then fall to products as bonds form — 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 of solution with a lid and a thermometer dipping in, used to measure the temperature change — 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

Exercise sheet

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.

A triangular Hess cycle linking reactants and products directly and through an intermediate, with the three enthalpy changes labelled — 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.