Chemical energetics

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 ($\Delta H > 0$).

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.

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:

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 $= 431$.

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:

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:

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

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.

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.

Worked example. Carbon, hydrogen and methane can all be burnt, but carbon and hydrogen will not react to make methane, so $\Delta H_f(\text{CH}_4)$ cannot be measured directly. Find it from these enthalpy changes of combustion (kJ mol⁻¹): C $= -394$, $\text{H}_2 = -286$, $\text{CH}_4 = -890$.

1. Write the reaction you want across the top: $\text{C(s)} + 2\text{H}_2\text{(g)} \rightarrow \text{CH}_4\text{(g)}$.
2. Both sides burn to the same products, $\text{CO}_2 + 2\text{H}_2\text{O}$, so draw that at the bottom of the cycle. Burning the elements goes down from the left; burning methane goes down from the right.
3. The two routes from the elements to $\text{CO}_2 + 2\text{H}_2\text{O}$ must have the same total:

The two short rules that come from these cycles are:

Common mistakes.

• Swapping the two rules: formation data is products minus reactants, combustion data is reactants minus products. If unsure, draw the cycle.
• Forgetting to multiply by the number in the equation. Above, there are two moles of $\text{H}_2$, so $\Delta H_c(\text{H}_2)$ is counted twice.
• Giving an element a non-zero enthalpy of formation. $\Delta H_f$ of an element in its standard state, such as $\text{O}_2\text{(g)}$, is zero.
• In $q = mc\Delta T$, using the mass of the fuel or of the solid added. $m$ is the mass of the water or solution that is heated.
• Forgetting that $q$ is in joules. Divide by 1000 to get kJ before dividing by moles.
• Expecting a bond-energy answer to match the data book exactly. Bond energies are average values over many compounds, so the answer is only approximate.

• 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 ($\text{kJ mol}^{-1}$).