Chemical Energetics

Thermodynamics is the science of the relationships between heat and other forms of energy. Thermochemistry is the area of thermodynamics concerned with the study of the quantity of heat energy absorbed or evolved during physical or chemical changes. That is why it is also called energetics, and it is largely based on the first law of thermodynamics. Hess's law, a special case of the first law, is a remarkable tool for finding enthalpy changes, and thermodynamics also lets us predict whether a reaction can occur under specified conditions — that is, it discusses the spontaneity of a reaction.

Enthalpy (6.1). The sum of all the possible potential and kinetic energies of a system is called its heat content or enthalpy, denoted by H. Every substance possesses a characteristic amount of enthalpy. The total enthalpy of products (H_P) is never equal to that of reactants (H_R), so during a chemical reaction, when reactants are converted into products, there is a change in enthalpy denoted as ΔH. In other words, enthalpy change is the net energy which is either evolved or absorbed in the form of heat. Mathematically: ΔH = (Heat content of products, H_P) − (Heat content of reactants, H_R).

There are two cases. (i) H_P < H_R: such reactions involve a lowering of enthalpy and always take place with the evolution of heat, which is equal to H_P − H_R, and ΔH carries a negative sign. “A chemical reaction or a physical change in which heat is evolved from the system to the surroundings is called an exothermic process.” (ii) H_P > H_R: such reactions involve an increase in enthalpy and always take place by the absorption of heat, which is equal to H_P − H_R, and ΔH carries a positive sign. “A chemical reaction or a physical change in which heat is absorbed by the system from the surroundings is known as an endothermic process.”

Examples: combustion of carbon in oxygen is exothermic: C_(s) + O_2(g) → CO_2(g), ΔH = −393.7 kJ mol⁻¹. The dissolution of ammonium chloride (NH₄Cl) in water is endothermic: NH₄Cl_(s) + H₂O → NH₄⁺_(aq) + Cl⁻_(aq), ΔH = +16.2 kJ mol⁻¹.

📌 Interesting information (book's own note): the dissolution of ammonium chloride in water is an endothermic process. This reaction is used in cold packs (or ice packs) to treat internal injuries. When the pack is kneaded, water and NH₄Cl crystals mix and energy is absorbed from the surroundings, producing a cooling sensation.

Energy profile diagram (6.2). All chemical reactions involve the breaking of bonds in the reactant molecules before the formation of new bonds. This can be achieved only if the reactant molecules collide with sufficient energy to overcome an energy barrier. The minimum amount of energy required by the reactant molecules just to cross that energy barrier is called energy of activation, denoted by E_a. The energy profile diagram of exothermic and endothermic reactions shows reactants, the peak (activated state), products, ΔH and E_a on one picture: in the exothermic diagram the products sit lower than the reactants (ΔH negative), while in the endothermic diagram the products sit higher (ΔH positive), and in both cases E_a is the height from the reactants up to the peak.

Quick Check 6.1 (book): draw the energy profile diagrams for (i) CH_4(g) + 2O_2(g) → CO_2(g) + 2H₂O_(l), ΔH = −890.3 kJ mol⁻¹ (exothermic — products lower, arrow down) and (ii) CaCO_3(s) → CaO_(s) + CO_2(g), ΔH = +572 kJ mol⁻¹ (endothermic — products higher, arrow up). In both, E_a is measured from the reactant level to the top of the hump.

The enthalpy of a substance depends not only upon its physical state but also on the pressure and temperature. Hence, we must specify these conditions while writing an equation. Therefore, when making accurate comparisons of enthalpy changes for various reactions, ΔH is determined under certain standard conditions: (a) Temperature: 25 °C or 298 K; (b) Pressure: 1 atm or 101 kPa. The superscript ° (degree sign) on ΔH means the change was measured under standard conditions.

📌 Did you know? (book's own note): the standard state of an element is its most stable form at 298 K and 1 atm pressure. For example, the standard state of C is graphite, not diamond. By definition, the standard enthalpy change of formation of any element in its standard state is zero.

6.3.1 Enthalpy change of reaction (ΔH°_r). The standard enthalpy of a reaction is the enthalpy change involved when stoichiometric amounts of reactants in their standard states react together completely to form products under standard conditions. Example: hydrogen and oxygen gases forming 1 mole of water: H_2(g) + ½O_2(g) → H₂O_(l), ΔH°_r = −286 kJ mol⁻¹.

6.3.2 Enthalpy change of combustion (ΔH°_c). The standard enthalpy change of combustion of a substance is the enthalpy change involved when one mole of the substance is completely burnt in excess of oxygen, under standard conditions. It is always exothermic. Example: standard enthalpy of combustion of ethanol is −1368 kJ mol⁻¹: C₂H₅OH_(l) + 3O_2(g) → 2CO_2(g) + 3H₂O_(l), ΔH°_c = −1368 kJ mol⁻¹. Enthalpy change of combustion is useful in calculating the caloric content of foods and fuels.

6.3.3 Enthalpy change of formation (ΔH°_f). Standard enthalpy change of formation of a compound is the enthalpy change involved when one mole of the compound is formed from its elements under standard conditions. It can be exothermic or endothermic. For instance, ΔH°_f for methane: C_(s) + 2H_2(g) → CH_4(g), ΔH°_f = −74.8 kJ mol⁻¹.

6.3.4 Enthalpy change of atomization (ΔH°_at). “The standard enthalpy change of atomization of an element is the enthalpy change involved when one mole of gaseous atoms is formed from the element, under standard conditions.” Example: H_2(g) → 2H_(g), bond energy E_H–H = +436 kJ mol⁻¹, so ½H_2(g) → H_(g), ΔH°_at = +218 kJ mol⁻¹.

6.3.5 Enthalpy change of neutralization (ΔH°_n). The standard enthalpy change of neutralization is the enthalpy change involved when one mole of water is formed by the reaction of an acid with an alkali under standard conditions. It is always exothermic. For example, the enthalpy of neutralization of NaOH by HCl is −57.1 kJ mol⁻¹. When these solutions are mixed, the only change is the formation of water molecules, leaving Na⁺ and Cl⁻ as spectator ions in solution. Thus the enthalpy of neutralization is merely the heat of formation of one mole of liquid water from its ionic components, and the net ionic neutralization reaction is: H⁺_(aq) + OH⁻_(aq) → H₂O_(l), ΔH°_n = −57.1 kJ mol⁻¹. For all strong acid–base reactions ΔH°_n is always near −57.1 kJ mol⁻¹.

6.3.6 Electron affinity (ΔH°_ea). The first electron affinity is the enthalpy change involved when 1 mole of electrons is added to 1 mole of gaseous atoms to form 1 mole of gaseous uni-negative ions under standard conditions. Electron affinity of chlorine atom: Cl_(g) + e⁻ → Cl⁻_(g), ΔH°_ea = −348.8 kJ mol⁻¹. Since energy is released, first electron affinity carries a negative sign.

Quick Check 6.2 (book). (a) Write equations, including state symbols, that represent the enthalpy change of atomization of: (i) Oxygen: ½O_2(g) → O_(g); (ii) Barium: Ba_(s) → Ba_(g); (iii) Bromine: ½Br_2(l) → Br_(g). (b) Name the enthalpy changes in: (i) C_(graphite) + O_2(g) → CO_2(g) — enthalpy of formation of CO₂ (also of combustion of carbon); (ii) HCl_(g) + NH_3(g) → NH₄Cl_(s) — enthalpy of formation of NH₄Cl; (iii) H_2(g) + ½O_2(g) → H₂O_(l) — enthalpy of formation (and of reaction) of water.

Bond energy / bond dissociation energy (6.4). When a bond is formed between two atoms, energy is released. The same amount of energy is absorbed when the bond is broken to form neutral atoms; we call this bond dissociation energy, defined as: “The average amount of energy required to break (dissociate) one mole of a particular bond in a substance.” It may be denoted as E.

If we are determining the bond energy of a particular bond in a particular substance, we call this exact bond energy. Note that bonds between the same pair of atoms usually have different bond energy values in different compounds, because bond energy is affected by other atoms in a molecule. For example, C–C bonds usually have bond energy values of approximately 350 to 380 kJ/mol: in ethane [H₃C–CH₃], E_C–C = 376 kJ mol⁻¹ (exact bond energy for ethane); in propane [H₃C–CH₂–CH₃], E_C–C ≈ 356 kJ mol⁻¹; in butane [H₃C–CH₂–CH₂–CH₃], E_C–C ≈ 352 kJ mol⁻¹. When we take the average of all E_C–C in different molecules, we obtain average C–C bond energy. Practically, average bond energy is used instead of exact bond energy. Average bond energies of some bonds are given in the book's Table 6.1 (kJ mol⁻¹), for example H–H 436, C–H 413, O=O 498, H–O 463, C=O 805, N≡N 945, H–Cl 431, Cl–Cl 242.

Enthalpy of reaction and chemical bonds (6.5). A chemical bond represents a form of energy known as chemical energy, which is interconvertible into all forms of energy. Bond breaking absorbs energy (ΔH = +ve) while bond formation releases it (ΔH = −ve), and their difference will decide the overall sign of ΔH. That is, whether a reaction is exothermic or endothermic is determined by the net change. Overall, a reaction is exothermic if the energy released in making new bonds is greater than that absorbed in bond breaking; the reverse is true for an endothermic reaction. For example, the formation of water is exothermic because more energy is released in forming the H–O bonds than is absorbed in breaking the H–H and O=O bonds.

Sample Problem 6.1 (book). With the bond energy data E_H–H = 436 kJ mol⁻¹, E_O=O = 495 kJ mol⁻¹ and E_H–O = 463 kJ mol⁻¹, calculate the enthalpy change of H_2(g) + ½O_2(g) → H₂O_(g). Solution: ΔH°_r = ΣE_R − ΣE_P = [E_H–H + ½E_O=O] − [2E_H–O] = [436 + 247.5] − [2(463)] = 683.5 − 926 = ΔH°_r = −242.5 kJ mol⁻¹. Note the book's value here (−242.5) is for gaseous water; the −286 kJ mol⁻¹ in section 6.3.1 is for liquid water — the difference is the enthalpy of condensation of water.

The amount of heat evolved or absorbed during a physical or chemical change can be measured by an instrument called a calorimeter. At its simplest, a calorimeter consists of an insulated vessel, a stirrer and a thermometer. There are various types of calorimeters but here we will describe only the “Glass Calorimeter”.

6.6.1 Glass calorimeter. A glass calorimeter is suitable for measuring heat-flow for reactions in solutions. However, it cannot be used for reactions involving gases which would escape from the vessel, nor would it be appropriate for reactions in which the products reach high temperatures. It consists of a beaker, a stirrer, a thermometer and a loose-fitting lid to keep the contents at atmospheric pressure. The outside walls of the calorimeter (beaker) are insulated using cotton wool to minimize the exchange of heat with the surrounding air. The reaction is carried out inside the beaker and the heat evolved or absorbed is measured by the temperature change. Since the pressure inside the calorimeter is constant, the temperature measurement makes it possible to calculate the enthalpy change ΔH during the reaction.

Calculations. ΔH is calculated as follows: (i) q = m_water × c_water × ΔT; (ii) convert into kJ by dividing with 1000; (iii) calculate ΔH for the reaction using the relation: ΔH = −q/n kJ mol⁻¹ (for an exothermic reaction), or = −mcΔT/n kJ mol⁻¹ (n = no. of moles). The solutions we are using here are so dilute that almost all of their mass consists of water, therefore we can simply use the specific heat capacity of water. Such a calorimeter could be used to measure the heat of neutralization (Δ) as explained in the following example.

Specific heat capacity. The amount of heat energy required to raise the temperature of a substance of mass 1 kg through 1 K (or 1 °C) is known as specific heat capacity, c, of the liquid. So the specific heat capacity of water is 4.18 J g⁻¹ K⁻¹ (or J g⁻¹ °C⁻¹), which is rounded off to 4.2 J g⁻¹ K⁻¹.

Sample Problem 6.2 (book). Neutralization of 100 cm³ of 0.5 mol dm⁻³ NaOH at 25 °C with 100 cm³ of 0.5 mol dm⁻³ HCl at 25 °C raised the temperature of the reaction mixture to 28.5 °C. Find the enthalpy of neutralization. Specific heat of water = 4.2 J g⁻¹ K⁻¹. Solution: density of H₂O is around 1 g cm⁻³, so total volume of solution 200 cm³ ≈ 200 g. Rise in temperature, ΔT = 28.5 − 25.0 = 3.5 °C = 3.5 K. Amount of total heat evolved, q = m × c × ΔT = 200 × 4.2 × 3.5 = 2940 J = 2.94 kJ. Number of moles of H₂O formed: n_(HCl) = n_(NaOH) = (0.5 × 100)/1000 = 0.05 mol, so n_(H2O) = 0.05 mol. Heat evolved in the formation of 0.05 mol of water, q = −2.94 kJ. ΔH°_n = q/n = −2.94 kJ / 0.05 mol = −58.8 kJ mol⁻¹ (close to the standard value −57.1 kJ mol⁻¹; the small difference comes from the rounded specific heat and experimental heat losses).

Quick Check 6.3 (book). Calculate ΔH°_n of the reaction of 50 cm³ of 1.5 mol dm⁻³ HNO₃ with 50 cm³ of 1.5 mol dm⁻³ NaOH. The change in temperature is 4 °C. Solution: total mass ≈ 100 g; q = 100 × 4.2 × 4 = 1680 J = 1.68 kJ; moles = (1.5 × 50)/1000 = 0.075 mol; ΔH°_n = −1.68/0.075 = −22.4 kJ mol⁻¹ — note this is far from −57.1, which shows how sensitive a school calorimeter result is to the temperature reading and heat losses; the method, not this particular number, is what the book is teaching.

In this section, we will look at foods as fuels. When food is digested, the chemical energy stored in the food (also called calorie content) is released as heat energy. In other words, digestion of food releases the same amount of energy as if it burned outside the body. So, the overall enthalpy of combustion is the same as the enthalpy of combustion, which can be determined in a calorimeter (typically in a bomb calorimeter).

Calorie content: the calorie content of food is a measure of the energy ‘released’ when the food is completely consumed in the body. This energy is typically expressed in units of kilocalories (k cal) or joules (J).

6.7.1 Relation between enthalpy change and calorie content. The enthalpy of combustion of a food (ΔH°_c) is the calorie content of that food when it is translated or converted into kilocalories per gram: Calorie content = ΔH° (kJ/g) / 4.184. Look at the energy provided from glucose: C₆H₁₂O_6(s) + 6O_2(g) → 6CO_2(g) + 6H₂O_(l), ΔH°_c = −2803 kJ mol⁻¹. The calorie content of glucose can be calculated as follows: first, we find ΔH per gram of glucose. Molar mass of glucose is 180 g mol⁻¹. The above equation shows that 1 mol (180.0 g) of glucose burns to produce energy = 2803 kJ; 1.0 g of glucose = 2803 kJ/g ÷ 180 g/mol = 15.57 kJ/g. Using the relation ΔH (kJ/g) = −4.184 × calorie content, calorie content = −ΔH (kJ/g) / 4.184 (kJ/kcal) = −(−15.57 kJ/g) / 4.184 (kJ/kcal) = −3.72 kcal/g. (The negative sign here only records that energy is released; food labels quote the magnitude, 3.72 kcal/g.)

Remember, the calorie content we take from food must be balanced by working, exercising and doing positive activities. Otherwise, our bodies will have imbalanced growth and maintenance — the book calls this negative energy balance when intake and expenditure are not matched in the healthy direction, and the picture in the book shows energy intake (food) balanced against energy expenditure (resting calories, activity, exercise).

The first law of thermodynamics is actually a manifestation of the law of conservation of energy, which states: “Energy can neither be created nor destroyed, but can be converted from one form to another.” Germain Henri Hess applied the law of conservation of energy to enthalpy changes. There are many reactions for which ΔH cannot be measured directly by a calorimetric method. For example, tetrachloromethane CCl₄ cannot be prepared directly by combining carbon and chlorine. Hess's law helps us to calculate the enthalpy changes for such reactions or processes. It states: “The total enthalpy change in a chemical reaction is independent of the route by which the chemical reaction takes place as long as the initial and final conditions are the same.”

Hess's law can be illustrated by drawing enthalpy cycles, often called energy cycles or Hess cycles. Let A be converted to ‘B’ directly in a single step, which is the direct route; or in a series of two or three steps designated as indirect route 1 and indirect route 2. The products formed in these routes (M, N and X, Y, Z) are called reaction intermediates. Then according to Hess's law: for indirect route 1, ΔH° = ΔH₁ + ΔH₂; for indirect route 2, ΔH° = ΔH₃ + ΔH₄ + ΔH₅. Because enthalpy is a state function, only the start and finish matter — the path does not.

(i) Calculating enthalpy of formation (ΔH°_f) using enthalpy of combustion (ΔH°_c) — Sample Problem 6.3. Calculate the enthalpy change of formation of CO using Hess cycle with the following combustion data: C_(graphite) + O_2(g) → CO_2(g), ΔH = −393.5 kJ mol⁻¹; CO_(g) + ½O_2(g) → CO_2(g), ΔH₂ = −283 kJ mol⁻¹; C_(graphite) + ½O_2(g) → CO_(g), ΔH₁ = ? Solution: applying Hess's law, ΔH = ΔH₁ + ΔH₂, so ΔH₁ = ΔH − ΔH₂ = −393.5 − (−283) = −110.5 kJ mol⁻¹.

(ii) Calculating enthalpy of reaction (ΔH°_r) using enthalpies of formation (ΔH°_f) — Sample Problem 6.4. Calculate the enthalpy change of C₂H_4(g) + HCl_(g) → C₂H₅Cl_(g) using Hess cycle with: ΔH°_f of C₂H₄ = +52.2 kJ mol⁻¹; ΔH°_f of HCl = −92.3 kJ mol⁻¹; ΔH°_f of C₂H₅Cl = −109 kJ mol⁻¹. Solution: ΔH°_r = ΣΔH°_f(products) − ΣΔH°_f(reactants) = −109 − [+52.2 + (−92.3)] = −109 − (−40.1) = −68.9 kJ mol⁻¹.

(iii) Calculating enthalpy of formation (ΔH°_f) of a substance using enthalpy of combustion and enthalpies of formation of other substances — Sample Problem 6.5. Propane (C₃H_8(g)) burns in oxygen: C₃H_8(g) + 5O_2(g) → 3CO_2(g) + 4H₂O_(g). When 14.64 g of propane is burned in an excess of oxygen in a calorimeter at 25 °C and 1 atm pressure, 678.6 kJ of heat is evolved. Calculate the standard enthalpy of formation of propane. The standard enthalpies of formation of CO_2(g) and H₂O_(g) are −393.51 kJ mol⁻¹ and −241.82 kJ mol⁻¹ respectively. Solution: molar mass of propane, C₃H₈ = 3(12.011) + 8(1.0079) = 44.096 g mol⁻¹. No. of moles of propane = 14.64/44.096 = 0.3320 mol. So 0.3320 mol propane evolves heat = 678.6 kJ; 1.0 mol propane evolves heat = 678.6/0.3320 kJ mol⁻¹ = −2044 kJ mol⁻¹ (this is ΔH°_c of propane). According to Hess's law: ΔH°_c = [3 × ΔH°_f(CO₂) + 4 × ΔH°_f(H₂O)] − [1 × ΔH°_f(C₃H₈) + 5 × ΔH°_f(O₂)]; −2044 = [3 × (−393.51) + 4 × (−241.82)] − [1 × ΔH°_f(C₃H₈) + 5 × (0)]; solving gives ΔH°_f(C₃H₈) = −104 kJ mol⁻¹. (Book's arithmetic: 3 × (−393.51) + 4 × (−241.82) = −2147.8; −2147.8 − (−2044) = −103.8 ≈ −104.)

(iv) Calculating enthalpy change of reactions (ΔH°_r) using bond energies. A special case of Hess's law involves the use of bond energies to estimate heats of reaction for gas-phase reactions. In the first step, the bonds in all the reactant molecules are broken to give free atoms in the gas phase; the enthalpy change for this step can be calculated by adding the bond enthalpies from Table 6.1. In the second step, the product molecules are formed; the enthalpy change for this step can again be estimated from the bond enthalpies of Table 6.1, but must now be taken with minus sign because bonds are being formed instead of broken. In general, the heat of reaction for any gaseous chemical reaction can be calculated from average bond energies by this version of Hess's law: ΔH°_rxn = ΣE_R − ΣE_P.

Sample Problem 6.6 (book; bond-energy version). In the formation of HCl_(g) from H_2(g) and Cl_2(g), use bond-energy data from Table 6.1 to estimate ΔH for H_2(g) + Cl_2(g) → 2HCl_(g), and finally calculate the heat of formation of HCl. First step: H_2(g) + Cl_2(g) → 2H_(g) + 2Cl_(g); E_H–H = 436 kJ mol⁻¹, E_Cl–Cl = 242 kJ mol⁻¹; ΣE_R = (1 mol × 436) + (1 mol × 242) = 678 kJ. Second step: 2H_(g) + 2Cl_(g) → 2HCl_(g); E_H–Cl = 431 kJ mol⁻¹; ΣE_P = 2 mol × 431 = 862 kJ. ΔH° = ΣE_R − ΣE_P = 678 − 862 = −184 kJ (for 2 mol of HCl); enthalpy of formation of HCl = −184/2 = −92 kJ mol⁻¹.

Quick Check 6.4 (book) — note the book's two slips here. The box says “formation of methane” but the reaction printed is 6C_(s) + 3H_2(g) → C₆H_6(l), which is the formation of benzene, not methane; and the value −3267 kJ mol⁻¹ labelled “enthalpy of formation of C₆H_6(l)” is in fact benzene's enthalpy of combustion (its true ΔH°_f is about +49 kJ mol⁻¹). Read correctly, the exercise is: find ΔH°_f of benzene from combustion data — ΔH°_c(C) = −394, ΔH°_c(H₂) = −286, ΔH°_c(C₆H₆) = −3267 kJ mol⁻¹. By Hess's law: ΔH°_f(C₆H₆) = [6 × ΔH°_c(C) + 3 × ΔH°_c(H₂)] − ΔH°_c(C₆H₆) = [6(−394) + 3(−286)] − (−3267) = (−2364 − 858) + 3267 = +45 kJ mol⁻¹ — the positive sign correctly shows benzene formation from its elements is endothermic. Quick Check 6.5 (book): draw the enthalpy cycle of Sample Problem 6.5 according to Hess's law to validate the calculation — the cycle runs reactants → CO₂ + H₂O directly (ΔH°_c = −2044) and indirectly via elements, closing with ΔH°_f(C₃H₈) = −104 kJ mol⁻¹. Quick Check 6.6 (book). (a) Haber process: N_2(g) + 3H_2(g) → 2NH_3(g); E_N≡N = 945, E_H–H = 436, E_N–H = 391 kJ mol⁻¹. ΣE_R = 945 + 3(436) = 2253; ΣE_P = 6(391) = 2346; ΔH = 2253 − 2346 = −93 kJ mol⁻¹. (b) C₂H₅OH_(l) + 3O_2(g) → 2CO_2(g) + 3H₂O_(l) with E_C–C = +347, E_C–H = +410, E_C–O = +336, E_O=O = +496, E_C=O = +805, E_O–H = +465 kJ mol⁻¹: bonds broken: 1 C–C (347) + 5 C–H (2050) + 1 C–O (336) + 1 O–H (465) + 3 O=O (1488) = 4686; bonds formed: 4 C=O (3220) + 6 O–H (2790) = 6010; ΔH = 4686 − 6010 = −1324 kJ mol⁻¹ (close to the experimental combustion value −1368; the gap is because average bond energies are approximate).

The process of dissolving a solute in a solvent is called dissolution. It is assumed that the formation of a solution takes place in three steps: (i) overcoming the intermolecular forces in the solvent to make room for the solute (expanding the solvent); (ii) breaking up the solute into individual components (expanding the solute); (iii) allowing the solute and solvent to interact to form the solution. In this process heat is either given out or taken in. Standard enthalpy change of solution is used to describe it.

Standard enthalpy change of solution (ΔH°_sol). “The standard enthalpy change of solution is the amount of heat absorbed or evolved when one mole of a substance is dissolved in a solvent to give an infinitely dilute solution.” It is denoted by ΔH°_sol and it may be exothermic or endothermic. Examples: Na₂CO_3(s) + aq → 2Na⁺_(aq) + CO₃²⁻_(aq), ΔH°_sol = −25 kJ mol⁻¹; NH₄Cl_(s) + aq → NH₄⁺_(aq) + Cl⁻_(aq), ΔH°_sol = +16.2 kJ mol⁻¹. The book's Table 6.2 gives heats of solution of some important ionic solids in water at infinite dilution (kJ mol⁻¹): NaCl 4.98, KCl 17.8, KBr 19.9, KI 21.4, NH₄Cl 16.2, NH₄NO₃ 26.0.

An ionic compound consists of oppositely charged ions, held together by strong electrostatic forces. Two factors govern the process of dissolution: (i) hydration energy (accounts for the attraction of solute ions with water molecules); (ii) lattice energy (controls the breaking of the ionic compounds).

6.9.1 Hydration. “The process in which water molecules surround and interact with the solute ions is called hydration.” The forces created between water molecules and the ions are called ion–dipole forces, and as a result all the ions in aqueous solution are said to be hydrated. The energy of attraction due to an ion–dipole force is known as enthalpy of hydration, defined as: “The enthalpy change involved when one mole of a solute is dissolved in excess of water to make infinitely dilute solution under standard conditions.” It is denoted by ΔH°_hyd. This energy is always released as it is formation of ion–dipole bond.

Factors affecting the magnitude of hydration energy. The heat of hydration depends on: (i) charge on the ion; (ii) size of the ion. We use a combined term for these two factors, i.e. “charge density”, defined as charge per unit area. If greater charge is present on a smaller ion, the charge density is large and vice versa. And large values of charge density mean an increase in the hydration energy values. For example, enthalpies of hydration of the following ions are in the order: Na⁺ < Mg²⁺ < Al³⁺ (charge increases, size decreases), and Mg²⁺ > Ca²⁺ > Sr²⁺ (same charge, size increases down the group). On the same basis we can explain the hydration energy of anions.

6.9.2 Lattice energy (ΔH°_latt). “The lattice energy of an ionic crystal is the enthalpy change involved when one mole of the ionic compound is formed from gaseous ions under standard conditions.” Example: Na⁺_(g) + Cl⁻_(g) → NaCl_(s), ΔH°_latt = −787 kJ mol⁻¹. The negative sign is because forming the lattice releases energy; breaking it apart costs the same amount with the opposite sign.

Factors affecting the magnitude of lattice energy. The lattice energy depends on: (i) charge on the ion; (ii) size of the ion. Lattice energy and size of the ions: with the increase in the size of the cation/anion, oppositely charged ions become less and less tightly packed, so lattice energy decreases (the book's Figure 6.7 shows lattice energies of alkali metal halides falling as ion size grows). Lattice energy and charge on the ions: lattice energy is directly proportional to the charges on the ions — the greater the magnitude of the charge on an ion, the greater its lattice energy and vice versa. Compare lithium fluoride, LiF, with magnesium oxide, MgO: these compounds have the same arrangement of ions in their lattice structure and comparative sizes of cations and anions are the same in both compounds. Magnesium oxide ΔH°_{latt [MgO]} = −3923 kJ mol⁻¹ is greater than lithium fluoride ΔH°_{latt [LiF]} = −1049 kJ mol⁻¹. The doubly-charged Mg²⁺ and O²⁻ ions in MgO attract each other more strongly than the singly-charged ions of the same size in LiF.

Solubility trends of Group 2 hydroxides and sulphates. ΔH°_sol depends on both lattice energy and hydration enthalpy. Group 2 hydroxides: both hydration energy and lattice enthalpy decrease down the group due to decrease in charge density of Group 2 cations. But it is observed that the energy of hydration decreases more rapidly in the series Mg(OH)₂, Ca(OH)₂, Sr(OH)₂ and Ba(OH)₂ than does the energy of hydration of ions Mg²⁺, Ca²⁺, Sr²⁺ and Ba²⁺. For this reason, the lattice energy factor dominates. ΔH°_sol becomes more exothermic, so solubility increases down the group. Group 2 sulphates: because the SO₄²⁻ is much larger than the OH⁻, the decrease in lattice energy going through the series of sulphates from MgSO₄ to BaSO₄ is less, but the energy of hydration of the cation decreases by a greater amount. Now the energy of hydration dominates and the solubility decreases from MgSO₄ to BaSO₄. ΔH°_sol becomes more endothermic, thus solubility decreases.

6.9.3 Calculating enthalpy change of solution (ΔH°_sol). We can calculate the enthalpy change of solution or the enthalpy change of hydration by constructing an enthalpy cycle and using Hess's law. From the cycle: ΔH°_latt + ΔH°_hyd = ΔH°_sol — with the sign convention that ΔH°_latt is taken as the positive energy needed to break the lattice apart (the reverse of lattice formation). Sample Problem 6.6 (book; solution version — note the book reuses the number 6.6 here). Determine the enthalpy change of solution (ΔH°_sol) of sodium fluoride (NaF) using: lattice energy of NaF = −902 kJ mol⁻¹ (formation value; +902 kJ mol⁻¹ to break the lattice); enthalpy of hydration of Na⁺ = −406 kJ mol⁻¹; enthalpy of hydration of F⁻ = −506 kJ mol⁻¹. Solution: ΔH°_sol = ΔH°_{latt(breaking)} + ΔH°_hyd = 902 + (−406 − 506) = 902 − 912 = +10 kJ mol⁻¹ (the book prints “ΔH°_hyd[NaF] = +10” as a slip for ΔH°_sol). So dissolving NaF is slightly endothermic.

It is impossible to determine the lattice energy of a compound by a single direct experiment. We can, however, calculate the value for ΔH°_latt using several experimental data and an energy cycle called the Born–Haber cycle. The Born–Haber cycle is an application of Hess's law. Let us consider the calculation of lattice energy of sodium chloride using Hess's law and the Born–Haber cycle.

In the simple energy triangle: Na_(s) + ½Cl_2(g) → Na⁺Cl⁻_(s) directly by ΔH°_f, or indirectly via gaseous ions Na⁺_(g) + Cl⁻_(g): first spend ΔH_x, the total energy involved in changing sodium and chlorine from their standard physical states to gaseous ions, then release ΔH°_latt on forming the lattice. Applying Hess's law to the cycle: ΔH°_f = ΔH_x + ΔH°_latt. Here ΔH°_f, the standard enthalpy of formation of NaCl, can be measured conveniently in a calorimeter.

The energy triangle is extended to show the various stages involved in finding ΔH_x. The complete energy cycle is called the Born–Haber cycle (the book's Figure 6.10, Born–Haber cycle of sodium chloride, not according to scale). Its steps for NaCl are: (1) atomization of sodium: Na_(s) → Na_(g), ΔH_at of Na_(s) = +108 kJ mol⁻¹; (2) ionization of sodium: Na_(g) → Na⁺_(g) + e⁻, ΔH_i1 of Na_(g) = +496 kJ mol⁻¹; (3) atomization of chlorine: ½Cl_2(g) → Cl_(g), ΔH_at of Cl_2(g) = +121 kJ mol⁻¹ (half the Cl–Cl bond energy); (4) electron affinity of chlorine: Cl_(g) + e⁻ → Cl⁻_(g), ΔH_ea of Cl = −349 kJ mol⁻¹. Calculation of ΔH_x: from the cycle, ΔH_x = ΔH_at(Na) + ΔH_i1(Na) + ΔH_at(Cl2) + ΔH_ea(Cl) = 108 + 496 + 121 − 349 = 376 kJ mol⁻¹. Using ΔH°_latt = ΔH°_f − ΔH_x = −411 − 376 = ΔH°_latt = −787 kJ mol⁻¹, matching the value in section 6.9.2.

Sample Problem 6.7 (book). Calculate the heat of formation of sodium fluoride, which crystallizes in the sodium chloride lattice. The heat of atomization of Na_(s) is 109 kJ/mol, half the bond energy of F_2(g) is 79 kJ/mol, the ionization energy of sodium atoms is 494 kJ/mol, the electron affinity of fluorine atoms is −328 kJ/mol, and the lattice energy is −939 kJ/mol. Solution: values are given for all the quantities in the Born–Haber cycle, so we can apply Hess's law: ΔH°_f = ΔH°_latt + ΔH°_at(Na) + ΔH°_i1(Na) + ΔH°_at(F2) + ΔH°_ea(F) = (−939 + 109 + 494 + 79 − 328) kJ mol⁻¹ = −585 kJ mol⁻¹.

Quick Check 6.7 (book). (a) Draw the Born–Haber cycle for Sample Problem 6.7 — the same ladder as NaCl, with NaF values: +109 (atomization Na), +494 (ionization Na), +79 (½F₂ atomization), −328 (electron affinity F), −939 (lattice), closing at ΔH°_f = −585 kJ mol⁻¹. (b) Calculate the heat of formation of lithium fluoride: heat of atomization of Li_(s) is 161 kJ/mol, half the bond energy of F_2(g) is 79 kJ/mol, ionization energy of lithium atoms 520 kJ/mol, electron affinity of fluorine atoms −328 kJ/mol, lattice energy −1107 kJ/mol. ΔH°_f(LiF) = −1107 + 161 + 520 + 79 − 328 = −675 kJ mol⁻¹.

Entropy is a measure of the number of ways energy can be distributed within a system at a specific temperature. When the energy is distributed in more ways, a system is more stable. Entropy can also be thought of as a measure of the randomness or disorder of a system. The higher the randomness or disorder, the greater the entropy of the system. The system becomes energetically more stable when it becomes more disordered.

6.11.1 Entropy, diffusion and number of ways of arrangement. We can show that the molecules in a vapour diffuse by chance by thinking about the probability of finding them at one place at any one time. Consider the simplified model of the book's Figure 6.11: gas jar A with 3 molecules and an empty gas jar B, separated by a partition. There are 8 different ways for these molecules to arrange themselves in two jars by diffusion from jar A to jar B. This is calculated as: number of molecules in A = 3; number of jars = 2; number of possible arrangements = 2³ = 8. Similarly, if there were 5 molecules initially in jar A, the possible ways of arrangement will be 2⁵, i.e. 32. And if there were 100 molecules, the probability will be 2¹⁰⁰. A general formula x^y can be written for the calculation of probability, where x is the number of places and y is the number of particles to be arranged. Diffusion happens because there is a large number of ways of arranging the molecules. The concept of the ‘number of ways’ of arrangement of either particles or the energy within these particles helps predict whether a change can happen or not.

6.11.2 Comparison of entropy values. Entropies of different substances can be compared based on the number of particles, physical properties, and state. Small number of particles means low entropy and vice versa: for example CaCO₃ has higher standard entropy (92.9 J K⁻¹ mol⁻¹) than CaO (39.7 J K⁻¹ mol⁻¹). This is because the number of possible arrangements is lower when the number of particles is smaller. The entropy of substances having similar chemical nature is dictated by their hardness. Harder substances have lower entropy than softer ones. Diamond has lower entropy than graphite because it is much harder. Stronger forces result in limited vibrations in harder substances which decrease the probability of disorder. A substance has lower entropy in solid state than in liquid and gaseous states. The entropy of ice near its melting point is 48.0 J K⁻¹ mol⁻¹, for water it is 69.9 J K⁻¹ mol⁻¹, whereas water vapour just above the boiling point is 188.7 J K⁻¹ mol⁻¹. (The book's picture note: dissolution of sugar in tea increases the entropy of both the solute and solvent.)

Quick Check 6.8 (book). Explain the difference in the entropy of each of the following pairs in terms of their states and structures: (i) Br_2(l) S° = 151.6 J K⁻¹ mol⁻¹ and I_2(s) S° = 116.8 J K⁻¹ mol⁻¹ — liquid bromine is more disordered than solid iodine, and its molecules are freer to move, so S° is higher despite iodine's greater mass. (ii) H_2(g) S° = 130.6 and CH_4(g) S° = 186.2 J K⁻¹ mol⁻¹ — methane has more atoms per molecule and more ways to distribute energy (more vibrations/rotations), so higher entropy. (iii) Hg_(l) S° = 76.00 and Na_(s) S° = 51.20 J K⁻¹ mol⁻¹ — liquid mercury is more disordered than solid sodium. (iv) SO_2(g) S° = 248.1 and SO_3(g) S° = 95.60 J K⁻¹ mol⁻¹ — as printed, this pair is anomalous (a larger molecule would normally have the higher entropy); the book's SO₃ value looks like a misprint, but the comparison rule the book is teaching is state first, then particle count and complexity.

6.11.3 Entropy changes in reactions. In a chemical reaction, if we compare the entropies of the reactants and products, we can try to explain the magnitude of the entropy change and whether or not it increases or decreases. The following rules must be followed: when a solid is converted to a liquid or a gas in the product, the entropy change is positive. If no. of moles in the products are more, the entropy change is positive. If there is a change in the number of gaseous molecules in a reaction, due to high values of entropy associated with gases, the more gas molecules there are, the greater the number of ways of arranging them and the higher the entropy. For example: CaCO_3(s) → CaO_(s) + CO_2(g) — there is an increase in entropy because a gas is being produced (high entropy) but the reactant, calcium carbonate, is a solid (low entropy). Such reactions are spontaneous. Also consider 2N₂O_5(g) → 4NO_2(g) + O_2(g): we should expect an increase of entropy because there are a greater number of moles of gas molecules in the products (5 molecules) than in the reactants (2 molecules). This is also a spontaneous process.

6.11.4 Calculating the entropy change of the system (reaction). In order to calculate the entropy change of the system we use: ΔS°_system = S°_products − S°_reactants. Example: 2Ca_(s) + O_2(g) → 2CaO_(s). Standard entropy values: S°_[Ca(s)] = 41.40 J K⁻¹ mol⁻¹; S°_[O2(g)] = 205.0 J K⁻¹ mol⁻¹; S°_[CaO(s)] = 39.70 J K⁻¹ mol⁻¹. ΔS°_system = S°_products − S°_reactants = 2 × S°_[CaO(s)] − {2 × S°_[Ca(s)] + S°_[O2(g)]} = 2 × 39.70 − {(2 × 41.40) + 205.0} = 79.40 − 287.8 = ΔS°_system = −208.4 J K⁻¹ mol⁻¹. The negative value shows the entropy of the system has decreased. We know, however, that calcium reacts spontaneously with oxygen — so the entropy of the surroundings must also play a part, because the total entropy change must be positive for the reaction to be feasible.

6.11.5 Total entropy change, ΔS°_total. ΔS°_total is the sum of the standard entropy change of the system and the standard entropy change of the surroundings. ΔS° for the system can be found out by subtracting the standard entropies of reactants from the standard entropies of products (similar to the way we obtained ΔH°): ΔS°_system = ΣS°_(products) − ΣS°_(reactants), where the respective no. of moles of reactants and products in a balanced chemical equation are multiplied with entropy of reactants and products. ΔS° for surroundings is calculated using: ΔS°_surroundings = −ΔH°_r / T, where: the value of ΔH°_r is in kJ/mol so it must be multiplied with 1000 (because entropy changes are measured in J/K·mol), and the negative sign is part of the equation and not the sign of ΔH°_r. Then: ΔS°_total = ΔS°_system + ΔS°_surroundings = [ΣS°_(products) − ΣS°_(reactants)] + [−ΔH°_r/T]. It is useful to predict spontaneity using the sign of ΔS°_total: ΔS°_total is +ve ⇒ entropy increases ⇒ reaction is spontaneous; ΔS°_total is −ve ⇒ entropy decreases ⇒ reaction is nonspontaneous.

Sample Problem 6.8 (book). Calculate the entropy change of the system ΔS°_system for 2Ca_(s) + O_2(g) → 2CaO_(s) — same calculation as 6.11.4: ΔS°_system = −208.4 J K⁻¹ mol⁻¹. Sample Problem 6.9 (book). Calculate the entropy change of the surroundings ΔS°_surroundings for the reaction at 298 K: 2Ca_(s) + O_2(g) → 2CaO_(s), ΔH°_reaction = −1270.2 kJ mol⁻¹. Step 1: convert the enthalpy change into J mol⁻¹ by multiplying by 1000: −1270.2 × 1000 = −1 270 200 J mol⁻¹. Step 2: apply ΔS°_surroundings = −ΔH°_reaction/T = −(−1 270 200)/298 = +4262.4 J K⁻¹ mol⁻¹. Sample Problem 6.10 (book). Calculate the total entropy change ΔS°_total for the reaction and check its spontaneity: ΔS°_system = −208.4, ΔS°_surroundings = +4262.4 J K⁻¹ mol⁻¹; ΔS°_total = −208.4 + 4262.4 = +4054.0 J K⁻¹ mol⁻¹. The total entropy change is positive and the reaction is feasible — the huge heat released to the surroundings more than compensates for the system's own entropy decrease.

In determining whether a chemical reaction is likely to be spontaneous, we use the new quantity Gibbs free energy change, ΔG. The Gibbs free energy change is given by the relationship: ΔG = −TΔS_total. Multiplying the ΔS°_total expression by −T, we can also get the expression without considering the entropy changes of the surroundings. This expression is called the Gibbs equation: −T × {ΔS°_total = ΔS°_system + [−ΔH°/T]} gives −TΔS_total = −TΔS°_system + ΔH°, i.e. ΔG° = ΔH° − TΔS°.

📌 Extend your knowledge (book's own note): the unit of temperature times entropy (TΔS) is Joule. ΔS° represents ΔS°_system.

And draw the following conclusion from it: (i) If ΔG < 0 (−ve), the given process may occur spontaneously; (ii) If ΔG > 0 (+ve), the indicated process cannot occur spontaneously; instead, the reverse of it may occur; (iii) If ΔG = 0, neither the indicated process nor the reverse of it can occur spontaneously — the system is in a state of equilibrium.

6.12.1 Calculating ΔG° for a reaction — Sample Problem 6.11. For the reaction CaSO_4(s) → Ca²⁺_(aq) + SO₄²⁻_(aq), calculate ΔH°, ΔS° and ΔG° at 25 °C using the following data, and discuss its spontaneity. Enthalpy of formation: ΔH°_f(CaSO4(s)) = −1432.7 kJ, ΔH°_{f(Ca²⁺(aq))} = −543.0 kJ, ΔH°_{f(SO4²⁻(aq))} = −907.5 kJ. Standard entropy: S°_(CaSO4(s)) = 106.7 J/K, S°_{(Ca²⁺(aq))} = −55.2 J/K, S°_{(SO4²⁻(aq))} = +17.2 J/K. Solution: (i) ΔH° = ΣΔH_f(products) − ΣΔH_f(reactants) = [−543.0 − 907.5] − [−1432.7] = −17.8 kJ. (ii) ΔS° = ΣS°_(products) − ΣS°_(reactants) = [(−55.2 + 17.2) − (106.7)] J/K = −144.7 J/K; ΔS° = −0.1447 kJ/K. (Note: the book prints “−1447 J/K” in one line, a misplaced decimal for −144.7.) (iii) ΔG° = ΔH° − TΔS° = (−17.8 kJ) − (298 K) × (−0.1447 kJ/K) = +25.3 kJ. Note that you substitute ΔS° in units of kJ/K in the formula of ΔG°. We conclude that this process is non-spontaneous at standard conditions at 25 °C.

Sample Problem 6.12 (book). What is the standard free-energy, ΔG°, for the following reaction (Haber's process) at 25 °C? Also discuss its spontaneity. N_2(g) + 3H_2(g) → 2NH_3(g). Enthalpy of formation: ΔH°_f(N2(g)) = 0 kJ, ΔH°_f(H2(g)) = 0 kJ, ΔH°_f(NH3(g)) = −45.9 kJ. Standard entropy: S°_(N2(g)) = 191.5 J/K, S°_(H2(g)) = 130.6 J/K, S°_(NH3(g)) = +193 J/K. Solution: ΔH° = ΣΔH_f(products) − ΣΔH_f(reactants) = [2 × (−45.9)] − [0 + 0] = −91.8 kJ. ΔS° = ΣS°_(products) − ΣS°_(reactants) = [2 × (193)] − [191.5 + 3 × 130.6] J/K = −197 J/K = −0.197 kJ/K. ΔG° = ΔH° − TΔS° = (−91.88 kJ) − (298 K) × (−0.197 kJ/K) = −33.1 kJ. ΔG° is a negative value, thus it is concluded that Haber's process is spontaneous at 25 °C.

Spontaneity and temperature change. Until now we obtained the free-energy for a reaction at 25 °C. How do we find ΔG° at another temperature? For this purpose, we use the same equation, i.e. ΔG° = ΔH° − TΔS°. We get the value of ΔG at any temperature by substituting values of ΔH° and ΔS° at 25 °C. But in this method, we assume that ΔH° and ΔS° are essentially constant with respect to temperature (this is only approximately true). Remember that the superscript degree sign (°) refers to substances in standard states, which are substances at 1 atm and at the specified temperature. Although until now this was 25 °C (298 K), we now consider other temperatures. Note that in general, ΔG depends strongly on temperature.

Table 6.3 (book) — effect of temperature on the spontaneity of reactions. In Table 6.3, all of the four possible choices of signs for ΔH° and ΔS° give different temperature behaviours for ΔG° (in all this discussion temperature is always positive and is taken on the Kelvin scale): ΔH° −, ΔS° + ⇒ ΔG° − ⇒ spontaneous at all T; ΔH° +, ΔS° − ⇒ ΔG° + ⇒ non-spontaneous at all T; ΔH° −, ΔS° − ⇒ ΔG° + or − ⇒ spontaneous at low T, non-spontaneous at high T; ΔH° +, ΔS° + ⇒ ΔG° + or − ⇒ non-spontaneous at low T, spontaneous at high T. (Footnote: the terms low temperature and high temperature are relative; for a particular reaction, high temperature could mean room temperature.)

Sample Problem 6.13 (book). Calculate ΔG° at 1000 °C for the following reaction and discuss its spontaneity: CaCO_3(s) → CaO_(s) + CO_2(g). Solution: for this reaction, ΔH° = +178.0 kJ and ΔS° = +160.4 J/K = 0.1604 kJ/K. Substituting T = 1000 + 273 = 1273 K: ΔG = ΔH − TΔS = (+178.0 kJ) − (1273 K) × (+0.1604 kJ/K) = +178.0 kJ − 204.2 kJ = −26.2 kJ. We conclude that this reaction is spontaneous at 1000 °C and 1 atm, since ΔG has negative sign. In other words, at 1000 °C, limestone (CaCO₃) will decompose in an open container.

Quick Check 6.9 (book). (a) What do you expect about the entropy value of the following reactions, whether it would be positive or negative? (i) N_2(g) + 3H_2(g) → 2NH_3(g) — negative: 4 gas moles become 2, fewer arrangements. (ii) 2HI_(g) → H_2(g) + I_2(g) — approximately zero/unchanged: 2 gas moles become 2 gas moles, no change in gas count. (b) Which of the following changes are likely to be spontaneous? (i) The smell from an open bottle of aqueous ammonia diffusing throughout a room — yes, diffusion increases entropy. (ii) Water turning to ice at −10 °C — yes, below 0 °C freezing is spontaneous (ΔH term wins). (iii) Ethanol vaporising at 20 °C — yes, ethanol is volatile and vaporises readily at room temperature. (iv) Water mixing completely with salt — yes, dissolving increases disorder. (v) Limestone (calcium carbonate) decomposing at room temperature — no: at 25 °C ΔG for CaCO₃ decomposition is positive (it needs high temperature, as in Sample Problem 6.13).