Intermediate Chemistry - Mono Mole

October 26, 2019December 3, 2024

Distribution coefficient (partition coefficient)

The distribution coefficient describes the relative concentrations of a chemical compound in two immiscible solvents.

When a solute, X, with different solubilities in two immiscible solvents, e.g., water and hexane, is shaken in a separating funnel containing both solvents and left to settle, a dynamic equilibrium is established such that the rate of the solute moving from the aqueous layer to the organic layer is the same as the rate of the solute moving from the organic layer to the aqueous layer.

$X(aq)\rightleftharpoons X(org)$

The equilibrium constant, called the distribution coefficient (or partition coefficient) is given by:

$K_d=\frac{[X(org)]}{[X(aq)]}$

If the compound is more soluble in the organic phase, the distribution coefficient will be greater than 1. If it is more soluble in the aqueous phase, the coefficient will be less than 1. This coefficient is important in various fields, like pharmacology, where it helps predict how a drug will behave in the body (its absorption, distribution, and excretion), and in environmental science, where it can indicate how a pollutant might move between water and soil.

Question

10.00 g of benzoic acid that is dissolved in 100 ml of water is shaken with 50 ml of chloroform and left to settle. 20 ml is then extracted from the aqueous layer and titrated with 6.40 ml of 0.1000 M of NaOH. Calculate the distribution coefficient for benzoic acid in the two solvents.

Answer

$C_6H_5COOH(l)+NaOH(aq)\rightarrow C_6H_5COO^-Na^+(aq)+H_2O(l)$

$K_d=\frac{\frac{\frac{10.00}{122.05}-(0.1000\times 0.00064\times 5)}{0.05}}{\frac{0.1000\times 0.0064\times 5}{0.1}}=49.2$

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October 26, 2019September 25, 2024

Le Chatelier’s principle

Henry Louis Le Chatelier, a French chemist, developed a principle of chemical equilibrium in the late 1800s, which states:

A change in concentration, pressure or temperature to a system at dynamic equilibrium causes the position of the equilibrium to shift in the direction that minimises the change.

The principle is based on the thermodynamic properties of the equilibrium constant, which is dependent only on temperature for a particular reaction. Let’s see how the factors (concentration, pressure and temperature) affect the equilibria of chemical reactions in the next few articles.

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October 26, 2019May 4, 2025

Solubility product

The solubility product of a chemical compound is the equilibrium constant for the dissolution of the compound in its solid state in an aqueous solution.

The solubility of a solute is the maximum amount of the solute in grammes that can dissolve in a 100 ml (or sometimes 1 dm³ or 1 kg) of a solvent at a particular temperature. For example, the solubility of AgCl in 100 ml of water at 25^oC is about 1.92×10^-4 g or 1.34×10^-5 moles.

To illustrate the concept of solubility product, let’s begin by adding 1.0×10^-5 moles of solid AgCl in 100 ml of water at 25^oC. The solid completely dissolves, producing 1.0×10^-5 moles each of Ag⁺ and Cl^– ions. As we increase the amount of solid AgCl to 1.34×10^-5 moles, it still dissolves completely, yielding the maximum concentration of Ag⁺ and Cl^–ions possible in 100ml of water. At this point, the solution is considered saturated. If we add more than 1.34×10^-5 moles of AgCl, any excess solid will remain undissolved. The concentrations of Ag⁺ and Cl^– ions stay constant at 1.34×10^-5 mol per 100 ml, and the system reaches equilibrium. At equilibrium, the rate at which solid AgCl dissolves into ions equals the rate at which the ions recombine to form the solid, establishing a dynamic balance:

$AgCl(s)\rightleftharpoons Ag^+(aq)+Cl^-(aq)$

Since the concentration of solid silver chloride is assumed to be the same as that of its pure state, the equilibrium constant is:

$K_{sp}=[Ag^+][Cl^-]$

with

K_spbeing the solubility product of AgCl, which is 1.8×10^-10at 25^oC.
[Ag⁺] is the maximum amount of Ag⁺ in 100 ml of water.
[Cl^–] is the maximum amount of Cl^– in 100 ml of water.

The solubility product of AgCl is therefore the mathematical product of the solubility of Ag⁺(with respect to Cl^–) and the solubility of Cl^–(with respect to Ag⁺) raised to the power of their respective stoichiometric coefficients in 100 ml of solvent at a particular temperature. In general, the solubility product of A_xB_yis

$K_{sp,A_xB_y}=[A^{y+}]^x[B^{x-}]^y$

Question

Calculate the K_sp for PbCl₂, given that its solubility is 0.0108 g/ml at 20^oC.

Answer

$PbCl_2(s)\rightleftharpoons Pb^{2+}(aq)+2Cl^-(aq)$

Solubility of Pb²⁺ with respect to Cl^–= $\frac{10.8}{\left [ 207.2+\left ( 2\times 35.45 \right ) \right ]}\; mol\, dm^{-3}$

Solubility of Cl^– with respect to Pb²⁺= $\frac{2\times 10.8}{\left [ 207.2+\left ( 2\times 35.45 \right ) \right ]}\; mol\, dm^{-3}$

$K_{sp}=\left [ Pb^{2+} \right ]\left [ Cl^- \right ]^2=\left ( \frac{10.8}{278.1} \right )\left ( \frac{2\times 10.8}{278.1} \right )^2=2.3\times 10^{-4}\; mol^{\: 3}\, dm^{-9}$

Just as K_a and K_b are only useful for comparing weak acids and weak bases respectively, K_sp is only useful for comparing sparing soluble salts, as highly soluble salts have a higher probability of forming ion pairs. An ion pair consists of a cation and an anion that are electrostatically attracted to each other, rather than being individually surrounded by solvent molecules. This interaction alters their physical properties—for example, their mobility—and can distort the measurement of ion concentrations in solution when using ionic or conductivity methods. As a result, the accuracy of the K_sp value may be compromised. In general, the higher the solubility of a solid, the greater the concentration of ions in the solvent, which increases the likelihood of ion pair formation.

Furthermore, just as K_w is constant at a particular temperature regardless of the source of H₃O⁺ and OH^–, K_spfor a solid remains constant at a particular temperature regardless of the source of the dissolved ions. For example, the presence of NaCl in a saturated solution of AgCl causes the latter to precipitate, as Cl^– is common to both species. The decrease in the solubility of a dissolved compound in the presence of an ion in common with the dissolved compound is called the common ion effect.

Question

Is K_sp dependent on the volume of the solution?

Answer

No. is a thermodynamic equilibrium constant that is governed by the formula

$\Delta_rG^{\: o}=-RTlnK$

Hence, K_spis only dependent on temperature. If we dilute a solution of PbCl₂ that is in equilibrium with solid PbCl₂, the increase in volume of the solution shifts the position of the equilibrium according to Le Chatelier’s principle to produce more aqueous Pb²⁺ and Cl^– such that the saturation concentrations (mole per volume) of Pb²⁺ and Cl^– remains unchanged when the new equilibrium is attained.

Another way to look at it is that K_sp is the mathematical product of the solubility of the ions of a compound, raised to the power of their respective stoichiometric coefficients in a particular volume of solvent at a particular temperature. Since solubility is an intensive property, K_spis independent of the volume of solvent.

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October 26, 2019December 31, 2024

Effect of concentration on equilibrium

What is the effect of concentration on equilibrium?

As mention in an earlier article, the equilibrium constant K of a reaction is:

$K=e^{-\frac{\Delta_rG^o}{RT}}\; \; \; \; \; \; \; \; 17$

Eq17 shows that the value of K for a particular reaction only varies with temperature and is independent of the concentration of reaction species. With that in mind, how does concentration affect the position of chemical equilibrium of a reaction? Consider the hydrolysis of ethyl ethanoate,

$CH_3COOC_2H_5(l)+H_2O(l)\rightleftharpoons CH_3COOH(l)+C_2H_5OH(l)$

The equilibrium constant for the reaction is

$K=\frac{[CH_3COOH][C_2H_5OH]}{[CH_3COOC_2H_5][H_2O]}$

When the reaction reaches dynamic equilibrium at a particular temperature, the ratio of the products and the reactants is a constant. If we increase the concentration of ethyl ethanoate at this stage, the excess ethyl ethanoate reacts with water to give more products to maintain the constant value of K, thereby reducing the concentration of ethyl ethanoate and shifting the position of the equilibrium to the right to attain a new dynamic equilibrium. This is consistent with Le Chatelier’s principle.

If we now increase the concentration of ethanoic acid or decrease the concentration of ethyl ethanoate, the position of the equilibrium will shift to the left to minimise the change, maintaining the same value of K.

We can also explain the above observation using chemical kinetics. At equilibrium, the rate of the forward reaction is equal to the rate of the reverse reaction. If we increase the concentration of ethyl ethanoate, the greater number of ethyl ethanoate molecules leads to a higher frequency of collisions between ethyl ethanoate and water molecules, which increases the rate of the forward reaction relative to the reverse reaction. As a result, the position of equilibrium shifts to the right to maintain the value of K.

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October 26, 2019October 19, 2024

Acid/base dissociation constant

An acid/base dissociation constant is a measure of the strength of an aqueous acid/base.

A weak acid dissociates partially in an aqueous solution to give the hydroxonium ion and a conjugate base.

$HA(aq)+H_2O(l)\rightleftharpoons H_3O^+(aq)+A^-(aq)$

Since water is the solvent, its activity approximately equals to one, and the equilibrium constant is given by:

$K_a=\frac{\left [ H_3O^+ \right ][A^-]}{[HA]}\; \; \; \; \; \; \; \; 12$

K_a is called the acid dissociation constant. The higher the value of K_a of an acid, the greater the extent of dissociation of the acid (i.e., the stronger the acid). It is difficult to determine accurately the K_a of an acid that completely ionises in water (a strong acid), since [HA] → 0. Hence, K_a is a useful measure only for weak acids.

A weak base dissociates partially in an aqueous solution to give a conjugate acid and the hydroxide ion.

$B(aq)+H_2O(l)\rightleftharpoons BH^+(aq)+OH^-(aq)$

Since water is the solvent, its activity approximately equals to one, and the equilibrium constant is given by:

$K_b=\frac{[BH^+][OH^-]}{[B]}\; \; \; \; \; \; \; \; 13$

K_b is called the base dissociation constant. The higher the value of K_b of a base, the greater the extent of dissociation of the base (i.e., the stronger the base). Similar to acids, it is difficult to determine accurately the K_b of a base that completely ionises in water (a strong base), since [B] → 0. Hence, K_b is a useful measure only for weak bases.

Note that BH⁺ is an acid with the following dissociation equilibrium:

$BH^+(aq)+H_2O(l)\rightleftharpoons B(aq)+H_3O^+(aq)$

$K_a=\frac{[B][H_3O^+]}{[BH^+]}\; \; \; \; \; \; \; \; 14$

Multiplying eq14 with eq13,

$K_aK_b=[H_3O^+][OH^-]\; \; \; \; \; \; \; \; 15$

Substituting eq11 from the previous article in eq15

$K_aK_b=K_w\; \; \; \; \; \; \; \; 16$

Eq16 expresses the relationship for all conjugate acid-base pairs.

Question

Given that K_a for HCN is 6.2×10^-10 at 25^oC, calculate the equilibrium constant for

$CN^-(aq)+H_2O(l)\rightleftharpoons HCN(aq)+OH^-(aq)$

Answer

$K_b=\frac{K_w}{K_a}=\frac{10^{-14}}{6.2\times 10^{-10}}=1.6\times 10^{-5}$

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October 26, 2019November 20, 2024

Effect of pressure on equilibrium

What is the effect of pressure on equilibrium?

Similar to concentration, a change in pressure does not change the value of the equilibrium constant K. So how does pressure affect the position of chemical equilibrium of a reaction? Consider the following reaction:

$2NO_2(g)\rightleftharpoons 2NO(g)+O_2(g)$

The equilibrium constant for the reaction is

$K=\frac{p_{NO}^{\; \; \; \; \; \; 2}p_{O_2}}{p_{NO_2}^{\; \; \; \; \; \; \; 2}}$

where p_i is the partial pressure of gas i.

We can also write the above equation as:

$K=\frac{\left ( x_{NO}p \right )^2\left ( x_{O_2}p \right )}{\left ( x_{NO_2}p \right )^2}=\frac{x_{NO}^{\; \; \; \; \; \; 2}\, x_{O_2}\, p}{x_{NO_2}^{\; \; \; \; \; \; \; 2}}=\frac{\left (\frac{n_{NO}}{n} \right )^2\left ( \frac{n_{O_2}}{n} \right )p}{\left ( \frac{n_{NO_2}}{n}\right )^2} =\left ( \frac{n_{NO}^{\; \; \; \; \; \; 2}\, n_{O_2}}{n_{NO_2}^{\; \; \; \; \;\; \;\, 2}} \right )\frac{p}{n}\; \; \; \; \; \; \; \; 18$

where x_i is the mole fraction of gas i, p is the total pressure of the system, n_i is the number of moles of gas i, and n is the total number of moles of the gas mixture.

For ideal gases, we can substitute pV = nRT in eq18 to give:

$K=\left ( \frac{n_{NO}^{\; \; \; \; \; \; 2}\, n_{O_2}}{n_{NO_2}^{\; \; \; \; \;\; \;\, 2}} \right )\frac{RT}{V}\; \; \; \; \; \; \; \; 19$

If we increase the total pressure by decreasing the volume of the system at constant temperature, the denominator of the term in brackets in eq19 must increase to maintain the value of K (or the numerator of the term in brackets must decrease). The position of the equilibrium of the reaction therefore shifts left to increase the amount of nitrogen dioxide. This is consistent with Le Chatelier’s principle, where an increase in pressure shifts the position of the equilibrium from right (3 moles of gases) to left (2 moles of gases) to minimise the change in pressure.

It is important to note that changes in pressure only affects the position of an equilibrium where species are gases, as solids and liquids are relatively incompressible.

Question

Consider a fixed-volume reaction vessel with the following reaction at a state of equilibrium at constant temperature:

$2NO_2(g)\rightleftharpoons 2NO(g)+O_2(g)$

If we increase the pressure in the vessel by adding an inert gas (assuming that all gases in the vessel are ideal gases), will the composition of the gases change?

Answer

No, because the partial pressures of the gases remain the same even though the total pressure of the system has increased.

$p_i=x_ip=\frac{n_i}{n}p=\frac{n_iRT}{V}$

Since the factor of RT/V is a constant, p_idoes not change if n_iremains the same.

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October 26, 2019September 25, 2024

Effect of temperature on equilibrium

What is the effect of temperature on equilibrium?

The change in temperature of a chemical reaction at equilibrium results in a shift in position of the equilibrium that opposes the temperature change. For example, the increase in temperature of an endothermic reaction encourages the reaction to occur because the increased energy is absorbed to facilitate the reaction, thereby lowering the temperature.

Quantitatively, the effect of temperature on a chemical reaction system is best explained using the van’t Hoff equation (see below for derivation):

$\frac{dlnK}{d\frac{1}{T}}=-\frac{\Delta_rH^o}{R}\; \; \; \; \; \; \; \; 20$

where Δ_rH^o is the enthalpy of a reaction at standard conditions, K is the equilibrium constant, T is the temperature of the system and R is the gas constant.

Eq20 can be rewritten as:

$ln\left ( \frac{K_f}{K_i} \right )=-\frac{\Delta_rH^o }{R}\left ( \frac{1}{T_f}-\frac{1}{T_i} \right )\: \: \: \: \: \: \: \:\: 20a$

where the subscripts f and i represent ‘final’ and ‘initial’ respectively.

Let’s consider an increase in T of an endothermic reaction (Δ_rH^o> 0). The RHS of eq20a under such conditions is positive, which means that K_f> K_i, i.e. the equilibrium constant increases with an increase in T for an endothermic reaction. Therefore, unlike concentration and pressure, temperature affects the value of K. Nevertheless, Le Chatelier’s principle still applies, as illustrated by the following example:

$N_2(g)+O_x(g)\rightleftharpoons 2NO(g)\; \; \; \; \; \; \; \Delta_rH^o=+180\: kJmol^{-1}$

$K=\frac{p_{NO}^{\; \; \; \; \; \; 2}}{p_{N_2}\, p_{O_2}}\; \; \; \; \; \; \; \; 21$

If K increases as a result of an increase in T, the numerator of eq21 must increase relative to the denominator. This means that the position of the equilibrium shifts to the right, which is consistent with the definition of Le Chatelier’s principle.

Using the same logic and eq20a, we can describe the effect of temperature on the position of an exothermic reaction’s equilibrium.

In summary,

	T	K	Equilibrium position
Endothermic	↑	↑	→
Endothermic	↓	↓	←
Exothermic	↑	↓	←
Exothermic	↓	↑	→

Question

Derive the van’t Hoff equation.

Answer

Substituting the definition of the standard Gibbs energy of reaction Δ_rG^o= Δ_rH^o – TΔ_rS^o in the Gibbs energy equilibrium constant relation Δ_rG^o = –RTlnK, we have

$lnK=-\frac{\Delta_r H^o}{RT}+\frac{\Delta_rS^o}{R}$

If we assume that Δ_rH^o and Δ_rS^o are constant for a reaction over a range of temperature, the change of lnK with respect to the change in temperature is:

$\frac{dlnK}{dT}=\frac{\Delta_r H^o}{RT^2}$

This is the called the van’t Hoff equation, which can also be written as:

$\frac{dlnK}{d\frac{1}{T}}=-\frac{\Delta_r H^o}{R}$

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October 26, 2019May 29, 2025

Effect of catalysts on equilibria

Catalysts have no effect on either the equilibrium constant or the position of equilibrium of a reaction. A catalyst offers a separate pathway for a reaction to proceed with lower activation energy, E_a, and does not affect the energy levels of the reactants and products. As a result, a catalyst increases the rates of both the forward and reverse reactions by the same factor and allows a reaction to reach its dynamic equilibrium faster without affecting either the equilibrium constant or the position of equilibrium of the reaction.

Mathematically, let’s consider the following reversible reaction

$A\; \begin{matrix} k_f\\\rightleftharpoons \\ k_r \end{matrix}\: B$

where k_f is the rate constant of the forward reaction and k_r is that for the reverse reaction. In chemical kinetics, the rates of formation of B and A are:

$rate_f=k_f[A]\; \; \; and\; \; \;rate_r=k_r[B]$

At equilibrium, rate_f = rate_r

$k_f[A]=k_r[B]\; \; \; \; \; \; \; \; 22$

Substitute the equilibrium constant, K = [B]/[A] in eq22

$K=\frac{k_f}{k_r}\; \; \; \; \; \; \; \; 23$

Substitute the Arrhenius equation where $k=Ae^{-\frac{E_a}{RT}}$ in eq23

$K=\frac{A_fe^{-\frac{E_a}{RT}}}{A_re^{-\frac{E_a\, '}{RT}}}\; \; \; \; \; \; \; \; 24$

For a catalysed reaction where the activation energy decreases by an amount, $\left | \Delta E \right |$ ,

$K_{cat}=\frac{k_{f,cat}}{k_{r,cat}}=\frac{A_{f,cat}e^{-\frac{E_a-\left | \Delta E \right |}{RT}}}{A_{r,cat}e^{-\frac{E_a\, '-\left | \Delta E \right |}{RT}}}=\frac{A_{f,cat}e^{-\frac{E_a}{RT}}}{A_{r,cat}e^{-\frac{E_a\, '}{RT}}}\frac{e^{\frac{\left | \Delta E \right |}{RT}}}{e^{\frac{\left | \Delta E \right |}{RT}}} =\frac{A_{f,cat}e^{-\frac{E_a}{RT}}}{A_{r,cat}e^{-\frac{E_a\, '}{RT}}}\; \; \; \; \; \; \; \; 25$

Assuming the ratio of the forward pre-exponential factor and the reverse pre-exponential factor for the uncatalysed reaction (A_f/A_r) is the same as that for the catalysed one (A_f,cat/A_r,cat), the equilibrium constant for an uncatalysed reaction, eq24, is the same as that for a catalysed one, eq25. This means that the catalytic pathway does not affect the value of the equilibrium constant. Furthermore, both the numerator and denominator of eq24 are unchanged despite a change in the activation energy in the catalysed reaction, which means that the position of equilibrium remains the same. Lastly, the forward rate of reaction, k_f[A], and reverse rate of reaction, k_r[B], increase by the same factor of $e^{\frac{\left | \Delta E \right |}{RT}}$ for the catalysed reaction, allowing the reaction to achieve equilibrium faster.

Question

If a catalyst increases the rates of both the forward and reverse reactions of a reversible reaction, does it mean that we will not achieve a greater amount of products than an uncatalysed reaction (assuming we begin both the catalysed and uncatalysed reactions consisting of only the same amount of reactants under the same temperature and pressure, and that k_f> k_r for the uncatalysed reaction)?

Answer

Since a catalysed reaction does not alter either the equilibrium constant or the position of the equilibrium of the reaction, we will obtain the same amount of product as the uncatalysed reaction, but at a faster rate.

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October 24, 2019September 25, 2024

Standard conditions (chemical energetics)

Standard conditions in chemical energetics provide a consistent reference framework that enables reliable comparison of thermodynamic properties and reaction feasibility across different chemical systems.

The value of the change in enthalpy of a reaction, ΔH_r, depends on the following physical properties of reactants and products:

1. Temperature
2. Pressure or concentration
3. State of matter

In other words, ΔH_rfor a reaction carried out, for example, at 298.15 K, is different from that at 373.15K.

To avoid confusion, thermodynamic calculations are often made using data derived under a specific set of conditions known as standard conditions, which are defined as:

1. Temperature: 298.15K (data are sometimes derived at other temperatures)
2. Pressure: 100 kPa or 1 bar
3. Concentration: 1 M
4. State of matter: Each substance is in its normal state (s, l or g) at 100 kPa and 298.15K. For example, the normal state of molecular oxygen at 100 kPa and 298.15K is O₂ (g).

Thermodynamic properties, e.g. ΔH_r, that are calculated using standard conditions data are given an additional symbol “o” as a superscript to their current symbols, i.e. ΔH_r^o. ΔH_r^ois therefore the standard enthalpy change of reaction. If ΔH_r^o is observed at a temperature other than 298.15K, the symbol ΔH_r^o(T) will be used, where T is the temperature at which ΔH_r^ois observed.

Question

The standard enthalpy change of the manufacture of ammonia via the Haber process, $\frac{1}{2}N_2(g)+\frac{3}{2}H_2(g)\rightleftharpoons NH_3(g)$ , is ΔH_r^o = -46.1 kJmol^-1. However, the process is too slow at 298.15 K and 1 bar. Instead, it takes place at around 720 K and 200 bar. How then is ΔH_r^oobtained?

Answer

The enthalpy change of the reaction, ΔH_r, is first measured at the optimum conditions of 720 K and 200 bar and subsequently converted to ΔH_r^o at 298.15 K and 1 bar using Kirchhoff’s law (the understanding of this law requires the knowledge of chemical thermodynamics at the advanced level).

The value of the standard enthalpy change of a reaction is quoted to match the stoichiometric coefficients of the written reaction equation. For example, if the equation for the Haber process is written as $2NH_3(g)\rightleftharpoons N_2(g)+3H_2(g)$ , the quoted standard enthalpy change of reaction is:

$2NH_3(g)\rightleftharpoons N_2(g)+3H_2(g)\; \; \; \; \; \; \; \; \Delta H=+92.2\: kJmol^{-1}$

Question

Why is water always in the liquid state in combustion equations like $CH_4(g)+2O_2(g)\rightarrow CO_2(g)+2H_2O(l)$ when it should be in vapour form considering the temperature of combustion?

Answer

Equations like the combustion of methane are usually written under standard conditions, i.e. with all reactants and products in their normal states at 100 kPa and 298.15K. Although it is not wrong to write the equation as $CH_4(g)+2O_2(g)\rightarrow CO_2(g)+2H_2O(g)$ , calculations are often carried out using thermodynamic properties that are quoted in standard conditions, which makes $CH_4(g)+2O_2(g)\rightarrow CO_2(g)+2H_2O(l)$ more relevant.

Next article: Types of enthalpy changes

As reactions are classified into different categories (e.g. combustion, neutralisation and hydration), standard enthalpy changes are similarly categorised as:

1. Standard enthalpy change of formation, ΔH_f^o
2. Standard enthalpy change of vaporisation, ΔH_vap^o
3. Standard enthalpy change of fusion, ΔH_fus^o
4. Standard enthalpy change of sublimation, ΔH_sub^o
5. Standard enthalpy change of combustion, ΔH_c^o
6. Standard enthalpy change of neutralisation, ΔH_n^o
7. Standard enthalpy change of atomisation, ΔH_at^o
8. Standard enthalpy change of ionisation, ΔH_ion^o
9. Standard enthalpy change of electron gain, ΔH_eg^o
10. Standard enthalpy change of hydration, ΔH_hyd^o
11. Standard enthalpy change of solution, ΔH_sol^o
12. Standard enthalpy change of lattice energy, ΔH_latt^o
13. Other standard enthalpy changes

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