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Standard enthalpy change of ionisation

The standard enthalpy change of ionisation, ΔHiono, is the change in enthalpy for removing one mole of electrons from one mole of atoms or ions in the gaseous state under standard conditions.

When one mole of electrons is removed from atoms to form one mole of monovalent cations, the change in enthalpy is called the first ionisation enthalpy, e.g.

Na(g)\rightarrow Na^+(g)+e^-\; \; \; \; \; \; \; \Delta H_{ion}^{\: o}=502.2\: kJmol^{-1}

When one mole of electrons is removed from monovalent cations to form one mole of divalent cations, the change in enthalpy is called the second ionisation enthalpy, e.g.

Mg^+(g)\rightarrow Mg^{2+}(g)+e^-\; \; \; \; \; \; \; \Delta H_{ion}^{\: o}=1457.2\: kJmol^{-1}

The standard enthalpy change of ionisation of a substance is calculated from the substance’s ionisation energy (the two are not the same, as ionisation energy is defined at absolute zero). Ionisation energies are determined experimentally using photoelectron spectroscopy (PES), where the known energy of an incident photon on an atom equals to the atom’s first ionisation energy plus the measured kinetic energy of the ionised electron. These ionisation energies are then converted to ionisation energies at absolute zero before converting to standard enthalpies of ionisation, with both conversions using Kirchhoff’s law.

 

Question

Does the standard enthalpy of ionisation apply to anions?

Answer

Yes, e.g.

Cl^-(g)\rightarrow Cl(g)+e^-\; \; \; \; \; \; \; \Delta H_{ion}^{\: o}=355.2\: kJmol^{-1}\; \; \; \; \; \; 3

 

 

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Standard enthalpy change of electron gain

The standard enthalpy change of electron gain, ΔHego, is the change in enthalpy when one mole of electrons is attached to atoms or anions in the gaseous state to form one mole of anions under standard conditions.

When one mole of electrons is attached to atoms to form one mole of monovalent anions, the change in enthalpy is called the first electron gain enthalpy, e.g.

Cl(g)+e^-\rightarrow Cl^-(g)\; \; \; \; \; \; \; \Delta H_{eg}^{\: o}=-355.2\: kJmol^{-1}

When one mole of electrons is attached to monovalent anions to form one mole of divalent anions, the change in enthalpy is called the second electron gain enthalpy, e.g.

O^-(g)+e^-\rightarrow O^{2-}(g)\; \; \; \; \; \; \; \Delta H_{eg}^{\: o}=837.8\: kJmol^{-1}

The standard enthalpy change of electron gain of a substance is calculated from the substance’s electron affinity (the two are not the same, as electron affinity is defined at absolute zero) using Kirchhoff’s law, or from the relationship:

\Delta H_{eg}^{\: o}\left [ X^n \right ]=-\Delta H_{ion}^{\: o}\left [ X^{n-1} \right ]\; \; \; \; \; \; 4

which states that the standard enthalpy change of electron gain of a species is the negative of the standard enthalpy change of ionisation of that species with an additional electron attached, e.g.

\Delta H_{eg}^{\: o}\left [ Cl(g) \right ]=-\Delta H_{ion}^{\: o}\left [Cl^-(g) \right ]\; \; \; \; \; \; 5

Substituting eq3 from the previous section in eq5,

\Delta H_{eg}^{\: o}\left [ Cl(g) \right ]=-355.2\: kJmol^{-1}

Electron affinities are determined experimentally using photoelectron spectroscopy (PES) where the known energy of an incident photon on an anion equals to the atom’s electron affinity plus the measured kinetic energy of the ionised electron. These electron affinities are then converted to electron affinities at absolute zero, which are then converted to standard enthalpies of electron gain, with both conversions using Kirchhoff’s law. It is easier to measure ionisation energies than electron affinities and therefore standard enthalpies of electron gain of substances are usually calculated using eq4.

 

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Standard enthalpy change of hydration

The standard enthalpy change of hydration, ΔHhydo, is the change in enthalpy when one mole of an ion (or molecule) in the gaseous state dissolves in water to form an infinitely dilute solution under standard conditions. This means that we need to dissolve the solute in excess water until there is no change in the energy absorbed or released by the system.

Some examples are:

Zn^{2+}(g)\rightarrow Zn^{2+}(aq)\; \; \; \; \; \; \; \Delta H_{hyd}^{\: o}=-2046\: kJmol^{-1}

ClO_4^-(g)\rightarrow ClO_4^-(aq)\; \; \; \; \; \; \; \Delta H_{hyd}^{\: o}=-238\: kJmol^{-1}

The standard enthalpy of hydration is usually exothermic (negative) and becomes more negative for ions with with higher charge-to-radius ratios and for molecules with greater polarity.

 

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Standard enthalpy change of solution

The standard enthalpy change of solution, ΔHsolo, is the change in enthalpy when one mole of a solute dissolves in a solvent to form an infinitely dilute solution under standard conditions. This means that we need to dissolve the solute in excess solvent until there is no change in the energy absorbed or released by the system.

Some examples are:

KOH(s)\rightarrow KOH(aq)\; \; \; \; \; \; \; \Delta H_{sol}^{\: o}=-57.6\: kJmol^{-1}

HCl(g)\rightarrow HCl(aq)\; \; \; \; \; \; \; \Delta H_{sol}^{\: o}=-74.8\: kJmol^{-1}

Since KOH(aq) and HCl(aq) are fully dissociated in water, we can also write the above equation as:

KOH(s)\rightarrow K^+(aq)+OH^-(aq)\; \; \; \; \; \; \; \Delta H_{sol}^{\: o}=-57.6\: kJmol^{-1}

HCl(g)\rightarrow H^+(aq)+Cl^-(aq)\; \; \; \; \; \; \; \Delta H_{sol}^{\: o}=-74.8\: kJmol^{-1}

ΔHsolo can be positive or negative. Compounds with large positive ΔHsolo are relatively insoluble.

Question

With reference to the diagram above showing the molecular structures of cisplatin (a chemotherapy drug) and transplatin, deduce which has a less positive standard enthalpy change of solution (in water), and hence, is more soluble in water.

Answer

According to Hess’s law,

where and are the standard enthalpy changes of sublimation and hydration respectively.

Both cisplatin and transplatin have square-planar molecule geometry. In the cis configuration, the two chloride ligands are adjacent to each other, and the two ammonia ligands occupy the remaining adjacent positions. Because Cl and NH3 have different electronegativities, their bond dipole moments do not cancel. This creates a polar molecule with a significant net electric dipole moment. In the trans configuration, identical ligands are positioned 180° apart. The bond dipoles of the two Pt–Cl bonds cancel each other, as do those of the two Pt–NH₃ bonds, resulting in a non-polar molecule with a net electric dipole moment of zero. Since water is a polar solvent, it interacts much more favourably with polar molecules. The stronger dipole–dipole interactions between water and gaseous cisplatin result in a more exothermic (more negative) standard enthalpy change of hydration than for transplatin.

In the solid phase, transplatin molecules are more symmetrical and can pack more efficiently into the crystal lattice. This more efficient packing increases intermolecular attractions in the solid, so the standard enthalpy change of sublimation of transplatin is expected to be more positive than that of cisplatin.

Combining the two terms, cisplatin has a more negative and a less positive . Therefore, its standard enthalpy change of solution is less positive than that of transplatin, consistent with cisplatin being more soluble in water.

 

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Standard enthalpy change of neutralisation

The standard enthalpy change of neutralisation, ΔHn o, is the change in enthalpy when one mole of water is formed from an acid reacting with an alkali, both in their most stable forms, under standard conditions.

For example,

\frac{1}{2}H_2SO_4(aq)+NaOH(aq)\rightarrow \frac{1}{2}Na_2SO_4(aq)+H_2O(l)\; \; \; \Delta H_n^{\: o}=-57.1\: kJmol^{-1}

or in the ionic form,

H^+(aq)+OH^-(aq)\rightarrow H_2O(l)\; \; \; \; \; \; \Delta H_n^{\: o}=-57.1\: kJmol^{-1}

Since acids and alkalis, by definition, are in the aqueous state, their most stable form is the aqueous form. The standard enthalpy change of neutralisation in the above example is also the standard enthalpy change of reaction between sulphuric acid and sodium hydroxide to give sodium sulphate and water, i.e., ΔHn o =ΔHr o.

 

Question

Is the standard enthalpy change of neutralisation, H+(aq) + OH–(aq) → H2O(l), the same as the standard enthalpy change of formation of water?

Answer

No. The definition of the standard enthalpy change of formation of water requires the reactants to be in their standard states, i.e., H2(g) and O2(g).

 

 

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Standard enthalpy change of lattice energy

The standard enthalpy change of lattice energy, ΔHlatto, is the change in enthalpy for breaking the bonds in one mole of a solid ionic compound and separating its gaseous ions to an infinite distance under standard conditions.

Since a large amount of energy is required to carry out the process, the standard enthalpy change of lattice energy of a compound is always positive (endothermic), e.g.

KCl(s)\rightarrow K^+(g)+Cl^-(g)\; \; \; \; \; \; \; \Delta H_{latt}^{\: o}=717\: kJmol^{-1}

However, it may be defined as the change in enthalpy when one mole of an ionic solid is formed from its gaseous ions that are initially infinitely apart under standard conditions. If so, it will always be a negative value (exothermic).

The magnitude of the standard enthalpy change of lattice energy increases for ions with higher charge densities, leading to stronger electrostatic forces of attraction between them in the ionic lattice. Hence,

The standard enthalpy change of lattice energy of an ionic compound is determined theoretically using a Born-Haber cycle, as it is very hard to measure it precisely through experiments.

 

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Other standard enthalpy changes

There are possibly as many standard enthalpy changes as there are types of reactions. Some common ones other than those mentioned in previous articles include:

    1. Standard enthalpy of hydrating an anhydrous salt (not to be confused with standard enthalpy of hydration), e.g.

CuSO_4(s)+5H_2O(l)\rightarrow CuSO_4\cdot 5H_2O(s)\; \; \; \; \; \; \; \Delta H_{hydt}^{\: o}=-78.2\: kJmol^{-1}

    1. Standard enthalpy of precipitation, e.g.

AgNO_3(aq)+NaCl(aq)\rightarrow AgCl(s)+NaNO_3(aq)\; \; \; \; \; \Delta H_{ppt}^{\: o}=-65.8\; kJmol^{-1}

Question

Calculate the enthalpy of precipitation of PbBr2 when 150.0 mL of 0.500 M Pb(NO3)2 is added to 80.0 mL of 1.000 M NaBr in a calorimeter with the temperature rising from 298.15 K to 299.28 K (assuming that the solution’s specific heat capacity is 4.200 Jg-1K-1 and its density is 1.0 g/ml).

Answer

0.0800 moles of NaBr precipitates 0.0400 moles of PbBr2. Using eq5 from a basic level article,

\Delta H_{ppt}=-(4.200)(150.0+80.0)(299.28-298.15)=-1092\: J

\Delta H_{ppt}=\frac{-1092}{0.0400}=-27.3\: kJmol^{-1}

 

 

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Hess’s law (chemical energetics)

Hess’s law is named after Germain Hess, a Russian chemist, who published it in 1840. It is based on the principle of conservation of energy, and states that:

The total enthalpy change in a chemical reaction is sum of enthalpy changes of all steps from the reactants to the products regardless of the route taken.

For example, there are two possible chemical routes (indicated by red arrows in the diagram below) for the combustion of methane to form carbon dioxide and water:

On closer scrutiny, ΔH1o is composed of the following standard enthalpy changes of formation:

C\left ( graphite \right )+2H_2\left ( g \right )\rightarrow CH_4\left ( g \right )\; \; \; \; \; \; \left ( \Delta H_{f}^{o} \right )_{R=1}

2O_2\left ( g \right )\rightarrow 2O_2\left ( g \right )\; \; \; \; \; \; 2\times \left ( \Delta H_{f}^{o} \right )_{R=2}

Similarly, ΔH2o is composed of the following standard enthalpy changes of formation:

C\left ( graphite \right )+O_2\left ( g \right )\rightarrow CO_2\left ( g \right )\; \; \; \; \; \; \left ( \Delta H_{f}^{o} \right )_{P=1}

2H_2\left ( g \right )+O_2\left ( g \right )\rightarrow 2H_2O\left ( g \right )\; \; \; \; \; \; 2\times \left ( \Delta H_{f}^{o} \right )_{P=2}

The choice of the indices R and P will be apparent shortly. Hence,

\Delta H_1^{\: o}=\sum _Rv_R\left ( \Delta H_f^{\: o} \right )_R

\Delta H_2^{\: o}=\sum _Pv_P\left ( \Delta H_f^{\: o} \right )_P

where vR and vP are the stoichiometric coefficients of the products of the respective standard enthalpy change of formation reactions.

In general, for a reaction:

where R denotes the reactants and P denotes the products. Hess’ law states that:

\Delta H_r^{\: o}=-\Delta H_1^{\: o}+\Delta H_2^{\: o}

\Delta H_r^{\: o}=\sum_Pv_P\left ( \Delta H_f^{\: o} \right )_P-\sum_Rv_R\left ( \Delta H_f^{\: o} \right )_R\; \; \; \; \; \; \; \; 6

Eq 6 is a very useful formula for calculating standard enthalpy changes.

 

Question

a) Calculate the standard enthalpy change of formation of CO2(g) given:

\Delta H_f^{\: o}\left [ C_2H_2(g) \right ]=226\: kJmol^{-1}

\Delta H_c^{\; o}\left [ C_2H_2(g) \right ]=-1300\; kJmol^{\: -1}

\Delta H_c^{\: o}\left [ H_2(g) \right ]=-286\: kJmol^{-1}

b) Deduce the relationship between ΔHsolo, ΔHhydo and ΔHlatto using MgCl2 as an example.

Answer

a)

2C(graphite)+H_2(g)\rightarrow C_2H_2(g)\; \; \; \; \; \; \Delta H_f^{\: o}=226\: kJmol^{-1}

C_2H_2(g)+\frac{5}{2}O_2(g)\rightarrow 2CO_2(g)+H_2O(l)\; \; \; \; \; \; \Delta H_c^{\; o}=-1300\: kJmol^{\: -1}

H_2(g)+\frac{1}{2}O_2(g)\rightarrow H_2O(l)\; \; \; \; \; \; \Delta H_f^{\: o}=-286\: kJmol^{-1}

Using eq6, the standard enthalpy change of C(graphite) + O2(g) → CO2(g) is:

\Delta H_f^{\: o}=\frac{226+(-1300)-(-286)}{2}=-394\: kJmol^{-1}

b)

\Delta H_{sol}^{\: o}=\Delta H_{latt}^{\: o}+\sum _iv_i\left ( \Delta H_{hyd}^{\: o} \right )_i

where vi is the stoichiometric coefficient of the reactant of the respective standard enthalpy change of hydration reaction. If ΔHlatto is defined as the change in enthalpy when one mole of an ionic solid is formed from its gaseous ions that are initially infinitely apart under standard conditions, the relation becomes:

\Delta H_{sol}^{\: o}=-\Delta H_{latt}^{\: o}+\sum _iv_i\left ( \Delta H_{hyd}^{\: o} \right )_i

 

 

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Born-Haber cycle (chemical energetics)

The Born-Haber cycle is a method to analyse reaction enthalpies, particularly to calculate the standard enthalpy change of lattice energy, ΔHlatto, which cannot be measured precisely via experiments. It is based on Hess’ law and was developed by the German scientists Max Born and Fritz Haber in 1916.

Born-Haber cycles are best represented in the form of energy diagrams. For example, if we want to calculate the standard enthalpy change of lattice energy of calcium fluoride ΔHlatto[CaF2(s)], we start by writing the equations for ΔHlatto[CaF2(s)] and the standard enthalpy change of the formation of calcium fluoride ΔHfo[CaF2(s)] :

CaF_2(s)\rightarrow Ca^{2+}(g)+2F^-(g)\; \; \; \; \; \; \; \; \Delta H_{latt}^{\: o}\; \; \; \; \; \; \; \; 7

Ca(s)+F_2(g)\rightarrow CaF_2(s)\; \; \; \; \; \; \; \; \Delta H_{f}^{\: o}

Applying Hess’ law, we can combine the two equations into a cycle:

ΔH1o is composed of a few standard enthalpy changes. If we rewrite the cycle to include every standard enthalpy change (starting with the reactants of ΔHfo), we have the Born-Haber cycle for CaF2(s):

The Born-Haber cycle (also known as the Born-Haber energy diagram) reveals that ΔH1o is composed of:

    1. the standard enthalpy change of atomisation of Ca (ΔHoat),
    2. the standard enthalpy change of the first ionisation of Ca (ΔHo1st ion),
    3. the standard enthalpy change of the second ionisation of Ca (ΔHo2nd ion),
    4. the standard enthalpy change of atomisation of F2 (ΔHoat), and
    5. the standard enthalpy change of electron gain of F (ΔHoeg).

It is also evident that the sum of the magnitudes of each enthalpy on the left-hand side of the energy diagram is equal to that on the right-hand side, i.e.

\Sigma_i\left | \Delta H^o_i\left ( LHS \right ) \right |=\Sigma_j\left | \Delta H^o_j\left ( RHS \right ) \right |

The above computation shows that ΔHolatt for CaF2 can be +2635 kJmol-1 or -2635 kJmol-1, depending on its definition. Since we have defined the standard enthalpy change of lattice energy as an endothermic process (see eq7, where ΔHolatt is the change in enthalpy to break the bonds in one mole of a solid ionic compound and separate its gaseous ions to an infinite distance under standard conditions),  ΔHlatto = +2635 kJmol-1.

In summary, the steps involved in generating the Born-Haber cycle are:

    1. Write the equations for the standard enthalpy change of lattice energy and the standard enthalpy change of formation of the ionic compound.
    2. Combine the two equations into a cycle using Hess’ law.
    3. Expand the cycle in a stepwise manner to include every standard enthalpy change.
    4. Draw the Born-Haber cycle energy diagram starting with the reactants of the standard enthalpy change of formation of the compound and ending with the ionic compound. There is no strict rule regarding what must be on the left or right side of the cycle, provided that the steps are shown sequentially and the arrow directions correctly reflect the signs of the enthalpy changes.
    5. Sum the magnitude of enthalpies on each side of the cycle and equate them to find ΔHlatto.

Question

Calculate the standard change in enthalpy of hydration ΔHohyd for F– given that ΔHolatt [CaF2 ] = +2635 kJmol-1, the standard enthalpy change of solution of CaF2 ( ΔHosol [CaF2 ]) is +11 kJmol-1 and ΔHohyd [Ca2+] = -1616 kJmol-1.

Answer

With reference to the Born-Haber cycle below,

\Sigma_i\left | \Delta H^o_i\left ( LHS \right ) \right |=\Sigma_j\left | \Delta H^o_j\left ( RHS \right ) \right |

 

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Cell notation

Cell notation is a shorthand representation of an electrochemical cell.

For example, the electrochemical cell depicted in the diagram above has a cell notation of

Pt\mid H_2\mid H^+ \parallel Cu^{2+}\mid Cu

A vertical line represents a phase boundary, while a double vertical line signifies a phase boundary with negligible junction potential, such as a salt bridge. The anode half-cell is denoted on the left and the cathode half-cell on the right, with the electrodes (Pt and Cu) positioned at both ends of the notation.

A gas is always written adjacent to the electrode. Furthermore, spectator ions are usually omitted and state symbols and concentrations may be included for emphasis, e.g.

Pt(s)\mid H_2(g)\mid H^+(aq,1\, M) \parallel Cu^{2+}(aq,1\, M)\mid Cu(s)

If the standard hydrogen electrode is undergoing reduction, its cell notation is:

Zn\mid Zn^{2+}\parallel H^+ \mid H_2\mid Pt

 

Question

What are the cell notations for a calomel reference electrode undergoing oxidation and a AgCl electrode undergoing reduction?

Answer

The calomel electrode cell notation can be written as

Hg\mid Hg_2Cl_2(sat'd),KCl(sat'd)\mid KCl(sat'd)\parallel

or

Hg\mid Hg_2Cl_2(sat'd),KCl(sat'd)\parallel

The comma separating Hg2Cl2 and KCl means that the two compounds are in the same phase. The double vertical line represents the porous frit, which has negligible junction potential.

The AgCl electrode cell notation is:

\parallel KCl(sat'd),AgCl(sat'd)\mid AgCl\mid Ag

or simply

\parallel KCl(sat'd),AgCl(sat'd)\mid Ag

 

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