Standard enthalpy change of formation

The standard enthalpy change of formation, ΔHf o, is the change in enthalpy when one mole of a substance is formed from its elements in their reference states, which are the most stable forms of the respective elements at 100 kPa and 298.15K.

For example:

Element

Most stable form

Notes

Oxygen

O2 (g)

The oxygen radical, O, is unstable

Carbon

Graphite

Graphite is thermodynamically more stable than diamond since it has delocalised electrons

Mercury

Liquid mercury

Phosphorous

White phosphorous

Although not the most thermodynamically stable form, it is easier to isolate in its pure form than red and black phosphorous. This is the only exception.

Examples of standard enthalpy change of formation are as follows:

2C(graphite)+3H_2(g)+\frac{1}{2}O_2(g)\rightarrow C_2H_5OH(l)\; \; \; \; \; \; \Delta H_f^{\: o}=-1367\; kJmol^{-1}

Mg(s)+\frac{1}{2}O_2(g)\rightarrow MgO(s)\; \; \; \; \; \; \Delta H_f^{\: o}=-602\; kJmol^{-1}

 

Question

What is the standard enthalpy change of formation of O2?

Answer

The standard enthalpy change of formation of O2 is written as:

O_2(g)\rightarrow O_2(g)

since the standard enthalpy change of formation of a substance is defined as the enthalpy change when 1 mol of the substance is formed from the most stable form of its elements in their standard states. Therefore, the standard enthalpy change of formation of O2 is zero, because there is no change involved when O2(g) is formed from itself. Similarly, \Delta H_f^{\: o}\left [ C(graphite) \right ]=0, \Delta H_f^{\: o}\left [ H_2(g) \right ]=0 and \Delta H_f^{\: o}\left [ Mg(s) \right ]=0.

 

We can also say that the standard enthalpy change of formation of a substance ΔHf o is the standard enthalpy change of the reaction ΔHo that forms the substance. For example,

Mg(s)+\frac{1}{2}O_2(g)\rightarrow MgO(s)\; \; \; \; \; \; \Delta H_f^{\: o}=\Delta H_r^{\: o}=-602\: kJmol^{-1}

 

Question

For the reaction MgCO_3(s)\rightarrow MgO(s)+CO_2(g), show that

\Delta H_r^{\: o}=\sum \Delta H_f^{\: o}<div class="woocommerce columns-4 "><ul class="products columns-4">
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</div>-\sum \Delta H_f^{\: o}[reactants]

Answer

As we know,

Mg(s)+\frac{1}{2}O_2(g)\rightarrow MgO(s)\; \; \; \; \; \; \Delta H_f^{\: o}[MgO]\; \; \; \; \; \; 1a

C(graphite)+O_2(g)\rightarrow CO_2(g)\; \; \; \; \; \; \Delta H_f^{\: o}[CO_2]\; \; \; \; \; \; 1b

Mg(s)+C(graphite)+\frac{3}{2}O_2(g)\rightarrow MgCO_3(s)\; \; \; \; \; \; \Delta H_f^{\: o}[MgCO_3]\; \; \; \; \; \; 1c

Reversing eq1c,

MgCO_3(s) \rightarrow Mg(s)+C(graphite)+\frac{3}{2}O_2(g)\; \; \; \; \; \; -\Delta H_f^{\: o}[MgCO_3]\; \; \; \; \; \; 1d

Taking the sum of eq1a, eq1b and eq1d, and simplifying,

Equation:\; MgCO_3(s)\rightarrow MgO(s)+CO_2(g)

Change\: in\: enthalpy:\:\Delta H_r^{\: o}=\Delta H_f^{\: o}[MgO]+\Delta H_f^{\: o}[CO_2]-\Delta H_f^{\: o}[MgCO_3]

Since \Delta H_f^{\: o}[MgO]+\Delta H_f^{\: o}[CO_2]=\sum \Delta H_f^{\: o}<div class="woocommerce columns-4 "><ul class="products columns-4">
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</div> and \Delta H_f^{\: o}[MgCO_3]= \sum \Delta H_f^{\: o}[reactants],

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</div> -\sum \Delta H_f^{\: o}[reactants]\; \; \; \; \; \; \; 1e

 

Eq1e is a very useful formula and is a consequence of Hess’ law. In general, for a reaction involving multiple products and multiple reactants, the relationship between ΔHr o, ΔHf o[product] and ΔHf o[reactants] is given by eq6 from the article on Hess’ law:

\Delta H_r^{\: o}=\sum _Pv_P\left (\Delta H_f^{\: o}\right )_P-\sum _Rv_R\left (\Delta H_f^{\: o}\right )_R

where P denotes the number of products and R denotes the number of reactants. vP and vR are the stoichiometric coefficients of the products of the respective standard enthalpy of formation reactions.

An issue arises when we want to determine the standard enthalpy change of formation of aqueous ions, which are always formed as pairs of cations and anions, e.g.

KOH(s)\; \begin{matrix} H_2O\\\rightarrow \end{matrix}\; K^+(aq)+OH^-(aq)

This implies that it is impossible to have a solution consisting of pure cations or anions. The problem is circumvented by defining the standard enthalpy change of formation of the aqueous hydrogen ion as zero:

\Delta H_f^{\: o}\left [ H^+(aq) \right ]=0\; \; \; \; \; \; \; \; 2a

The consequence of this can be illustrated by considering the ion pair of H+(aq) and Br(aq) where:

\frac{1}{2}H_2(g)+\frac{1}{2}Br_2(g)\rightarrow HBr(g)\; \; \; \; \; \; \Delta H_f^{\: o}=-36\: kJmol^{-1}\; \; \; \; \; 2b

HBr(g)\rightarrow HBr(aq)\; \; \; \; \; \; \Delta H_r^{\: o}=-85\: kJmol^{-1}\; \; \; \; \; 2c

ΔHr o for the above reaction is also known as ΔHsol o, the standard enthalpy change of solution of HBr(g). Since HBr(aq) is fully dissociated to H+(aq) and Br(aq), we can also write eq2c as:

HBr(g)\rightarrow H^+(aq)+Br^-(aq)\; \; \; \; \; \; \Delta H_r^{\: o}=-85\: kJmol^{-1}\; \; \; \; \; 2d

Using eq1e on eq2d,

\Delta H_r^{\: o}=\Delta H_f^{\: o}\left [ H^+(aq) \right ]+\Delta H_f^{\: o}\left [ Br^-(aq) \right ]-\Delta H_f^{\: o}\left [ HBr(g) \right ]

-85=\Delta H_f^{\: o}\left [ H^+(aq) \right ]+\Delta H_f^{\: o}\left [ Br^-(aq) \right ]-(-36 )

Since \Delta H_f^{\: o}\left [ H^+(aq) \right ]=0

\Delta H_f^{\: o}\left [ Br^-(aq) \right ]=-121\: kJmol^{-1}

The standard enthalpy change in formation of other aqueous ions, e.g. Cl(aq), can be determined using the same approach. This results in all standard enthalpy changes in formation of aqueous ions being adjusted by the real value of \Delta H_f^{\: o}\left [ H^+(aq) \right ].

 

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Standard enthalpy change of fusion

The standard enthalpy change of fusion, ΔHfus o, is the change in enthalpy when one mole of a substance in the liquid state is formed from the same substance in its solid state under standard conditions.

The process of melting is always endothermic, e.g.

CH_4(s)\rightarrow CH_4(l)\; \; \; \; \; \; \; \Delta H_{fus}^{\: o}[91.1K]=943.8\: kJmol^{-1}

Most data for the standard enthalpy change of fusion of substances are quoted at the melting points of those substances instead of at 298.15 K. The standard enthalpy change of solidification of a substance, ΔHsolid o, is the negative value of the standard enthalpy change of fusion of that substance.

 

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Standard enthalpy change of sublimation

The standard enthalpy change of sublimation, ΔHsub o, is the change in enthalpy when one mole of a substance in the gaseous state is formed from the same substance in its solid state, without going through the liquid state, under standard conditions.

The process of sublimation is always endothermic, e.g.

CO_2(s)\rightarrow CO_2(g)\; \; \; \; \; \; \; \Delta H_{sub}^{\: o}=26.1\: kJmol^{-1}

The standard enthalpy change of sublimation is greater than that of vaporisation and hence greater than that of fusion. The standard enthalpy change of deposition of a substance, ΔHdep o, is the negative value of the standard enthalpy change of sublimation of that substance.

 

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Standard enthalpy change of vaporisation

The standard enthalpy change of vaporisation, ΔHvap o, is the change in enthalpy when one mole of a substance in the gaseous state is formed from the same substance in its liquid state under standard conditions.

Vaporisation reactions are always endothermic, e.g.

H_2O(l)\rightarrow H_2O(g)\; \; \; \; \; \; \; \Delta H_{vap}^{\; o}=44\: kJmol^{-1}

Even though ΔHvap o = 44.0 kJmol-1 is the standard enthalpy change of vaporisation of water at 298.15 K, many data sets of the standard enthalpy change of vaporisation of substances are quoted at the boiling points of those substances. In general, the standard enthalpy change of vaporisation of a substance is higher than the standard enthalpy change of fusion of that substance since molecules are separated further apart from one another from the liquid state to the gaseous state (more energy required) compared to the separation from the solid state to the liquid state. The standard enthalpy change of condensation of a substance, ΔHcon o, is the negative value of the standard enthalpy change of vaporisation of that substance.

 

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Standard enthalpy change of combustion

The standard enthalpy change of combustion ΔHc o is the change in enthalpy when one mole of a substance, in its most stable form, is burnt in excess oxygen under standard conditions.

Combustion reactions are always exothermic, e.g.

CH_4(g)+2O_2(g)\rightarrow CO_2(g)+2H_2O(l)\; \; \; \; \; \; \; \; \Delta H_c^{\: o}-890.4\: kJmol^{-1}

2Fe(s)+\frac{3}{2}O_2(g)\rightarrow Fe_2O_3(s)\; \; \; \; \; \; \; \; \Delta H_c^{\: o}-824.0\: kJmol^{-1}

The standard enthalpy change of combustion of iron is also the standard enthalpy change of formation of iron (III) oxide and the standard enthalpy change of reaction between iron and oxygen to give iron (III) oxide.

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Standard enthalpy change of atomisation

The standard enthalpy change of atomisation, ΔHat o, is the change in enthalpy when one mole of atoms in the gaseous state is formed from the most stable form of its element under standard conditions.

The atomisation process is always endothermic. Some examples are:

C(graphite)\rightarrow C(g)\; \; \; \; \; \; \; \Delta H_{at}^{\: o}=717\: kJmol^{-1}

\frac{1}{2}O_2(g)\rightarrow O(g)\; \; \; \; \; \; \; \Delta H_{at}^{\: o}=249\: kJmol^{-1}

The standard enthalpy of atomisation of an atom is the same as the standard enthalpy of formation of that atom and the standard enthalpy of the reaction to form that atom. The standard enthalpy of atomisation of a species that exist as a homonuclear diatomic molecule under standard conditions, e.g. O2, is also the bond enthalpy of that species. So,

\Delta H_{at}^{\: o}[O]=\Delta H_{be}[O=O]=249\: kJmol^{-1}

 

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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 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 always exothermic (negative) and increases (more negative) for ions with higher charge densities, i.e. higher charge-to-radius ratios.

 

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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.

 

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