Intermediate Chemistry - Mono Mole

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.

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October 24, 2019March 27, 2025

Types of enthalpy changes (chemical energetics)

Types of enthalpy changes, including enthalpy of formation, enthalpy of combustion and enthalpy of reaction, play a vital role in thermodynamics by quantifying the heat absorbed or released during chemical processes.

ΔH_r^ois a thermodynamic property that describes the standard enthalpy change of any reaction.

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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October 24, 2019December 1, 2024

Standard enthalpy change of formation

The standard enthalpy change of formation, ΔH_f^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 O₂is 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 ΔH_f^o is the standard enthalpy change of the reaction ΔH_r^othat 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

$\Delta H_r^{\: o}=\sum \Delta H_f^{\: o}<div class="woocommerce columns-4 "><ul class="products columns-4"> <li class="product type-product post-9170 status-publish first instock product_cat-ebook has-post-thumbnail downloadable virtual taxable purchasable product-type-simple"> <a href="https://monomole.com/product/science-around-the-house-discovering-the-hidden-science-in-daily-life/" class="woocommerce-LoopProduct-link woocommerce-loop-product__link"></a></li> <li class="product type-product post-8651 status-publish instock product_cat-ebook has-post-thumbnail downloadable virtual taxable purchasable product-type-simple"> <a href="https://monomole.com/product/the-mole-theories-behind-the-experiments-that-define-the-avogadro-constant/" class="woocommerce-LoopProduct-link woocommerce-loop-product__link"></a></li> </ul> </div>-\sum \Delta H_f^{\: o}[reactants]$

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 $\Delta H_f^{\: o}[MgO]+\Delta H_f^{\: o}[CO_2]=\sum \Delta H_f^{\: o}<div class="woocommerce columns-4 "><ul class="products columns-4"> <li class="product type-product post-9170 status-publish first instock product_cat-ebook has-post-thumbnail downloadable virtual taxable purchasable product-type-simple"> <a href="https://monomole.com/product/science-around-the-house-discovering-the-hidden-science-in-daily-life/" class="woocommerce-LoopProduct-link woocommerce-loop-product__link"></a></li> <li class="product type-product post-8651 status-publish instock product_cat-ebook has-post-thumbnail downloadable virtual taxable purchasable product-type-simple"> <a href="https://monomole.com/product/the-mole-theories-behind-the-experiments-that-define-the-avogadro-constant/" class="woocommerce-LoopProduct-link woocommerce-loop-product__link"></a></li> </ul> </div>$ and $\Delta H_f^{\: o}[MgCO_3]= \sum \Delta H_f^{\: o}[reactants]$ ,

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 ΔH_r^o, ΔH_f^o[product] and ΔH_f^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. v_Pand v_Rare 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:

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

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

ΔH_r^o for the above reaction is also known as ΔH_sol^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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October 24, 2019September 25, 2024

Standard enthalpy change of fusion

The standard enthalpy change of fusion, ΔH_fus^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, ΔH_solid^o, is the negative value of the standard enthalpy change of fusion of that substance.

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October 24, 2019September 25, 2024

Standard enthalpy change of sublimation

The standard enthalpy change of sublimation, ΔH_sub^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, ΔH_dep^o, is the negative value of the standard enthalpy change of sublimation of that substance.

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October 24, 2019September 25, 2024

Standard enthalpy change of vaporisation

The standard enthalpy change of vaporisation, ΔH_vap^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 ΔH_vap^o= 44.0 kJmol^-1is 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, ΔH_con^o, is the negative value of the standard enthalpy change of vaporisation of that substance.

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October 24, 2019September 25, 2024

Standard enthalpy change of combustion

The standard enthalpy change of combustion ΔH_c^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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October 24, 2019September 25, 2024

Standard enthalpy change of atomisation

The standard enthalpy change of atomisation, ΔH_at^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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October 24, 2019January 4, 2025

Standard enthalpy change of ionisation

The standard enthalpy change of ionisation, ΔH_ion^o, 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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