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 the most stable form of its elements in their standard states. Furthermore, the reactant elements must be in their reference states, which are the most stable states 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 that forms the substance, ΔH_r^o, e.g.

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

Question

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

$\Delta H_r^{\: o}=\sum \Delta H_f^{\: o}[products]-\sum \Delta H_f^{\: o}[reactants]$

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 $\Delta H_f^{\: o}[MgO]+\Delta H_f^{\: o}[CO_2]=\sum \Delta H_f^{\: o}[products]$ and $\Delta H_f^{\: o}[MgCO_3]= \sum \Delta H_f^{\: o}[reactants]$ ,

$\Delta H_r^{\: o}=\sum \Delta H_f^{\: o}[products] -\sum \Delta H_f^{\: o}[reactants]\; \; \; \; \; \; \; 1e$

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 relation 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 $\Delta H_f^{\: o}\left [ H^+(aq) \right ]$ .

Standard enthalpy change of formation

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