Chemistry · Ch 5 — Thermodynamics

Standard Enthalpy of Formation

5.4(c)

Standard Enthalpy of Formation

The standard enthalpy of formation is the standard enthalpy change for the formation of one mole of a compound from its elements in their most stable states of aggregation (also called reference states). Its symbol is $\Delta_f H^\circ$ , where the subscript $f$ indicates formation.

The reference state of an element is its most stable state of aggregation at 25°C and 1 bar pressure. For example:

Dihydrogen: $\text{H}_2(g)$
Dioxygen: $\text{O}_2(g)$
Carbon: C(graphite, $s$ )
Sulphur: S(rhombic, $s$ )

\Delta_f H^\circ

Some reactions with standard molar enthalpies of formation:

\text{H}_2(g) + \frac{1}{2}\text{O}_2(g) \rightarrow \text{H}_2\text{O}(l); \quad \Delta_f H^\circ = -285.8\ \text{kJ mol}^{-1}

\text{C(graphite, }s) + 2\text{H}_2(g) \rightarrow \text{CH}_4(g); \quad \Delta_f H^\circ = -74.81\ \text{kJ mol}^{-1}

2\text{C(graphite, }s) + 3\text{H}_2(g) + \frac{1}{2}\text{O}_2(g) \rightarrow \text{C}_2\text{H}_5\text{OH}(l); \quad \Delta_f H^\circ = -277.7\ \text{kJ mol}^{-1}

A standard molar enthalpy of formation $\Delta_f H^\circ$ is a special case of $\Delta_r H^\circ$ , where one mole of a compound is formed from its constituent elements.

Watch out

Not every reaction that produces a compound is a formation reaction. For example:

$\text{CaO}(s) + \text{CO}_2(g) \rightarrow \text{CaCO}_3(s); \quad \Delta_r H^\circ = -178.3\ \text{kJ mol}^{-1}$

This is not $\Delta_f H^\circ$ for calcium carbonate, because calcium carbonate is formed from other compounds, not from its constituent elements.

Similarly, for the reaction:

\text{H}_2(g) + \text{Br}_2(l) \rightarrow 2\text{HBr}(g); \quad \Delta_r H^\circ = -72.8\ \text{kJ mol}^{-1}

Here two moles of product are formed, not one. So $\Delta_r H^\circ = 2\Delta_f H^\circ$ . To get the formation reaction, divide all coefficients by 2:

\frac{1}{2}\text{H}_2(g) + \frac{1}{2}\text{Br}_2(l) \rightarrow \text{HBr}(g); \quad \Delta_f H^\circ = -36.4\ \text{kJ mol}^{-1}

Important

By convention, the standard enthalpy of formation $\Delta_f H^\circ$ of an element in its reference state (most stable state of aggregation) is taken as zero.

Calculating Reaction Enthalpy from Formation Enthalpies

The following general equation can be used to calculate the enthalpy change for a reaction using standard enthalpies of formation:

\Delta_r H^\circ = \sum_i a_i \Delta_f H^\circ(\text{products})_i - \sum_i b_i \Delta_f H^\circ(\text{reactants})_i

Here $a_i$ and $b_i$ are the stoichiometric coefficients of products and reactants in the balanced equation.

Let us apply this to the decomposition of calcium carbonate: …

Table 5.2Standard Molar Enthalpies of Formation ($\Delta_f H^\ominus$) at 298 K of a Few Selected Substances

Substance	$\Delta_f H^\ominus$ /(kJ mol $^{-1}$ )	Substance	$\Delta_f H^\ominus$ /(kJ mol $^{-1}$ )
$Al_2O_3$ (s)	-1675.7	HI(g)	+26.48
$BaCO_3$ (s)	-1216.3	KCl(s)	-436.75
$Br_2$ (l)	0	KBr(s)	-393.8
$Br_2$ (g)	+30.91	MgO(s)	-601.70
$CaCO_3$ (s)	-1206.92	$Mg(OH)_2$ (s)	-924.54
C (diamond)	+1.89	NaF(s)	-573.65
C (graphite)	0	NaCl(s)	-411.15
CaO(s)	-635.09	NaBr(s)	-361.06
$CH_4$ (g)	-74.81	NaI(s)	-287.78
$C_2H_4$ (g)	+52.26	$NH_3$ (g)	-46.11
$CH_3OH$ (l)	-238.86	NO(g)	+90.25
$C_2H_5OH$ (l)	-277.69	$NO_2$ (g)	+33.18
$C_6H_6$ (l)	+49.0	$PCl_3$ (l)	-319.70
CO(g)	-110.53	$PCl_5$ (s)	-443.5
$CO_2$ (g)	-393.51	$SiO_2$ (s, quartz)	-910.94
$C_2H_6$ (g)	-84.68	$SnCl_2$ (s)	-325.1
$Cl_2$ (g)	0	$SnCl_4$ (l)	-511.3
$C_3H_8$ (g)	-103.85	$SO_2$ (g)	-296.83
n- $C_4H_{10}$ (g)	-126.15	$SO_3$ (g)	-395.72
HgS(s, red)	-58.2	$SiH_4$ (g)	+34

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