Standard Enthalpy of Formation (HSC SSCE Chemistry): Revision Notes

Standard Enthalpy of Formation

Introduction

When studying chemical reactions, we need to measure energy changes under consistent conditions. This is where the concept of standard enthalpy of formation becomes essential. It provides a reference point for calculating energy changes in chemical reactions.

Standard state and standard enthalpy change

What is a standard state?

A standard state is a set of reference conditions used to measure thermodynamic properties. The enthalpy change for a reaction can vary slightly depending on pressure and concentration, so we need a common reference point.

Standard state is defined as follows:

• For pure substances: The stable physical form (solid, liquid, or gas) at a pressure of $100.0 \text{ kPa}$ and at the specified temperature
• For gases in a mixture: The gas present at a pressure of $100.0 \text{ kPa}$
• For solutes in solution: The substance present at a concentration of exactly $1 \text{ mol L}^{-1}$

Standard enthalpy change

The standard enthalpy change ($\Delta H°$) for a reaction is the enthalpy change that occurs when all reactants and products are in their standard states.

What is standard enthalpy of formation?

Definition

The standard enthalpy of formation ($\Delta H_f°$), also called the heat of formation, is the enthalpy change when one mole of a compound in its standard state forms from its elements in their standard states.

Examples of standard enthalpies of formation

Water:

The standard enthalpy of formation of water at $298 \text{ K}$ is $-285 \text{ kJ} \cdot \text{mol}^{-1}$. This refers to the equation:

$\text{H}_2(g) + \frac{1}{2}\text{O}_2(g) \rightarrow \text{H}_2\text{O}(l) \quad \Delta H° = -285 \text{ kJ} \cdot \text{mol}^{-1}$

Notice that we use $\frac{1}{2}$ mole of oxygen to form exactly one mole of water.

Ammonia:

The standard enthalpy of formation of ammonia at $298 \text{ K}$ is $-46 \text{ kJ} \cdot \text{mol}^{-1}$:

$\frac{1}{2}\text{N}_2(g) + \frac{3}{2}\text{H}_2(g) \rightarrow \text{NH}_3(g) \quad \Delta H_f° = -46 \text{ kJ} \cdot \text{mol}^{-1}$

Again, fractional coefficients ensure we form exactly one mole of ammonia.

Different physical states

The standard enthalpy of formation refers to the stable form at the specified temperature and $100.0 \text{ kPa}$. However, we can also discuss the standard enthalpy of formation for substances in alternative physical states, provided we specify the state clearly.

Standard enthalpy of formation for elements

Example: At $298 \text{ K}$, $\Delta H_f°(\text{I}_2) = 0$ because this refers to solid iodine, which is iodine's stable form at $298 \text{ K}$.

Elements not in their standard state

We can discuss the standard enthalpy of formation of elements in non-standard states, but the state must be clearly specified.

Similarly, the standard enthalpy of formation of bromine at $298 \text{ K}$ is for liquid bromine (its stable form). If we want the value for gaseous bromine, we must specify this clearly.

Table of standard enthalpies of formation

Below are standard enthalpies of formation for common substances at $298 \text{ K}$:

Inorganic substances

Substance	$\Delta H_f°$ (kJ mol⁻¹)	Substance	$\Delta H_f°$ (kJ mol⁻¹)	Substance	$\Delta H_f°$ (kJ mol⁻¹)
Al₂O₃(s)	−1670	HF(g)	−271	NO₂(g)	+33
BaSO₄(s)	−1465	HI(g)	+26	N₂O₄(g)	+9
Br₂(g)	+31	H₂O(g)	−242	NH₃(g)	−46
CaO(s)	−636	H₂O(l)	−285	NH₄Cl(s)	−314
Ca(OH)₂(s)	−987	H₂O₂(g)	−136	NaCl(s)	−411
CaCO₃(s)	−1207	H₂O₂(l)	−188	NaOH(s)	−425
CaCl₂(s)	−795	H₂O₂(aq)	−191	Na₂SO₄(s)	−1385
CO(g)	−111	H₂S(g)	−21	O₃(g)	+143
CO₂(g)	−393	H₂SO₄(l)	−814	PBr₃(l)	−185
Fe₂O₃(s)	−823	I₂(g)	+62	SO₂(g)	−297
HBr(g)	−36	I₂(aq)	+23	SO₃(g)	−396
HCl(g)	−93	NO(g)	+90

Organic substances (carbon compounds)

Substance	$\Delta H_f°$ (kJ mol⁻¹)	Substance	$\Delta H_f°$ (kJ mol⁻¹)
Methane, CH₄(g)	−75	Ethanol, C₂H₅OH(g)	+235
Ethane, C₂H₆(g)	−85	Dichloromethane, CH₂Cl₂(g)	+92
Ethylene, C₂H₄(g)	+52	Chloromethane, CH₃Cl(g)	−86
Acetylene, C₂H₂(g)	+227	Bromomethane, CH₃Br(g)	+35

Ions in aqueous solution

Ion	$\Delta H_f°$ (kJ mol⁻¹)	Ion	$\Delta H_f°$ (kJ mol⁻¹)	Ion	$\Delta H_f°$ (kJ mol⁻¹)
Ag⁺	+106	Fe²⁺	−89	Mg²⁺	−467
Ba²⁺	−538	Fe³⁺	−49	NO₃⁻	−207
Br⁻	−122	H⁺	0	Na⁺	−240
Cl⁻	−167	I⁻	−55	OH⁻	−230
CO₃²⁻	−677	I₃⁻	−52	SO₄²⁻	−909
F⁻	−333	K⁺	−283	Zn²⁺	−154

Calculating standard enthalpy changes from formation enthalpies

General equation

Standard enthalpy changes for reactions can be calculated from the standard enthalpies of formation of the reactants and products using:

$\Delta H° = \sum \Delta H_f°(\text{products}) - \sum \Delta H_f°(\text{reactants})$

Understanding the calculation with enthalpy diagrams

Let's consider the reaction:

$\text{C}_2\text{H}_4(g) + \text{HCl}(g) \rightarrow \text{C}_2\text{H}_5\text{Cl}(g)$

The diagram shows how we can use Hess's law to calculate $\Delta H°$ for this reaction:

• The formation enthalpies take us from elements to reactants (upward arrows)
• The reaction enthalpy takes us from reactants to products (dashed arrow)
• The formation enthalpy of the product takes us from elements to product (upward arrow on right)

By Hess's law: $\Delta H° = \Delta H_f°(\text{C}_2\text{H}_5\text{Cl}) - \Delta H_f°(\text{C}_2\text{H}_4) - \Delta H_f°(\text{HCl})$

Calculating enthalpies of formation from heats of combustion

For many compounds, especially organic compounds, the heat of combustion is easier to measure experimentally than the enthalpy of formation. If we know the heats of formation of carbon dioxide and water, we can calculate the enthalpy of formation of the compound.

Remember!