# Heat of Formation Table for Common Compounds Acetylene torches create high combustion needed for welding. 1001slide / Getty Images

Also, called standard enthalpy of formation, the molar heat of formation of a compound (ΔHf) is equal to its enthalpy change (ΔH) when one mole of a compound is formed at 25 degrees Celsius and one atom from elements in their stable form. You need to know the values of the heat of formation to calculate enthalpy, as well as for other thermochemistry problems.

This is a table of the heats of formation for a variety of common compounds. As you can see, most heats of formation are negative quantities, which implies that the formation of a compound from its elements is usually an exothermic process.

## Table of Heats of Formation

Reference: Masterton, Slowinski, Stanitski, Chemical Principles, CBS College Publishing, 1983.

## Points to Remember for Enthalpy Calculations

When using this heat of formation table for enthalpy calculations, remember the following:

• Calculate the change in enthalpy for a reaction using the heat of formation values of the reactants and products.
• The enthalpy of an element in its standard state is zero. However, allotropes of an element not in the standard state typically do have enthalpy values. For example, the enthalpy values of O2 is zero, but there are values for singlet oxygen and ozone. The enthalpy values of solid aluminum, beryllium, gold, and copper are zero, but the vapor phases of these metals do have enthalpy values.
• When you reverse the direction of a chemical reaction, the magnitude of ΔH is the same, but the sign changes.
• When you multiply a balanced equation for a chemical reaction by an integer value, the value of ΔH for that reaction must also be multiplied by the integer.

## Sample Heat of Formation Problem

As an example, heat of formation values are used to find the heat of reaction for acetylene combustion:

2C2H2(g) + 5O2(g) → 4CO2(g) + 2H2O(g)

### 1: Check to Make Sure the Equation Is Balanced

You'll be unable to calculate enthalpy change if the equation isn't balanced. If you're unable to get a correct answer to a problem, it's a good idea to go back and check the equation. There are many free online equation-balancing programs that can check your work.

### 2: Use Standard Heats of Formation for the Products

ΔHºf CO2 = -393.5 kJ/mole

ΔHºf H2O = -241.8 kJ/mole

### 3: Multiply These Values by the Stoichiometric Coefficient

In this case, the value is four for carbon dioxide and two for water, based on the numbers of moles in the balanced equation:

vpΔHºf CO2 = 4 mol (-393.5 kJ/mole) = -1574 kJ

vpΔHºf H2O = 2 mol ( -241.8 kJ/mole) = -483.6 kJ

### 4: Add the Values to Get the Sum of the Products

Sum of products (Σ vpΔHºf(products)) = (-1574 kJ) + (-483.6 kJ) = -2057.6 kJ

### 5: Find Enthalpies of the Reactants

As with the products, use the standard heat of formation values from the table, multiply each by the stoichiometric coefficient, and add them together to get the sum of the reactants.

ΔHºf C2H2 = +227 kJ/mole

vpΔHºf C2H2 = 2 mol (+227 kJ/mole) = +454 kJ

ΔHºf O2 = 0.00 kJ/mole

vpΔHºf O2 = 5 mol ( 0.00 kJ/mole)= 0.00 kJ

Sum of reactants (Δ vrΔHºf(reactants)) = (+454 kJ) + (0.00 kJ) = +454 kJ

### 6: Calculate the Heat of Reaction by Plugging the Values Into the Formula

ΔHº = Δ vpΔHºf(products) - vrΔHºf(reactants)

ΔHº = -2057.6 kJ - 454 kJ

ΔHº = -2511.6 kJ