# Heat of Formation Table for Common Compounds

The molar heat of formation (also called standard enthalpy of formation) of a compound (ΔHf) is equal to its enthalpy change (ΔH) when one mole of compound is formed at 25°C and 1 atm from elements in their stable form. You need to know the heat of formation values to calculate enthalpy and 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 usually is 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 of solid aluminum, beryllium, gold, and copper are zero. The vapor phases of these metal 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 be multiplied by the integer, too.

## Sample Heat of Formation Problem

For 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 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 4 for carbon dioxide and 2 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