Standard Enthalpy of Formation (Standard Heat of Formation) Chemistry Tutorial

⚛ Standard Enthalpy of Formation is also known as Standard Heat of Formation.

⚛ Δ_fH^o, or Δ_fH°, is the symbol used for standard enthalpy of formation (standard heat of formation) ⁽¹⁾

⚛ Standard Enthalpy of Formation (standard heat of formation), Δ_fH^o, of any compound is defined as the enthalpy change of the reaction by which it is formed from its elements in their standard state.

⚛ By definition, the standard enthalpy of formation (standard heat of formation) of an element in its standard state is zero: ⁽²⁾

for an element in its standard state: Δ_fH^o = 0

⚛ Standard Molar Enthalpy of Formation (standard molar heat of formation) may be given the symbol Δ_fH^o_m.

⚛ Standard Molar Enthalpy of Formation (standard molar heat of formation) of a compound is defined as the enthalpy change that occurs when one mole of the compound in its standard state is formed from its elements in their standard states.

⚛ Values of the standard molar enthalpy of formation of some substances can be found in tables (usually at a temperature of 25°C and pressure of 100 kPa). ⁽³⁾

⚛ Standard Enthalpy Change of Reaction (standard heat of reaction), Δ_rH^o, is the difference between the standard enthalpies of formation of the products and the reactants.

⚛ Δ_rH^o = the sum of the enthalpy of formation of products − the sum of the enthalpy of formation of reactants:

Δ_rH^o = ΣΔ_fH^o(products) − ΣΔ_fH^o(reactants)

Σ is the symbol used for "sum of"

⚛ To calculate the Standard Enthalpy Change of Reaction (standard heat of reaction) using standard enthalpy (heat) of formation data:

Step 1: Write the balanced chemical equation for the chemical reaction

Remember to include the state (solid, liquid, gas, or aqueous) for each reactant and product!

Step 2: Write the general equation for calculating the standard enthalpy of reaction:

Δ_rH^o = ΣΔ_fH^o(products) − ΣΔ_fH^o(reactants)

Step 3: Substitute the values for the standard enthalpy (heat) of formation of each product and reactant into the equation.

Remember, if there are 2 moles of a reactant or product, you will need to multiply the enthalpy term by 2, if molar enthalpies (heats) of formation are used.

Step 4: Solve the equation to find the standard enthalpy change of reaction (standard heat of reaction) in kJ mol^-1.

Please do not block ads on this website.
No ads = no money for us = no free stuff for you!

The standard enthalpy of formation (standard heat of formation) of a compound is defined as the enthalpy change for the reaction in which elements in their standard states produce products.

At 25°C and 100 kPa, the standard state of any element is solid with the following exceptions:

Consider a water molecule, H₂O.
A molecule of water contains the elements hydrogen (H) and oxygen (O).

The standard enthalpy of formation of liquid water would be defined as the enthalpy change when the elements hydrogen and oxygen in their standard states react to produce liquid water. This is represented by the word equation below:

hydrogen in its standard state + oxygen in its standard state → liquid water

If the conditions set are 25°C and 100 kPa, then the standard states for reactants and products are:

• hydrogen exists as a diatomic molecule and is a gas, H_2(g)
• oxygen exists as a diatomic molecule and is a gas, O_2(g)
• water exists as a molecule and is a liquid, H₂O_(l)

The balanced chemical equation representing the standard enthalpy of formation per mole of liquid water (standard heat of formation of liquid water) is therefore:

H_2(g) + ½O_2(g) → H₂O_(l)

If you looked up the standard enthalpy of formation of liquid water in tables (at 25°C and 100 kPa), the value would be given as:

Δ_fH^o = -285.8 kJ mol^-1

This means that when molecular hydrogen gas reacts with molecular oxygen gas, 285.8 kJ of energy will be released for every mole of liquid water that is produced.

If 10 moles of liquid water was produced from molecular hydrogen gas and molecular oxygen gas, then 10 × 285.8 = 2858 kJ of energy would be released.

Similarly, if 0.1 moles of liquid water was produced from molecular hydrogen gas and molecular oxygen gas, then 0.1 × 285.8 = 28.58 kJ of energy would be released.

The values for the standard enthalpy of formation for a number of different compounds at 25°C is given in the table below:

Note that the standard enthalpy of formation of liquid water is NOT the same as the standard enthalpy of formation of gaseous water. Therefore, you must always include the states of matter when writing your chemical equations because states of matter are very important when using standard enthalpy of formation values.

Also note that the standard enthalpy of formation (heat of formation) of some compounds is positive and for others it is negative:

⚛ Δ_fH^o = + (a positive value)

Overall, energy is absorbed when the elements in their standard states form the compound.
Formation of the compound is an endothermic reaction.

⚛ Δ_fH^o = − (a negative value)

Overall, energy is released when the elements in their standard states form the compound.
Formation of the compound is an exothermic reaction.

↪ Back to top

We can use standard enthalpy of formation values (Δ_fH^o) to calculate the standard enthaly change for a chemical reaction (Δ_rH^o).

For example, consider the reaction in which gaseous hydrogen chloride (HCl_(g)) reacts with gaseous ammonia (NH_3(g)) to produce solid ammonium chloride (NH₄Cl_(s)) at 25°C.

NH_3(g) + HCl_(g) → NH₄Cl_(s)

From the table of values for Standard Enthalpy of Formation at 25°C given in the previous section, we find that the standard enthalpy of formation of ammonium chloride (NH₄Cl_(s)) is −314.4 kJ mol^-1.
That is, the formation of 1 mole of solid ammonium chloride (NH₄Cl_(s)) from the elements nitrogen, hydrogen and chlorine in their standard states releases 314.4 kJ of energy.
Therefore we can write a chemical equation to represent this reaction as shown below:

½N_2(g) + 2H_2(g) + ½Cl_2(g) → NH₄Cl_(s)     Δ_fH^o = −314.4 kJ mol^-1

But where will these elements come from in order to react?
They can come from the reactant molecules breaking apart.

Hydrogen chloride molecules could break apart to provide molecules of hydrogen and chlorine gas according to the following chemical equation:

HCl_(g) → ½H_2(g) + ½Cl_2(g)

But how much energy will be absorbed or released to do this?

From the table of values for Standard Enthalpy of Formation at 25°C we find that the enthalpy of formation of HCl_(g) is −92.3 kJ mol^-1.
We can represent this formation reaction as:

½H_2(g) + ½Cl_2(g) → HCl_(g)     Δ_fH^o = −92.3 kJ mol^-1

This reaction is just the reverse of the one we want! That is, we want to break up HCl_(g) molecules!
So we can reverse the reaction, AND, reverse the sign of the enthalpy change as well!

HCl_(g) → ½H_2(g) + ½Cl_2(g)     Δ_rH^o = +92.3 kJ mol^-1

Similarly, ammonia gas (NH_3(g)) could break apart to provide the nitrogen and hydrogen we need for the overall reaction to occur:

NH_3(g) → ½N_2(g) + ³/₂H_2(g)

This reaction is the reverse of the heat of formation (enthalpy of formation reaction) shown below:

½N_2(g) + ³/₂H_2(g) → NH_3(g)

From the table of values for Standard Enthalpy of Formation at 25°C we find that the enthalpy of formation of NH_3(g) is −46.1 kJ mol^-1.
That is:

½N_2(g) + ³/₂H_2(g) → NH_3(g)     Δ_fH^o = −46.1 kJ mol^-1

So, the reaction to break apart ammonia molecules into hydrogen gas and ammonia gas is the reverse of this equation, AND, we must remember to reverse the sign of the enthalpy change as well!

NH_3(g) → ½N_2(g) + ³/₂H_2(g)     Δ_rH^o = +46.1 kJ mol^-1

The enthalpy change required to produce the elements hydrogen, nitrogen and chlorine in their standard states is the sum of the enthalpy change for breaking apart hydrogen chloride molecules and for breaking apart ammonia molecules.
We can Hess's Law to determine the overall enthalpy change for this reaction:

Now we can use this H_2(g), Cl_2(g) and N_2(g) to produce NH₄Cl_(s).
Recall from the beginning of this section that this reaction, the formation of NH₄Cl_(s) from its elements in their standard states, releases 314.4 kJ mol^-1 of energy.
So now we can add together two chemical equations and their associated enthalpy terms:

(i) equation for the breaking apart of reactant molecules into elements

added to

(ii) equation for the elements coming together to form product molecules

=

(ii) equation for the standard enthalpy change for the overall chemical reaction

This is shown below:

Look at what we have done here:

(a) We found the value of the standard enthalpy of formation of the product

Δ_fH^o(product)

(b) We have added together the standard enthalpy of formation of each reactant molecule, AND, reversed the sign:

Δ_rH^o(reactants) = −ΣΔ_fH^o(reactants)

(c) We added the two values together to find the enthalpy change for the reaction, Δ_rH^o(reaction):

Δ_rH^o(reaction) = Δ_fH^o(product) + [−ΣΔ_fH^o(reactants)]

This is a very quick way to use standard enthalpy of formation data to calculate the enthalpy change of a chemical reaction.
The standard enthalpy change of a chemical reaction = sum of the enthalpy of formation of products − sum of the enthalpy or formation of reactants
In chemistry short-hand we represent this is as shown below:

Δ_rH^o(reaction) = ΣΔ_fH^o(products) −ΣΔ_fH^o(reactants)

or, we can also write:

Δ_rH^o(reaction) = −ΣΔ_fH^o(reactants) + ΣΔ_fH^o(products)

Using the formation of solid ammonium chloride (NH₄Cl_(s)) from the reactants hydrogen chloride gas (HCl_(g)) and ammonia gas (NH_3(g)) as an example of the application of this equation:

Note that this is the same result as we achieved above using Hess's Law, Δ_rH^o = −176.0 kJ mol^-1
Therefore we can write the following balanced thermochemical equation:

HCl_(g) + NH_3(g) → NH₄Cl_(s)     Δ_rH^o = −176.0 kJ mol^-1

↪ Back to top

Below is a worked example of finding the standard enthalpy change for a reaction using the standard enthalpy of formation data.

Solving the Problem using the StoPGoPS approach to problem solving in chemistry.

↪ Back to top

Methanol, CH₃OH_(l), undergoes complete combustion in the presence of excess oxygen gas to produce carbon dioxide gas and gaseous water.

Determine the standard enthalpy change for this combustion reaction using the tabulated values for enthalpy of formation given in the table in this tutorial.

↪ Back to top

↪ Back to top