General Chemistry

Standard enthalpy of formation (or heat of formation), ΔHof , is the enthalpy change when 1 mol of the substance is formed from its constituent elements in their standard states.

For example, the formation of 1 mol ammonia from H2 and N2 gases releases 46.0 kJ heat:

 

1.5H2(g) + 0.5N2(g) ⇆ NH3(g) ΔHo= 46.0 kJ

 

A reminder about the standard states; depending on the topic, standard states are defined as follows:

 

  • Gases: It is the pure gas at a pressure of exactly 1 atm.
  • Liquids and Solids: It is the pure substance in its most stable form at a pressure of 1 atm and at the temperature of interest (usually at 25 oC).
  • A Substance in Solution: It is when the concentration is exactly 1 M.

 

Let’s answer one question before moving on.

Why do we not talk about the enthalpy of a compound? Why enthalpy of formation of CO2 and not enthalpy of CO2?

Well, the problem is it is impossible to measure the absolute value of the enthalpy of a substance. So, what we do, is take a reference point like the sea level for the height of mountains, and measure the enthalpy change when a given compound or a substance is formed from its elements.

This reference point is the standard enthalpy of formation of any element in its most stable form which is set to zero.

 

For example, the standard enthalpy of formation for O2 is zero, and that of Na, C (graphite), Br2, etc. is zero.

 

Why did we mention graphite for carbon? Remember, carbon has a few allotropes, and the most common ones are graphite and diamond. Graphite is the most stable (standard state) of carbon and therefore, its enthalpy of formation is zero (), while that of diamond is not. Diamond is less stable and its conversion to graphite is an exothermic (downhill – thermodynamically favorable) process:

 

C(diamond) → C(graphite) ΔH° = – 1.9 kJ/mol

 

Fortunately, this process does not happen overnight, and we get to keep our diamonds.

 

Going back to the enthalpy of formation. So, imagine the elements in their standard form are lined up on the sea level (ΔH °f = 0). Now, when a compound is formed by combining any elements, there is an enthalpy change which is the standard enthalpy of formation for that compound:

 

 

For example, the standard enthalpy of formation for methane from carbon, and hydrogen is -74.6 kJ/mol:

 

C(graphite) + 2H2(g) → CH4(g) ΔH °f = -74.6 kJ/mol

 

The comprehensive list of the heat of formation can be found in your textbook or by a quick google search.

 

How do We use the Enthalpies of Formation?

There are countless reactions that can be carried out and it is not always that we can or need to measure their enthalpy change. However, by having a library of values for the enthalpies of formations, we can determine the enthalpy of a given reaction. There are two approaches to doing this.

1) The Direct Approach

If the enthalpies of formations for all the components of ac chemical reaction are known, then the following formula is used to determine the enthalpy of this reaction:

 

 

We will discuss this approach in the next post called “Enthalpy of Reaction from Enthalpies of Formation”.

 

2) The Indirect Approach

If the enthalpies of formation of all the components in chemical reactions are not known, then Hess’s law is used. You can go to the link for detailed coverage of the Hess’s law, but let’s use it here to address one more question you may be wondering about:

 

How is Enthalpy of Formation Measured?

We talked about the enthalpy of formation and, as an example, showed that the enthalpy of formation of methane is -74 kJ/mol, but how do we measure this?

 

C(graphite) + 2H2(g) → CH4(g) ΔH °f = -74.6 kJ/mol

 

When we say the “reaction” forming a substance from its constituent elements, it is, very often, a hypothetical reaction. We can’t combine carbon and hydrogen in the laboratory to make methane. So, to determine its enthalpy of formation, we measure the enthalpy of the combustion reaction of methane using bomb calorimetry and then, apply the Hess’s law to obtain the ΔH °f of methane.

The combustion reaction of methane is the sum of three reactions of formations of methane, CO2, and H2O:

 

C(s) + O2(g) → CO2(g) ΔH °f (CO2) = -393 kJ

+

2H2(g) + O2(g) → 2H2O(l) ΔH °f (H2O) = -572 kJ

+

C(graphite) + 2H2(g) → CH4(g) ΔH °f (CH4) = ?

____________________________________________

CH4(g) + 2O2(g) → CO2(g) + 2H2O(l) ΔH ° = -891 kJ

 

Notice that there is a negatice sign in from of ΔH °f (CH4) becasue the reaction is reversed.

Therefore, 

ΔH ° = ΔH °f (CO2) + ΔH °f (H2O) + (ΔH °f (CH4)) = -891 kJ

Solving for ΔH °f (CH4), we get:

ΔH °f (CH4) = ΔH °f (CO2) + ΔH °f (H2O) + 891

ΔH °f (CH4) = -393 – 572 + 891 = -74 kJ/mol

 

Check Also

 

 

Practice

1.

Using the standard heats of formation given below, calculate the heat of reaction for the combustion of ammonia:

4NH3(g) + 5O2(g) → 4NO(g) + 6H2O(g), ΔH°rxn = ?

ΔHf° for NH3(g) = –46.2 kJ/mol
ΔHf° for NO(g) = 90.4 kJ/mol
ΔHf° for H2O(g) = –241.8 kJ/mol

answer

ΔH°rxn = -904 kJ

Solution

To calculate 𝚫H°rxn, we need to subtract the enthalpies of formation of the reactants multiplied by their stoichiometric coefficients from the enthalpies of formation of products multiplied by their stoichiometric coefficients:

ΔH°rxn = ΣnpΔHof (products) – ΣnrΔH°f (reactants)

Where np and nr are the molar coefficients of the products and reactants in the balanced chemical equation.

Remember, the standard enthalpy for the formation of elements or their molecules in standard states is, by definition, equal to zero.

We can now enter the numbers to determine the heat of our reaction:

ΔH°rxn = [4 x ΔHof (NO) + 6 x ΔHof (H2O)] – [4 x ΔHof (NH3) + 5 x ΔHof (O2)]

ΔH°rxn = [4 x 90.4 kJ + 6 x (–241.8)] – [4 x (–46.2 kJ) + 5 x 0]

ΔH°rxn = -904.4 kJ

ΔH°rxn is negative which means it is an exothermic reaction releasing 904.40kJ of heat.

2.

Combustion of butane (C4H10) releases 5755 kJ of energy according to the following chemical equation.

2C4H10(g) + 13O2(g) → 8CO2(g) + 10H2O(l), ΔH°rxn = -5755 kJ

Calculate the molar enthalpy of formation of butane using the information given below:

ΔHf° for CO2(g) = –393.5 kJ/mol
ΔHf° for H2O(l) = –285.8 kJ/mol

answer

-126 kJ

Solution

To calculate ΔHof (C4H10), we are going to use the equation for the heat of reaction based on the standard enthalpies of formation:

ΔH°rxn = ΣnpΔHof (products) – ΣnrΔH°(reactants)

Where np and nr are the molar coefficients of the products and reactants in the balanced chemical equation.

Remember, the standard enthalpy for the formation of elements or their molecules in standard states is, by definition, equal to zero.

We can now enter the numbers to determine the heat of our reaction:

Let’s add the components for our reaction:

ΔH°rxn = [8 x ΔHof (CO2) + 10 x ΔHof (H2O)] – [2 x ΔHof (C4H10) + 13 x ΔHof (O2)] = -5755 kJ

ΔHof (C4H10) is the unknown that we need to solve for in this equation:

ΔH°rxn = [8 x (–393.5 kJ) + 10 x ΔHof (–285.8 kJ)] – [2 x ΔHof (C4H10) + 13 x 0] = -5755 kJ

-3148 kJ – 2858 kJ – 2 ΔHof (C4H10) = -5755 kJ

-3148 kJ – 2858 kJ + 5755 kJ = 2 ΔHof (C4H10)

ΔHof (C4H10) = -125.5 kJ/mol

Rounding off to three significant figures, we get -126 kJ/mol.

3.

Zinc is recovered from ZnS by first oxidizing it to ZnO. Calculate the enthalpy of this oxidation reaction using the data given below:

2ZnS(s) + 3O2(g) → 2ZnO(s) + 2SO2(g)

ΔHf° for SO2(g) = -296.8 kJ/mol
ΔHf° for ZnS(s) = -206.0 kJ/mol
ΔHf° for ZnO(s) = -350.5 kJ/mol

answer

-883 kJ

Solution

To calculate 𝚫H°rxn, we need to subtract the enthalpies of formation of the reactants multiplied by their stoichiometric coefficients from the enthalpies of formation of products multiplied by their stoichiometric coefficients:

ΔH°rxn = ΣnpΔHof (products) – ΣnrΔH°(reactants)

Where np and nr are the molar coefficients of the products and reactants in the balanced chemical equation.

Remember, the standard enthalpy for the formation of elements or their molecules in standard states is, by definition, equal to zero.

We can now enter the numbers to determine the heat of our reaction:

ΔH°rxn = [2 x ΔHof (ZnO) + 2 x ΔHof (SO2)] – [2 x ΔHof (ZnS) + 3 x ΔHof (O2)]

ΔH°rxn = [2 x (-350.5 kJ) + 2 x (-296.8 kJ)] – [2 x (-206.0 kJ) + 3 x 0]

ΔH°rxn = -701 kJ – 593.6 kJ + 412 kJ = -883 kJ

ΔH°rxn is negative which means it is an exothermic reaction releasing 883 kJ of heat.

4.

Using the standard heats of formation given below, calculate the heat of reaction between barium carbonate (BaCO3) and sulfuric acid (H2SO4).

BaCO3(s) +  H2SO4(aq) → BaSO4(s) + CO2(g) + H2O(l)

ΔHf° for BaSO4(s) = -1473.2 kJ/mol
ΔHf° for CO2(g) = –393.5 kJ/mol
ΔHf° for H2O(l) = –285.8 kJ/mol
ΔHf° for BaCO3(s) = -1213.0 kJ/mol
ΔHf° for 3H2SO4(aq) = -814.0kJ/mol

answer

-126 kJ

Solution

To calculate 𝚫H°rxn, we need to subtract the enthalpies of formation of the reactants multiplied by their stoichiometric coefficients from the enthalpies of formation of products multiplied by their stoichiometric coefficients:

ΔH°rxn = ΣnpΔHof (products) – ΣnrΔH°(reactants)

Where np and nr are the molar coefficients of the products and reactants in the balanced chemical equation.

Remember, the standard enthalpy for the formation of elements or their molecules in standard states is, by definition, equal to zero.

We can now enter the numbers to determine the heat of our reaction:

ΔH°rxn = [1 x ΔHof (BaSO4(s)) + 1 x ΔHof (CO2(g)) + 1 x ΔHof (H2O(l))] – [1 x ΔHof (BaCO3(s)) + 1 x ΔHof (H2SO4(aq))]

ΔH°rxn = [-1473.2 kJ –393.5 kJ –285.8 kJ] – [-1213.0 kJ -814.0kJ]

ΔH°rxn = -2,152.5 kJ + 2,027 kJ = -125.5 kj

ΔH°rxn is negative which means it is an exothermic reaction releasing 126 kJ of heat.

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