General Chemistry

Having a library of values for standard enthalpies of formation allows calculating the enthalpy of reaction without doing calorimetric analysis.

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:

 

 

where m and n are the stoichiometric coefficients for the reactants and products, and Σ (sigma) means “the sum of.”

So, the enthalpy of a reaction is the difference between the sum of the enthalpies of the products and the sum of the enthalpies of the reactants.

 

2) The Indirect Approach

If the enthalpies of formation of all the components in chemical reactions are not known, then the Hess’s law is used.

The Hess’s law is covered in a separate post, and our focus today is going to be on the direct method for determining the enthalpy of a reaction.

For example,

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

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 = [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.

 

Let’s another example, where the heat of formation needs to be calculated from the enthalpy of the reaction.

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

 

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.

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