Enthalpies of Formation

Core Concept

Enthalpy of Formation (ΔH∘f​) is the enthalpy change when 1 mole of a compound is formed from its elements in their standard states under standard conditions (298 K and 1 atm). Standard enthalpies of formation are essential for calculating the enthalpy changes of reactions.

Key Tips

  • Use Standard Enthalpies: Always use ΔH∘f​ values from standard tables to ensure accuracy.

  • Check Physical States: Ensure each substance’s physical state (solid, liquid, gas) matches its enthalpy of formation, as ΔH∘f​ values vary by state.

  • Remember Elements Have DeltaH∘f=0: Pure elements in their standard states have a formation enthalpy of zero.

  • Multiply by Stoichiometric Coefficients: Don’t forget to multiply each ΔH∘f∘​ by its coefficient from the balanced chemical equation.

Test Yourself

Assorted Multiple Choice
Using standard enthalpies of formation ($\Delta H_f^\circ$), which of the following expressions correctly calculates the standard enthalpy change ($\Delta H^\circ$) for the reaction:$$\text{CH}_4(g) + 2\text{O}_2(g) \rightarrow \text{CO}_2(g) + 2\text{H}_2\text{O}(l)$$

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Enthalpies of Formation

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Calculating Enthalpy of a Reaction Using Enthalpies of Formation

The enthalpy change of a reaction can be calculated using the enthalpies of formation of the reactants and products:

$\Delta H_{\text{reaction}}^\circ = \sum \Delta H_f^\circ(\text{products}) - \sum \Delta H_f^\circ(\text{reactants})$

  1. Identify the Products and Reactants:

    • Write the balanced chemical equation for the reaction.

  2. Find the Enthalpies of Formation:

    • Look up the $\Delta H_f^\circ$​ values for each reactant and product in a table.

  3. Multiply by Coefficients:

    • Multiply each $\Delta H_f^\circ$​ value by its coefficient in the balanced equation.

  4. Subtract the Sum of Reactants from Products:

    • Sum the enthalpies of formation for the products and reactants separately, then subtract the total for the reactants from the total for the products.

Key Concepts

Standard Enthalpy of Formation ($\Delta H_f^\circ$​):

  • The enthalpy change when 1 mole of a compound is formed from its elements in their most stable forms at 1 atm pressure and 298 K.

  • Symbol: $\Delta H_f^\circ$​ (the degree symbol $\circ$∘ denotes standard conditions).

  • Measured in kilojoules per mole (kJ/mol).

Standard State:

  • Elements are in their most stable physical forms at 1 atm and 298 K (e.g., $\text{O}_2$​ for oxygen, $\text{N}_2$​ for nitrogen, and graphite for carbon).

Enthalpy of Formation of Elements:

  • The standard enthalpy of formation of any pure element in its standard state is zero.

  • Example: $\Delta H_f^\circ$​ for $\text{O}_2\text{(g)}$ and $\text{N}_2\text{(g)}$ is zero.

Applications:

  • Enthalpies of formation are used to calculate the enthalpy change for chemical reactions.

  • Hess’s Law and standard enthalpies of formation can be combined to find $\Delta H_{\text{reaction}}^\circ$​.

Example Problem

Calculate the enthalpy change (ΔHrxn​) for the combustion of methane ($\text{CH}_4$​) given the following reaction:

$\text{CH}_4(g) + 2\text{O}_2(g) \rightarrow \text{CO}_2(g) + 2\text{H}_2\text{O}(l)$

Use the standard enthalpies of formation ($\Delta H^\circ_f$):

  • $\Delta H^\circ_f (\text{CH}_4(g)) = -74.8 \, \text{kJ/mol}$

  • $\Delta H^\circ_f (\text{O}_2(g)) = 0 \, \text{kJ/mol}$ (element in its standard state)

  • $\Delta H^\circ_f (\text{CO}_2(g)) = -393.5 \, \text{kJ/mol}$

  • $\Delta H^\circ_f (\text{H}_2\text{O}(l)) = -285.8 \, \text{kJ/mol}$

Step-by-Step Solution:

  1. Write the formula for ΔHrxn∘​:

    $\Delta H^\circ_{\text{rxn}} = \sum (\Delta H^\circ_f \text{ of products}) - \sum (\Delta H^\circ_f \text{ of reactants})$

  1. Substitute the values for the products:

    Products: $\text{CO}_2(g)$ and $2\text{H}_2\text{O}(l)$

    $\sum (\Delta H^\circ_f \text{ of products}) = [1 \cdot \Delta H^\circ_f (\text{CO}_2(g))] + [2 \cdot \Delta H^\circ_f (\text{H}_2\text{O}(l))]$

    =[1⋅(−393.5)]+[2⋅(−285.8)]

    −393.5+(−571.6)=−965.1kJ

  1. Substitute the values for the reactants:

    Reactants: $\text{CH}_4(g)$ and 2$\text{O}_2(g)$

    $\sum (\Delta H^\circ_f \text{ of reactants}) = [1 \cdot \Delta H^\circ_f (\text{CH}_4(g))] + [2 \cdot \Delta H^\circ_f (\text{O}_2(g))]$

    = [ 1⋅(−74.8) ] + [ 2⋅(0) ]

    = −74.8 + 0

    = −74.8 kJ

  1. Calculate $\Delta H^\circ_{\text{rxn}}$​:

    $\Delta H^\circ_{\text{rxn}} = \sum (\Delta H^\circ_f \text{ of products}) - \sum (\Delta H^\circ_f \text{ of reactants})$

    = −965.1 − (−74.8)

    = −965.1 + 74.8

    = − 890.3 kJ

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