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Standard Heat (Enthalpy) of Formation and Reaction

Key Concepts

  • Standard Heat (Enthalpy) of Formation, ΔHfo, of any compound is the enthalpy change of the reaction by which it is formed from its elements, reactants and products all being in a given standard state.

  • By definition, the standard enthalpy (heat) of formation of an element in its standard state is zero, ΔHfo = 0.

  • Standard Molar Enthalpy (Heat) of Formation, ΔHmo, of a compound is the enthalpy change that occurs when one mole of the compound in its standard state is formed from its elements in their standard states.

  • Standard Enthalpy (Heat) of Reaction, ΔHo, is the difference between the standard enthalpies (heats) of formation of the products and the reactants.

  • ΔHo(reaction) = the sum of the enthalpy (heat) of formation of products - the sum of the enthalpy (heat) of formation of reactants:

    ΔHo(reaction) = ΣHof(products) - ΣHof(reactants)

  • To calculate an Enthalpy (Heat) of Reaction:
  1. Write the balanced chemical equation for the reaction
    Remember to include the state (solid, liquid, gas, or aqueous) for each reactant and product.

  2. Write the general equation for calculating the enthalpy (heat) of reaction:
    ΔHo(reaction) = ΣHof(products) - ΣHof(reactants)

  3. Substitute the values for the 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.

  4. Solve the equation to find the enthalpy (heat) of reaction.

Standard Enthalpy (Heat) of Formation

Example: Standard Enthalpy (Heat) of Formation of Water

The standard enthalpy (heat) of formation for liquid water at 298K (25o) is -286 kJ mol-1.

This means that 286 kJ of energy is released when liquid water, H2O(l), is produced from its elements, hydrogen and oxygen, in their standard states, ie, H2(g) and O2(g).

This reaction is written as:

H2(g) + ½O2(g) → H2O(l)     ΔHfo = -286 kJ mol-1

The standard enthalpy (heat) of formation of water vapour at 298K (25o) is -242 kJ mol-1.

This means that 242 kJ of energy is released when gaseous water (water vapour), H2O(g), is produced from its elements, hydrogen and oxygen, in their standard states, ie, H2(g) and O2(g).

This reaction is written as:

H2(g) + ½O2(g) → H2O(g)     ΔHfo = -242 kJ mol-1

Standard Enthalpy (Heat) of Reaction

Example : Oxidation of Ammonia

Calculate the standard enthalpy (heat) of reaction for the oxidation of ammonia gas to produce nitrogen dioxide gas and water vapour given the following standard enthalpies (heats) of formation:

NH3(g)     ΔHfo = -46 kJ mol-1

NO2(g)     ΔHfo = +34 kJ mol-1

H2O(g)     ΔHfo = -242 kJ mol-1

O2(g)     ΔHfo = 0 kJ mol-1 (by definition)

  1. Write the balanced chemical equation for the reaction:

    NH3(g) + 7/4O2(g) → NO2(g) + 3/2H2O(g)

  2. Write the general equation for calculating the enthalpy (heat) of reaction:

    ΔHo(reaction) = ΣHof(products) - ΣHof(reactants)

    ΔHo = [ΔHof(NO2(g)) + 3/2 x ΔHof(H2O(g))] - [ΔHof(NH3(g)) + 7/4 x ΔHof(O2(g))]

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

    ΔHo = [+34 + (3/2 x -242)] - [-46 + (7/4 x 0)]

  4. Solve the equation to find the enthalpy (heat) of reaction.

    ΔHo = [+34 + -363] - [-46 + 0] = -329 + 46 = -283 kJ


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