What is the basic difference between enthalpy of formation?

Answer 1

• Standard Heat (Enthalpy) of Formation, Hfo, 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, Hfo = 0.
• Standard Molar Enthalpy (Heat) of Formation, Hmo, 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, Ho, is the difference between the standard enthalpies (heats) of formation of the products and the reactants.
• Ho(reaction) = the sum of the enthalpy (heat) of formation of products - the sum of the enthalpy (heat) of formation of reactants:

  Ho(reaction) = Hof(products) - Hof(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:

   Ho(reaction) = Hof(products) - Hof(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.

Standard Enthalpy (Heat) of FormationExample: Standard Enthalpy (Heat) of Formation of WaterThe 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) Hfo = -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) Hfo = -242 kJ mol-1

Answer 2

The enthalpy of formation refers to the energy change when one mole of a compound is formed from its elements in their standard states, while the standard enthalpy of formation refers to the enthalpy change under standard conditions (usually at 25°C and 1 atm pressure). The enthalpy of formation is specific to the compound being formed, whereas the standard enthalpy of formation allows for comparison between different compounds.