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Standard Heat (Enthalpy) of Formation and Reaction

Key Concepts

Standard Heat (Enthalpy) of Formation, H_f^o, 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, H_f^o = 0.
Standard Molar Enthalpy (Heat) of Formation, H_m^o, 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, H^o, is the difference between the standard enthalpies (heats) of formation of the products and the reactants.
H^o_(reaction) = the sum of the enthalpy (heat) of formation of products - the sum of the enthalpy (heat) of formation of reactants:
H^o_(reaction) = H^o_f(products) - H^o_f(reactants)
To calculate an Enthalpy (Heat) of Reaction:

Write the balanced chemical equation for the reaction
Remember to include the state (solid, liquid, gas, or aqueous) for each reactant and product.
Write the general equation for calculating the enthalpy (heat) of reaction:
H^o_(reaction) = H^o_f(products) - H^o_f(reactants)
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.
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 (25^o) 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, H_2(g) and O_2(g).

This reaction is written as:

H_2(g) + ½O_2(g) -----> H₂O_(l)

H_f^o = -286 kJ mol^-1

The standard enthalpy (heat) of formation of water vapour at 298K (25^o) 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, H_2(g) and O_2(g).

This reaction is written as:

H_2(g) + ½O_2(g) -----> H₂O_(g)

H_f^o = -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:

NH_3(g)

H_f^o = -46 kJ mol^-1

NO_2(g) H_f^o = +34 kJ mol^-1

H₂O_(g) H_f^o = -242 kJ mol^-1

O_2(g) H_f^o = 0 kJ mol^-1 (by definition)

Write the balanced chemical equation for the reaction:
NH_3(g) + ⁷/₄O_2(g) -----> NO_2(g) + ³/₂H₂O_(g)
Write the general equation for calculating the enthalpy (heat) of reaction:
H^o_(reaction) = H^o_f(products) - H^o_f(reactants)
H^o = [H^o_{f(NO_2(g))} + ³/₂ x H^o_{f(H₂O_(g))}] - [H^o_{f(NH_3(g))} + ⁷/₄ x H^o_{f(O_2(g))}]
Substitute the values for the enthalpy (heat) of formation of each product and reactant into the equation.
H^o = [+34 + (³/₂ x -242)] - [-46 + (⁷/₄ x 0)]
Solve the equation to find the enthalpy (heat) of reaction.
H^o = [+34 + -363] - [-46 + 0] = -329 + 46 = -283 kJ