Standard Enthalpy of Reaction

When all the participating substances are in their standard state then the enthalpy of reaction is called the standard enthalpy of reaction. Standard state of a substance is its purest state at 1 bar pressure and 298K temperature. For example standard state of iron is pure solid iron at 1 bar pressure and 298K. Standard state of oxygen is O₂ gas, standard state of water is H₂O liquid and standard state of hydrogen is H₂ gas. Standard conditions are denoted by adding symbol ө to the superscript, e.g. standard enthalpy can be denoted by H^ө.

Phase transfer reactions can be categorized into three categories

1. Fusion or Melting: Involves phase transfer of solid to liquid and enthalpy change is known as enthalpy of fusion Δ_fus H. When 1mole solid substance melts at constant temperature and 1bar pressure then enthalpy change is known as molar or standard enthalpy of fusion Δ_fusH^ө.

H₂O(s) ⟶ H₂O(l)      Δ_fusH^ө = 6kJ mol^-1

The fusion of 1mol of water requires 6kJ mol^-1 heat, this is an endothermic reaction. And same amount of heat will release if we reverse the reaction.

H₂O(l) ⟶ H₂O(s)      Δ_melting H^ө = -6kJ mol^-1

2. Vaporization: involves phase transfer of liquid to gas and enthalpy change is known as enthalpy of vaporization Δ_vap H. Amount of heat required to vaporize 1mole of a liquid substance at constant temperature and 1bar pressure is known as molar or standard enthalpy of vaporization Δ_vap H^ө.

H₂O(l) ⟶ H₂O(g)     Δ_vapH^ө = 40.79kJ mol^-1

3. Sublimation: involves phase transfer of solid to gas and enthalpy change is known as enthalpy of sublimation Δ_sub H. When 1mole solid substance sublimates at constant temperature and 1bar pressure then the enthalpy change is known as molar or standard enthalpy of sublimation Δ_sub H^ө.

CO2(s) ⟶ CO2 (g)       Δ_subH^ө = 25.2kJ mol^-1

You know that enthalpy is an extensive property. Its value depends on external parameters like mass, intermolecular interactions, polarity of bonds etc. For example let's compare the Δ_vap H^ө of water and acetone. Water (H₂O) and acetone (CH₃COCH₃) both are polar molecules, but H bonding in water molecules holds them tightly, so water requires more heat to get vaporized than acetone.

Phase transfer reactions 

Standard enthalpy of combustion Δ_cH^ө

Combustion reactions are very important. These are exothermic reactions. We use them to get energy to run vehicles, machinery, plane and rockets. Our body also uses this reaction to get energy from food. When one mole of a substance undergoes combustion and all the reactant and products are being in their standard state at a specific temperature then the enthalpy change of reaction is called as Standard enthalpy of combustion Δ_cH^ө. Let’s see how much energy we get by combustion of 1mole of glucose:

C₆H₁₂O₆(g) + 6O₂(g) ⟶ 6CO₂(g) + 6H₂O(l)    Δ_cH^ө = -2802 kJ mol^-1

Standard enthalpy of formation Δ_fH^ө

Formation of something means being formed by its constituent elements.  Standard enthalpy of formation Δ_f H^ө is the enthalpy change for the formation of one mole of a compound from its constituent elements in their standard state. For example, reaction for the formation of water will be:

H₂(g) + 1/2O₂(g) ⟶ H₂O(l)     Δ_f H^ө = -285.8 kJ mol^-1

Reaction for the formation of hydrogen bromide will be:

1/2H₂(g) + 1/2Br₂(l) ⟶ HBr(g)    Δ_f H^ө = -36.4 kJ mol^-1

I hope you have understood the concept of formation. Reactant must be in their standard state and only one mole of product should be formed. Only then you get the Δ_f H^ө otherwise it will be just enthalpy of reaction Δ_r H.

H₂(g) + Br₂(l) ⟶ 2HBr(g)     Δ_r H^ө = -72.8 kJ mol^-1

Here 2mole of HBr is formed. So the change in enthalpy you will get here is the Δ_r H^ө not Δ_f H^ө. If you divide the whole reaction by 2 then you will get the Δ_f H^ө = Δ_r H^ө /2.

By convention, standard enthalpy of formation of an element in its standard state is taken as zero. For example Δ_f H^ө of H₂(g), Δ_f H^ө of Br₂(l) and  Δ_f H^ө of O₂(g) all will be zero.

How to calculate standard enthalpy of reaction Δ_rH^өwith the help of Δ_f H^ө?

Enthalpy change of a reaction gives us valuable information about the reaction. With the help of enthalpy change of formation you can calculate exact value of Δ_rH^ө. Let’s see how to calculate the energy needed to decompose calcium carbonate (CaCO₃) into lime (CaO). First write the balance equation for it:

CaCO₃(s) ⟶ CaO(s) + CO₂(g)

Δ_r H^ө = Ʃ ai Δ_f H^ө (products) - Ʃ bi Δ_f H^ө (reactants)

Here a and b are coefficients of the products and reactants for a balanced chemical equation.

Get the values of Δ_f H^ө of compounds and put them in formula,

Δ_f H^ө (CaO) = -635.1 kJ mol^-1

Δ_f H^ө (CO₂) = -393.5 kJ mol^-1

Δ_f H^ө (CaCO₃) = -1206.9 kJ mol^-1

Δ_rH^ө = [1(-635.1 kJ mol^-1) + 1(-393.5 kJ mol^-1)] -1(-1206.9 kJ mol^-1)

Δ_rH^ө = + 178.3 kJ mol^-1

Positive value of Δ_rH^ө shows that it is an endothermic reaction and you will have to give 178.3 kJ mol^-1 heat to decompose calcium carbonate into lime.

Let’s take another example:

Fe₂O₃(s) + 3H₂(g) ⟶ 2Fe(s) + 3H₂O(l)

Δ_f H^ө (Fe) and Δ_f H^ө (H2) is taken as 0 kJ mol^-1by convention as they are elements in their standard state.

Δ_f H^ө (H₂O) = -285.83 kJ mol^-1

Δ_f H^ө (Fe₂O₃) = -824.2 kJ mol^-1

Δ_r H^ө = [2(Δ_f H^ө Fe) + 3(Δ_f H^ө H₂O)] – [3(Δ_f H^ө H₂) + (Δ_f H^ө Fe₂O₃)]

Δ_r H^ө = [2(0) + 3(-285.83 kJ mol^-1)] – [3(0) + (-824.2 kJ mol^-1)]

Δ_r H^ө = -857.49 kJ mol^-1 - (-824.2 kJ mol^-1)

Δ_r H^ө = -33.29 kJ mol^-1

Negative value of Δ_r H^өshows that it is an exothermic reaction and 33.29 kJ mol^-1 heat will be released from this reaction.

Enthalpy is a state function and it is independent of path taken. On the basis of this fact scientist Germain Henri Hess gave a law which is known as Hess’s law of heat summation. We will discuss it in this next post.

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