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 Hө.
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 Δfus H. When 1mole solid substance melts at constant temperature and 1bar pressure then enthalpy change is known as molar or standard enthalpy of fusion ΔfusHө.
H2O(s) ⟶ H2O(l) ΔfusHө = 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) Δmelting Hө = -6kJ mol-1
2. Vaporization: involves phase transfer of liquid to gas and enthalpy change is known as enthalpy of vaporization Δvap 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 Δvap Hө.
H2O(l) ⟶ H2O(g) ΔvapHө = 40.79kJ mol-1
3. Sublimation: involves phase transfer of solid to gas and enthalpy change is known as enthalpy of sublimation Δsub 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 Δsub Hө.
CO2(s) ⟶ CO2 (g) ΔsubHө = 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 Δvap Hө 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.
Standard enthalpy of combustion ΔcHө
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 ΔcHө. Let’s see how much energy we get by combustion of 1mole of glucose:
C6H12O6(g) + 6O2(g) ⟶ 6CO2(g) + 6H2O(l) ΔcHө = -2802 kJ mol-1
Standard enthalpy of formation ΔfHө
Formation of something means being formed by its constituent elements. Standard enthalpy of formation Δf Hө 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) Δf Hө = -285.8 kJ mol-1
Reaction for the formation of hydrogen bromide will be:
1/2H2(g) + 1/2Br2(l) ⟶ HBr(g) Δf Hө = -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 Δf Hө otherwise it will be just enthalpy of reaction Δr H.
H2(g) + Br2(l) ⟶ 2HBr(g) Δr Hө = -72.8 kJ mol-1
Here 2mole of HBr is formed. So the change in enthalpy you will get here is the Δr Hө not Δf Hө. If you divide the whole reaction by 2 then you will get the Δf Hө = Δr Hө /2.
By convention, standard enthalpy of formation of an element in its standard state is taken as zero. For example Δf Hө of H2(g), Δf Hө of Br2(l) and Δf Hө of O2(g) all will be zero.
How to calculate standard enthalpy of reaction ΔrHөwith the help of Δf Hө?
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 ΔrHө. 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)
Δr Hө = Ʃ ai Δf Hө (products) - Ʃ bi Δf Hө (reactants)
Here a and b are coefficients of the products and reactants for a balanced chemical equation.
Get the values of Δf Hө of compounds and put them in formula,
Δf Hө (CaO) = -635.1 kJ mol-1
Δf Hө (CO2) = -393.5 kJ mol-1
Δf Hө (CaCO3) = -1206.9 kJ mol-1
ΔrHө = [1(-635.1 kJ mol-1) + 1(-393.5 kJ mol-1)] -1(-1206.9 kJ mol-1)
ΔrHө = + 178.3 kJ mol-1
Positive value of ΔrHө 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)
Δf Hө (Fe) and Δf Hө (H2) is taken as 0 kJ mol-1by convention as they are elements in their standard state.
Δf Hө (H2O) = -285.83 kJ mol-1
Δf Hө (Fe2O3) = -824.2 kJ mol-1
Δr Hө = [2(Δf Hө Fe) + 3(Δf Hө H2O)] – [3(Δf Hө H2) + (Δf Hө Fe2O3)]
Δr Hө = [2(0) + 3(-285.83 kJ mol-1)] – [3(0) + (-824.2 kJ mol-1)]
Δr Hө = -857.49 kJ mol-1 - (-824.2 kJ mol-1)
Δr Hө = -33.29 kJ mol-1
Negative value of Δr Hөshows 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.
This work is licensed under the Creative Commons Attribution-Non Commercial-No Derivatives 4.0 International License. To view a copy of this license, visit http://creativecommons.org/licenses/by-nc-nd/4.0/.
No comments:
Post a Comment