Please do not block ads on this website.
No ads = no money for us = no free stuff for you!
Standard Enthalpy of Formation or Standard Heat of Formation
The standard heat of formation (standard enthalpy of formation) of a compound is defined as the enthalpy change for the reaction in which elements in their standard states produce products.
At 25°C and 1 atm (101.3 kPa), the standard state of any element is solid with the following exceptions:
| monatomic gases |
(Group 18 or Noble Gases)
| bromine, Br2(l)|
| helium, Heg|
| hydrogen, H2(g)|
Consider a water molecule, H2O.
A molecule of water contains the elements hydrogen (H) and oxygen (O).
The standard heat of formation of liquid water would be defined as the enthalpy change when the elements hydrogen and oxygen in their standard states to produce liquid water.
If the conditions set are 25°C and 101.3 kPa (1 atm), then the standard states for reactants and products are:
- hydrogen exists as a diatomic molecule and is a gas, H2(g)
- oxygen exists as a diatomic molecule and is a gas, O2(g)
- water exists as a molecule and is a liquid, H2O(l)
The chemical reaction for the standard heat of formation per mole of liquid water (standard enthalpy of formation of liquid water) is:
H2(g) + ½O2(g) → H2O(l)
If you looked up the standard enthalpy of formation of liquid water in tables (at 25°C and 1 atm), the value would be given as:
ΔHfo = -285.8 kJ mol-1
This means that when molecular hydrogen gas reacts with molecular oxygen gas, 285.8 kJ of energy will be released for every mole of liquid water that is produced.
If 10 moles of liquid water was produced from molecular hydrogen gas and molecular oxygen gas, then 10 × 285.8 = 2858 kJ of energy would be released.
Similarly, if 0.1 moles of liquid water was produced from molecular hydrogen gas and molecular oxygen gas, then 0.1 × 285.8 = 28.58 kJ of energy would be released.
The values for the standard enthalpy of formation for a number of different compounds at 25°C is given in the table below:
Note that the standard heat of formation (enthalpy of formation) of some compounds is positive and for others it is negative:
- ΔHfo = + (a positive value)
Overall, energy is absorbed when the elements in their standard states form the compound.
Formation of the compound is an endothermic reaction.
- ΔHfo = - (a negative value)
Overall, energy is released when the elements in their standard states form the compound.
Formation of the compound is an exothermic reaction.
Calculating Heat of Reaction Using Standard Heat of Formation Data
Consider the reaction in which gaseous hydrogen chloride (HCl(g)) reacts with gaseous ammonia (NH3(g)) to produce solid ammonium chloride (NH4Cl(s)) at 25°C.
NH3(g) + HCl(g) → NH4Cl(s)
From the table of values for Standard Enthalpy of Formation at 25°C given in the previous section, we find that the standard enthalpy of formation of ammonium chloride (NH4Cl(s)) is -314.4 kJ mol-1.
That is, the formation of 1 mole of solid ammonium chloride (NH4Cl(s)) from the elements nitrogen, hydrogen and chlorine in their standard states releases 314.4 kJ of energy.
Therefore we can write a chemical equation to represent this reaction as shown below:
½N2(g) + 2H2(g) + ½Cl2(g) → NH4Cl(s) ΔHfo = -314.4 kJ mol-1
But where will these elements come from in order to react?
They can come from the reactant molecules breaking apart.
Hydrogen chloride molecules could break apart to provide molecules of hydrogen and chlorine gas according to the following chemical equation:
HCl(g) → ½H2(g) + ½Cl2(g)
But how much energy will be absorbed or released to do this?
From the table of values for Standard Enthalpy of Formation at 25°C we find that the enthalpy of formation of HCl(g) is -92.3 kJ mol-1.
We can represent this formation reaction as:
½H2(g) + ½Cl2(g) → HCl(g) ΔHfo = -92.3 kJ mol-1
This reaction is just the reverse of the one we want!
So we can reverse the reaction, AND, reverse the sign of the enthalpy change as well!
HCl(g) → ½H2(g) + ½Cl2(g) ΔHo = +92.3 kJ mol-1
Similarly, ammonia gas (NH3(g)) could break apart to provide the nitrogen and hydrogen we need for the overall reaction to occur:
NH3(g) → ½N2(g) + 3/2H2(g)
This reaction is the reverse of the heat of formation (enthalpy of formation reaction) shown below:
½N2(g) + 3/2H2(g) → NH3(g)
From the table of values for Standard Enthalpy of Formation at 25°C we find that the enthalpy of formation of NH3(g) is -46.1 kJ mol-1.
½N2(g) + 3/2H2(g) → NH3(g) ΔHfo = -46.1 kJ mol-1
So, the reaction to break apart ammonia molecules into hydrogen gas and ammonia gas is the reverse of this equation, AND, we must remember to reverse the sign of the enthalpy change as well!
NH3(g) → ½N2(g) + 3/2H2(g) ΔHo = +46.1 kJ mol-1
The enthalpy change required to produce the elements hydrogen, nitrogen and chlorine in their standard states is the sum of the enthalpy change for breaking apart hydrogen chloride molecules and for breaking apart ammonia molecules:
|HCl(g)||→||½H2(g) + ½Cl2(g)||ΔHo = +92.3 kJ mol-1|
|NH3(g)||→||½N2(g) + 3/2H2(g)||ΔHo = +46.1 kJ mol-1|
|HCl(g) + NH3(g)||→||2H2(g) + ½Cl2(g) + ½N2(g)|| ΔHo = 92.3 + 46.1 |
= +138.4 kJ mol-1
Now we can use this H2(g), Cl2(g) and N2(g) to produce NH4Cl(s).
Recall from the beginning of this section that this reaction, the formation of NH4Cl(s) from its elements in their standard states, releases 314.4 kJ mol-1 of energy.
So now we can add together two chemical equations and their associated enthalpy terms; 1 equation for the breaking apart of reactant molecules into elements, and, 1 equation for the elements coming together to form product molecules.
This is shown below:
|HCl(g) + NH3(g)||→||+ +||ΔHo = +138.4 kJ mol-1|
|+ +||→||NH4Cl(s)||ΔHo = -314.4 kJ mol-1|
|HCl(g) + NH3(g)||→||NH4Cl(s)|| ΔHo = +138.4 + -314.4|
= -176.0 kJ mol-1
Look at what we have done here:
(a) We found the value of the standard heat of formation of the product
(b) We have added together the standard heat of formation of each reactant molecule, AND, reversed the sign:
ΔHo(reactants) = -ΣΔHfo(reactants)
(c) We added the two values together to find the enthalpy change for the reaction, ΔHo(reaction):
ΔHo(reaction) = ΔHfo(product) + -ΣΔHfo(reactants)
This is a very quick way to use standard heat of formation data to calculate the enthalpy change of a chemical reaction:
ΔHo(reaction) = ΣΔHfo(products) -ΣΔHfo(reactants)
Using the formation of solid ammonium chloride (NH4Cl(s)) from the reactants hydrogen chloride gas (HCl(g)) and ammonia gas (NH3(g)) as an example of the application of this equation:
|= -(-92.3 + -46.1)||+||-314.4|
- What is Stable on the Strand
- What is a full blown tribal war
- How safe is Ferndale Michigan
- Why are Grammy Awards given
- Is CN Tower in Toronto worth visiting
- Can a Canadian hold dual citizenship
- What is interstitial fluid
- What are the four dominant CMOS technology
- What are the most secure Linux distributions
- Do you think about after life 1
- Do you have Balenciaga Speed Trainer Shoes
- What is economics honors all about
- Which education is better Indian or American
- Is the Labour party finished
- Who is the owner of Aaj Tak
- How can you develop a positive attitude