Hess's Law: Additivity of Heats of Reaction |
Key Concepts
- For any chemical reaction, the change in enthalpy in going from reactants to products is constant, regardless of the set of reaction steps used to bring the overall reaction about.
That is, the enthalpy change for a particular chemical reaction is independent of the path followed to get from reactants to products.
- To calculate the overall enthalpy change for a chemical reaction, the enthalpy changes for each individual reaction during the process are added together.
- Steps involved in calculating ΔH for a reaction:
- Rearrange each equation so that it is balanced, with the reactants and products on the correct side.
Check that you have the correct states, enthalpy changes will be different for species in the solid, liquid and gas states.
- Assign each rearranged equation the correct ΔH value.
Remember, if you need to multiply each species in the chemical equation by 2, then the enthalpy change must also be multiplied by 2.
If you reverse a chemical equation, then the sign of the enthalpy term also changes.
- Add the rearrranged equations together to give the overall equation for the reaction.
- Add the ΔH values for these equations to calculate ΔH for the overall reaction.
Demonstration of Hess's Law
When carbon combusts in an excess of oxygen, carbon dioxide is formed and 393.5 kJ of heat is released per mole of carbon.
C(s) + O2(g) → CO2(g) ΔH = -393.5 kJ
This overall reaction can also be produced as a two stage process:
- Carbon combusts in limited oxygen producing carbon monoxide:
C(s) + ½O2(g) → CO(g) ΔH = -110.5 kJ
- Carbon monoxide then combusts in additional oxygen:
CO(g) + ½O2(g) → CO2(g) ΔH = -283.0 kJ
These two equations can be added together to calculate ΔH for the overall reaction:
| C(s) + ½O2(g) |
→ |
CO(g) |
ΔH = -110.5 kJ |
CO(g) + ½O2(g) |
→ |
CO2(g) |
ΔH = -283.0 kJ |
|
| C(s) + O2(g) |
→ |
CO2(g) |
ΔH = -393.5 kJ |
A graphical representation of both reaction pathways is shown to the right.
ΔH for the reaction producing carbon dioxide gas from solid carbon and oxygen gas is the same irrespective of whether the reaction is the result of a one step process from the combustion of carbon in excess oxygen, or, as a two step process of the combustion of carbon in limited oxygen followed by the combustion of the resulting carbon monoxide.
Example
Calculate ΔH for the reaction:
NH3(g) + HCl(g) → NH4Cl(s)
Given that:
| ½N2(g) + 1½H2(g) → NH3(g) | ΔH = -46.1 kJ/mol |
| ½H2(g) + ½Cl2(g) → HCl(g) | ΔH = -92.3 kJ/mol |
| ½N2(g) + 2H2(g) + ½Cl2(g) → NH4Cl(s) | ΔH = -314.4 kJ/mol |
- Rearrange each balanced equation with reactants and products on the correct side, ie, NH3(g) and HCl(g) on the left, NH4Cl(s) on the right hand side.
| NH3(g)
| → |
½N2(g) + 1½H2(g) |
|
| HCl(g) |
→ |
½H2(g) + ½Cl2(g) |
|
| ½N2(g) + 2H2(g) + ½Cl2(g) |
→ |
NH4Cl(s) |
|
- Assign correct
 H values to each equation.
| NH3(g) |
→ |
½N2(g) + 1½H2(g) |
ΔH = +46.1 kJ/mol |
| HCl(g) |
→ |
½H2(g) + ½Cl2(g) |
ΔH = +92.3 kJ/mol |
| ½N2(g) + 2H2(g) + ½Cl2(g) |
→ |
NH4Cl(s) |
ΔH = -314.4 kJ/mol |
- Add the rearranged equations together.
| NH3(g) |
→ |
½N2(g) + 1½H2(g) |
ΔH = +46.1 kJ/mol |
| HCl(g) |
→ |
½H2(g) + ½Cl2(g) |
ΔH = +92.3 kJ/mol |
½N2(g) + 2H2(g) + ½Cl2(g) |
→ |
NH4Cl(s) |
ΔH = -314.4 kJ/mol |
|
| NH3(g) + HCl(g) |
→ |
NH4Cl(s) |
|
|
- Add the ΔH values.
| NH3(g) |
→ |
½N2(g) + 1½H2(g) |
ΔH = +46.1 kJ/mol |
| HCl(g) |
→ |
½H2(g) + ½Cl2(g) |
ΔH = +92.3 kJ/mol |
½N2(g) + 2H2(g) + ½Cl2(g) |
→ |
NH4Cl(s) |
ΔH = -314.4 kJ/mol |
|
| NH3(g) + HCl(g) |
→ |
NH4Cl(s) |
ΔH = -176.0 kJ/mol |
|
|
| Practice Questions |
|
|---|
For AUS-e-TUTE members:
- Click on the Hess' Law drill link:
Hess' Law drill
- Enter your username and password if prompted.
- Click the "New Question" button to begin the drill.
- Worked solutions are provided if you need some help!
Not an AUS-e-TUTE Member?
|
|
|
|
|
to
Bookmark this site!