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 H for a reaction:

1. 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.

2. Assign each rearranged equation the correct H 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.

3. Add the rearrranged equations together to give the overall equation for the reaction.

4. Add the H values for these equations to calculate H for the overall reaction.

Demonstration of Hess's Law

C_(s) + O_2(g) -----> CO_2(g)     H = -393.5 kJ

This overall reaction can also be produced as a two stage process:

1. Carbon combusts in limited oxygen producing carbon monoxide:
   C_(s) + ½O_2(g) -----> CO_(g)     H = -110.5 kJ

2. Carbon monoxide then combusts in additional oxygen:
   CO_(g) + ½O_2(g) -----> CO_2(g)     H = -283.0 kJ

C(s) + ½O2(g)	--->	CO(g)	H = -110.5 kJ
CO(g) + ½O2(g)	--->	CO2(g)	H = -283.0 kJ
C(s) + O2(g)	--->	CO2(g)	H = -393.5 kJ

A graphical representation of both reaction pathways is shown to the right.

H 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

Given that:

½N_2(g) + 1½H_2(g) -----> NH_3(g)     H = -46.1 kJ/mol

½H_2(g) + ½Cl_2(g) ------> HCl_(g)     H = -92.3 kJ/mol

½N_2(g) + 2H_2(g) + ½Cl_2(g) -----> NH₄Cl_(s)     H = -314.4 kJ/mol

1. Rearrange each balanced equation with reactants and products on the correct side, ie, NH_3(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)

2. Assign correct H values to each equation.

   NH3(g)	----->	½N2(g) + 1½H2(g)	H = +46.1 kJ/mol
   HCl(g)	----->	½H2(g) + ½Cl2(g)	H = +92.3 kJ/mol
   ½N2(g) + 2H2(g) + ½Cl2(g)	----->	NH4Cl(s)	H = -314.4 kJ/mol

3. Add the rearranged equations together.

   NH3(g)	----->	½N2(g) + 1½H2(g)	H = +46.1 kJ/mol
   HCl(g)	----->	½H2(g) + ½Cl2(g)	H = +92.3 kJ/mol
   ½N2(g) + 2H2(g) + ½Cl2(g)	----->	NH4Cl(s)	H = -314.4 kJ/mol
   NH3(g) + HCl(g)	----->	NH4Cl(s)

4. Add the H values.

   NH3(g)	----->	½N2(g) + 1½H2(g)	H = +46.1 kJ/mol
   HCl(g)	----->	½H2(g) + ½Cl2(g)	H = +92.3 kJ/mol
   ½N2(g) + 2H2(g) + ½Cl2(g)	----->	NH4Cl(s)	H = -314.4 kJ/mol
   NH3(g) + HCl(g)	----->	NH4Cl(s)	H = -176.0 kJ/mol

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