General Chemistry/Enthalpy and Bond Strength/Hess’s Law

Enthalpy and Bond Strength

Contents

Hess’s Law

Hess's Law

Hess's law states that if a reaction is carried out in multiple steps, the reaction enthalpy is equal to the sum of the reaction enthalpies of each state.

A chemical reaction can occur in multiple steps and through different pathways. The exact path taken for a reaction or the order of the steps does not affect the reaction enthalpy. Enthalpy is a state function, which means it is independent of the path taken.

Hess's law, named for Swiss-Russian chemist Germain Hess (1802–50), states that in a multistep reaction, the reaction enthalpy is equal to the sum of the reaction enthalpies for each individual step. Hess's law means that enthalpies of reaction can be added to each other.

Consider the reaction for the combustion of methane:
CH4(g)+2O2(g)→CO2(g)+2H2O(l){\rm{CH}}_{4}(g)+2{\rm{O}}_{2}(g)\rightarrow{\rm{CO}}_{2}(g)+2{\rm{H}}_{2}{\rm{O}}(l)
At room temperature water (H2O) is a liquid, not a gas, so the enthalpy of water changing from a liquid to a gas needs to be factored in. It is possible to calculate the enthalpy of reaction for this reaction using two separate reactions used before:
CH4(g)+2O2(g)→CO2(g)+2H2O(g)ΔH=−802.3kJH2O(l)→H2O(g)ΔH=44kJ\begin{aligned}{\rm{CH}}_{4}(g)+2{\rm{O}}_{2}(g)&\rightarrow{\rm{CO}}_{2}(g)+2{\rm H}_{2}{\rm{O}}(g)&&\Delta H=-802.3\;{\rm{kJ}}\\{\rm{H}}_{2}{\rm{O}}(l)&\rightarrow{\rm{H}}_{2}{\rm{O}}(g)&&\Delta H=44\;{\rm{kJ}}\end{aligned}
The second reaction provides the energy needed to convert 1 mol of liquid water into 1 mol of water vapor. The reverse reaction has the same amount of energy with the opposite sign.
H2O(g)→H2O(l)ΔH=−44kJ{\rm{H}}_{2}{\rm{O}}(g)\rightarrow{\rm{H}}_{2}{\rm{O}}(l)\;\;\;\;\;\Delta H=-44\;{\rm{kJ}}
Furthermore, 2 mol water vapor must be converted to liquid water.
2H2O(g)→2H2O(l)ΔH=−88kJ2{\rm{H}}_{2}{\rm{O}}(g)\rightarrow2{\rm{H}}_{2}{\rm{O}}(l)\;\;\;\;\;\Delta H=-88\;{\rm{kJ}}
Now, add the reaction enthalpies of these two reactions.
CH4(g)+2O2(g)→CO2(g)+2H2O(g)ΔH=−802.3kJ2H2O(g)→2H2O(l)ΔH=−88kJCH4(g)+2O2(g)+2H2O(g)→CO2(g)+2H2O(g)+2H2O(l)ΔH=−802.3kJ+−88kJ=−890.3kJ\begin{aligned}&{\rm{CH}}_{4}(g)+2{\rm O}_{2}(g)\rightarrow{\rm{CO}}_{2}(g)+2{\rm H}_{2}{\rm O}(g)\;\;\;\;\;&&\Delta H=-802.3\;{\rm{kJ}}\\&2{\rm H}_{2}{\rm O}(g)\rightarrow2{\rm H}_{2}{\rm O}(l)\;\;\;\;\;&&\Delta H=-88\;{\rm{kJ}}\\&{\rm{CH}}_{4}(g)+2{\rm O}_{2}(g)+2{\rm H}_{2}{\rm O}(g)\rightarrow{\rm{CO}}_{2}(g)+2{\rm H}_{2}{\rm O}(g)+2{\rm H}_{2}{\rm O}(l)\;\;\;\;\;&&\Delta H=-802.3\;{\rm{kJ}}+-88\;{\rm{kJ}}=-890.3\;{\rm{kJ}}\end{aligned}
Because 2H2O(g) is located on both sides of the yields arrow, it can be removed from the equation.
CH4(g)+2O2(g)→CO2(g)+2H2O(l)ΔH=−890.3kJ{\rm{CH}}_{4}(g)+2{\rm O}_{2}(g)\rightarrow{\rm{CO}}_{2}(g)+2{\rm H}_{2}{\rm O}(l)\;\;\;\;\;\Delta H=-890.3\;{\rm{kJ}}
Hess's law allows scientists to use known reaction enthalpies to calculate the reaction enthalpy for an unknown reaction. This means scientists do not need to measure the reaction enthalpy of each reaction separately. As long as there is a reaction path with known reaction enthalpies, the enthalpy for the reaction can be calculated.

Hess's law also allows scientists to calculate enthalpies for reactions that do not normally occur.

Step-By-Step Example
Calculation of Enthalpy Change for the Formation of Carbon Monoxide

The reaction C(s)+12O2(g)→CO(g){\rm{C}}(s)+{\textstyle\frac12}{\rm{O}}_{2}(g)\rightarrow{\rm{CO}}(g) does not occur under normal conditions.

Carbon and carbon monoxide both undergo combustion with oxygen to form carbon dioxide. These reactions occur, and their reaction enthalpies are measured:
C(s)+O2(g)→CO2(g)ΔH=−393.5kJ(Reaction 1)CO(g)+12O2(g)→CO2(g)ΔH=−283.0kJ(Reaction 2)\begin{gathered}{\rm{C}}(s)+{\rm{O}}_{2}(g)\rightarrow{\rm{CO}}_{2}(g)\;\;\;\;\;\Delta H=-393.5\;{\rm{kJ}}\;\;\;\;\;\text{(Reaction 1)}\\\\{\rm{CO}}(g)+{\textstyle\frac12}{\rm{O}}_{2}(g)\rightarrow{\rm{CO}}_{2}(g)\;\;\;\;\;\Delta H=-283.0\;{\rm{kJ}}\;\;\;\;\;\text{(Reaction 2)}\end{gathered}
Use these reactions to calculate ΔH\Delta H:
C(s)+12O2(g)→CO(g)(Reaction 3){\rm{C}}(s)+{\frac12}{\rm{O}}_{2}(g)\rightarrow{\rm{CO}}(g)\;\;\;\;\;\text{(Reaction 3)}
Step 1
The goal is to construct reaction 3 using reactions 1 and 2. Check the products. Reaction 3 has CO(g) as a product. Reaction 2 has CO(g) as a reactant. The first step is to reverse reaction 2 so that carbon monoxide becomes a product, as in reaction 3. Note that this reverses the sign of ΔH\Delta H as well.
CO2(g)→CO(g)+12O2(g)ΔH=283.0kJ(Reaction 4){\rm{CO}}_{2}(g)\rightarrow{\rm{CO}}(g)+{\textstyle\frac12}{\rm {O}}_{2}(g)\;\;\;\;\;\Delta H=283.0\;{\rm{kJ}}\;\;\;\;\;\text{(Reaction 4)}
Step 2
Now add reaction 1 and reaction 4.
C(s)+O2(g)+CO2(g)→CO2(g)+CO(g)+12O2(g)ΔH=−393.5kJ+283.0kJ=−110.5kJ\begin{gathered}{\rm{C}}(s)+{\rm{O}}_{2}(g)+{\rm{CO}}_{2}(g)\rightarrow{\rm{CO}}_{2}(g)+{\rm{CO}}(g)+{\textstyle\frac12}{\rm{O}}_{2}(g)\\\\{\Delta H}=-393.5\;{\rm{kJ}}+283.0\;{\rm{kJ}}=-110.5\;{\rm{kJ}}\end{gathered}
Solution
The term CO2(g) appears on both sides of the reaction arrow, so it can be eliminated. There is 1 mol of O2(g) on the reactant side and half a mole of O2(g) on the product side. We can eliminate half a mole of O2(g) from both sides as well. The remainder is the net reaction.
C(s)+12O2(g)→CO(g)ΔH=−110.5kJ{\rm{C}}(s)+{\textstyle\frac12}{\rm{O}}_{2}(g)\rightarrow{\rm{CO}}(g)\;\;\;\;\;\Delta H=-110.5\;{\rm{kJ}}
<Enthalpies of Reaction>Enthalpies of Formation