Hess’s Law (Cambridge International A-Level Chemistry)

Hess Cycles

• Hess’s Law states that:

"The total enthalpy change in a chemical reaction is independent of the route by which the chemical reaction takes place as long as the initial and final conditions are the same."

• This means that whether the reaction takes place in one or two steps, the total enthalpy change of the reaction will still be the same

Illustration of Hess's Law

According to Hess’ Law, the enthalpy change of the direct route, going from reactants (A+B) to product (C), is equal to the enthalpy change of the indirect routes

• Hess’ Law is used to calculate enthalpy changes which can’t be found experimentally using calorimetry, e.g.:

3C (s) + 4H₂ (g) → C₃H₈(g)

• ΔH_f (propane) can’t be found experimentally as hydrogen and carbon don’t react under standard conditions

Calculating ΔH_r from ΔH_f using Hess’s Law energy cycles

• The products can be directly formed from the elements = ΔH₂ OR The products can be indirectly formed from the elements = ΔH₁ + ΔH_r

Applying Hess's Law

The enthalpy change from elements to products (direct route) is equal to the enthalpy change of elements forming reactants and then products (indirect route)

• Equation

ΔH ₂ = ΔH₁ + ΔH_r

Therefore,

ΔH _r = ΔH₂– ΔH ₁

Worked Example Calculate ΔH_r* *for the following reaction:

2NaHCO₃ (s) → Na₂CO₃ (s) + CO₂ (g) + H₂O (l)

The table shows the standard enthalpy of formations, ΔH^θ_f, relevant to this reaction.

Molecule	ΔH θf kJ mol-1 **
NaHCO3 (s)	–950.8
Na2CO3 (s)	–1130.7
CO2 (g)	–393.5
H2O (l)	–285.8

Answer

-** Step 1: **Write the balanced equation at the top

-** Step 2: **Draw the cycle with the elements at the bottom

-** Step 3: **Draw in all arrows, making sure they go in the correct directions. Write the standard enthalpy of formations

-** Step 4: Apply Hess’s Law:

• ΔH_r= ΔH₂– ΔH₁
• ΔH_r= ( ΔH_f* [Na₂CO₃ (s)] + ΔH*_f* [CO₂ (g)]+* ΔH_f* [H₂O (l)] ) – 2ΔH*_f* *[NaHCO₃ (s)]
• ΔH_r* *= ( (–1130.7) + (–393.5) + (–285.8) ) – ( 2 x –950.8)
• ΔH_r* *= +91.6 kJ mol^-1

Calculating ΔH_f from ΔH_c using Hess’s Law energy cycles

• The combustion products can be formed directly from elements to combustion products =ΔH₁OR The combustion products can be formed indirectly from elements to compounds to combustion products = ΔH_f + ΔH₂

Hess's Law and combustion enthalpies

The enthalpy change going from elements to products (direct route) is equal to the enthalpy change of elements forming reactants and then products (indirect route)

• Equation

ΔH ₁ = ΔH_f + ΔH₂

• Therefore,

ΔH _f = ΔH₁ – ΔH₂

Worked Example Calculate, using the standard enthalpy change of combustion values given, the enthalpy of formation of ethane.

Reaction	ΔH c kJ mol-1 **
C (graphite) + O2** (g) → CO 2 (g)**	–393.5
H2 (g) + ½O2 (g) → H2O (l)	–285.8
C2H6 (g) + 3½O2 (g) → 2CO2 (g) + 3H2O (l)	–1559.7

Answer

-** Step 1: **Write the equation for the enthalpy change of formation at the top and add oxygen on both sides

-** Step 2: **Draw the cycle with the combustion products at the bottom

-** Step 3: **Draw all arrows in the correct direction

-** Step 4: Apply Hess’s Law:

• ΔH_f= ΔH₁– ΔH₂
• ΔH_f* *= ( (2 x –393.5) + (3 x –285.8) ) – ( –1559.7)
• ΔH_f* *= –84.7 kJ mol^-1

Calculating average bond energies using Hess's cycles

• Bond energies cannot be found directly so enthalpy cycles are used to find the average bond energy

• This can be done using enthalpy changes of atomisation and combustion or formation

• The enthalpy change of atomisation(ΔH^θ_at ) is the enthalpy change when one mole of gaseous atoms is formed from its elements under standard conditions.

  • E.g.* ΔH*^θ_at [H₂] relates to the equation:

½H₂ (g) → H (g)

Worked Example Calculate the average bond enthalpy of the C–H bond, using the relevant ΔH^θ_f and ΔH^θ_at values shown in the table.

	Energy kJ mol-1
ΔHθf [CH4 (g)]	–74.8
ΔHθat [½H2 (g)]	+218
ΔHθat [C (graphite)]	+717.7

Answer

• Step 1: Write down the equation for the dissociation of methane at the top

• Step 2: Write down the elements at the bottom

• Step 3: Draw all arrows in the correct direction

• Step 4: Apply Hess’s Law:

  • ΔH = ΔH^θ_at– ΔH_f
  • ΔH = ( (+717.7) + (4 x 218) ) – ( –74.8)
  • ΔH_f* *= +1664.5 kJ mol^-1

• Step 5: Calculate the bond energy for one C–H bond

  • There there are 4 C-H bonds in methane
  • Average bond enthalpy (C–H) = $\frac{+1664.5}{4}$
  • Average bond enthalpy (C–H) = +416 kJ mol^-1 (to 3 significant figures)

Exam Hint

• Remember to take into account the number of moles of each reactant and product.
• For example, there are two moles of NaHCO₃ (s) so the ΔH_f value is multiplied by 2.