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R1.2.4 Hess's Law with combustion and formation data (Higher Level Only)

Using Hess’s Law to Calculate Enthalpy Changes with Combustion and Formation Data

Applying Hess’s Law with Combustion Data

  • Hess’s law can be used to calculate the enthalpy change of a reaction (ΔHr∘) from combustion data using the formula:

ΔHr∘=∑ΔHc∘(reactants)−∑ΔHc∘(products)

  • This equation works because the energy released during combustion represents the total energy change of breaking and forming bonds.
Example question

Calculating Enthalpy Change from Combustion Data

Let’s calculate the enthalpy change for the reaction:

C2H4(g)+H2(g)→C2H6(g)

Using the following combustion data:

  • ΔHc∘(C2H4)=−1411kJ mol−1
  • ΔHc∘(C2H6)=−1560kJ mol−1
  • ΔHc∘(H2)=−286kJ mol−1

Solution

  1. Write the combustion equation for each species:
    • C2H4(g)+3O2(g)→2CO2(g)+2H2O(l)
    • C2H6(g)+7/2O2(g)→2CO2(g)+3H2O(l)
    • H2(g)+1/2O2(g)→H2O(l)
  2. Apply the formula:
    ΔHr∘=ΔHc∘(C2H4)+ΔHc∘(H2)−ΔHc∘(C2H6)
  3. Substitute the values:
    ΔHr∘=(−1411)+(−286)−(−1560)
  4. Simplify:
    ΔHr∘=−1411−286+1560=−137kJ mol−1

Thus, the enthalpy change for the reaction is ΔHr∘=−137kJ mol−1.

Common Mistake

Be careful with the signs! Subtracting a negative value is equivalent to adding it.

Applying Hess’s Law with Formation Data

  • Hess’s law can also be applied to formation data using the formula:

ΔHr∘=∑ΔHf∘(products)−∑ΔHf∘(reactants)

  • This approach is particularly useful when combustion data is unavailable.
Example question

Calculating Enthalpy Change from Formation Data

Calculate the enthalpy change for the reaction:

N2(g)+3H2(g)→2NH3(g)

Using the following formation data:

  • ΔHf∘(NH3)=−46.1kJ mol−1
  • ΔHf∘(N2)=0kJ mol−1
  • ΔHf∘(H2)=0kJ mol−1

Solution

  1. Write the formation equation for each species:
    • N2(g)+3H2(g)→2NH3(g)
  2. Apply the formula:
    ΔHr∘=∑ΔHf∘(products)−∑ΔHf∘(reactants)
  3. Substitute the values:
    ΔHr∘=[2(−46.1)]−[1(0)+3(0)]
  4. Simplify:
    ΔHr∘=−92.2−0=−92.2kJ mol−1

Thus, the enthalpy change for the reaction is ΔHr∘=−92.2kJ mol−1.

Tip

When using formation data, always ensure your chemical equation is balanced. The coefficients in the balanced equation determine how many times you multiply each ΔHf∘ value.

Calculating Enthalpy Change Using ΔHf Values

The overall enthalpy change (ΔH∘) for a reaction can be calculated as:

ΔH∘=∑ΔHf∘(products)−∑ΔHf∘(reactants)

Steps:

  1. Write the balanced chemical equation.
  2. Identify the ΔHf values for all reactants and products from the data booklet.
  3. Substitute the values into the formula.
Example question

Combustion of Methane

Calculate the enthalpy change for the combustion of methane:

CH4(g)+2O2(g)→CO2(g)+2H2O(l)

From the data booklet:

  • ΔHf∘ of CH₄(g) = -74.8 kJ mol⁻¹
  • ΔHf∘f of CO₂(g) = -393.5 kJ mol⁻¹
  • ΔHf∘f of H₂O(l) = -286 kJ mol⁻¹

Solution

  • ΔHf∘ of O₂(g) = 0 (element in standard state)
  • Using the formula:ΔH∘=[(−393.5)+2(−286)]−[(−74.8)+2(0)]ΔH∘=−965.5kJ mol−1

The reaction releases 965.5 kJ of energy per mole of methane.

Common Mistake

Students often forget that ΔHf∘ for elements in their standard states is zero. Always verify standard states using the data booklet!

Calculating Enthalpy Change Using ΔHc Values

The overall enthalpy change (ΔH∘) for a reaction can also be calculated using ΔHc values:

ΔH∘=∑ΔHc∘(reactants)−∑ΔHc∘(products)

Steps:

  1. Write the balanced chemical equation.
  2. Identify the ΔHc values for all reactants and products from the data booklet.
  3. Substitute the values into the formula.
Example question

Formation of Ethanol

Calculate the enthalpy change for the formation of ethanol (C₂H₅OH) using combustion data:

2C(s)+3H2(g)+12O2(g)→C2H5OH(l)

From the data booklet:

  • ΔHc∘ of C(s) = -393.5 kJ mol⁻¹
  • ΔHc∘ of H₂(g) = -286 kJ mol⁻¹
  • ΔHc∘ of C₂H₅OH(l) = -1367 kJ mol⁻¹

Solution

  • Using the formula:

ΔH∘=[2(−393.5)+3(−286)]−[(−1367)]

ΔH∘=−1367kJ mol−1

The reaction releases 1367 kJ of energy per mole of ethanol formed.

Practice Problems

Self review

  1. Can you clearly differentiate between ΔHc∘ and ΔHf∘? How do their definitions guide their use in Hess’s law calculations?
  2. Use the following combustion data to calculate the enthalpy change for the reaction:
    C2H6(g)+Cl2(g)→C2H5Cl(g)+HCl(g)
    • ΔHc∘(C2H6)=−1560kJ mol−1
    • ΔHc∘(C2H5Cl)=−1550kJ mol−1
    • ΔHc∘(HCl)=−92kJ mol−1
  3. Calculate the enthalpy change for the reaction:
    CO(g)+H2(g)→CH3OH(l)
    Using the following formation data:
    • ΔHf∘(CO)=−110.5kJ mol−1
    • ΔHf∘(H2)=0kJ mol−1
    • ΔHf∘(CH3OH)=−238.7,kJ mol−1
  4. Explain why enthalpy changes calculated using bond enthalpy data often differ from those calculated using formation or combustion data.

Theory of Knowledge

How does the concept of energy conservation in Hess’s law reflect broader natural laws, such as the conservation of mass and energy in physics?

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Questions

Recap questions

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Question 1

Consider the reaction:

C2H4(g)+H2(g)→C2H6(g)\text{C}_2\text{H}_4(g) + H_2(g) \to \text{C}_2\text{H}_6(g)

Using the following combustion data:

  • ΔHc∘(C2H4)=−1411 kJ mol−1\Delta H_c^\circ (\text{C}_2\text{H}_4) = -1411 \, \text{kJ mol}^{-1}
  • ΔHc∘(C2H6)=−1560 kJ mol−1\Delta H_c^\circ (\text{C}_2\text{H}_6) = -1560 \, \text{kJ mol}^{-1}
  • ΔHc∘(H2)=−286 kJ mol−1\Delta H_c^\circ (\text{H}_2) = -286 \, \text{kJ mol}^{-1}

What is the enthalpy change for the reaction?

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What is the standard enthalpy of combustion (ΔHc∘\Delta H_c^\circ)?

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Note

Hess's Law

  • States that the total enthalpy change for a reaction is the same, no matter how many steps the reaction is carried out in.
  • This is because enthalpy is a state function, meaning it depends only on the initial and final states, not the path taken.
  • Allows us to calculate enthalpy changes for reactions that are difficult to measure directly by using alternative pathways.
Definition

State Function

A property that depends only on the current state of a system, not on how that state was reached.

Analogy

Think of Hess's Law like taking a trip up a mountain. Whether you take a direct path or a winding trail, the total change in altitude is the same.

Example

Calculating the enthalpy change for converting graphite to diamond using Hess's Law:

C(graphite)+O2(g)→CO2(g)\text{C(graphite)} + \text{O}_2(g) \rightarrow \text{CO}_2(g)

C(diamond)+O2(g)→CO2(g)\text{C(diamond)} + \text{O}_2(g) \rightarrow \text{CO}_2(g)

C(graphite)→C(diamond)\text{C(graphite)} \rightarrow \text{C(diamond)}

Note

The enthalpy change for the direct conversion of graphite to diamond is difficult to measure, but we can calculate it using the combustion reactions of both forms of carbon.