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 ($ \Delta H_r^\circ $) from combustion data using the formula: $$
\Delta H_r^\circ = \sum \Delta H_c^\circ (\text{reactants}) - \sum \Delta H_c^\circ (\text{products})
$$ - 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:
$$
\text{C}_2\text{H}_4(g) + H_2(g) \to \text{C}_2\text{H}_6(g)
$$
Using the following combustion data:
- $ \Delta H_c^\circ (\text{C}_2\text{H}_4) = -1411 \, \text{kJ mol}^{-1} $
- $ \Delta H_c^\circ (\text{C}_2\text{H}_6) = -1560 \, \text{kJ mol}^{-1} $
- $ \Delta H_c^\circ (\text{H}_2) = -286 \, \text{kJ mol}^{-1} $
Solution
- Write the combustion equation for each species:
- $ \text{C}_2\text{H}_4(g) + 3\text{O}_2(g) \to 2\text{CO}_2(g) + 2\text{H}_2\text{O}(l) $
- $ \text{C}_2\text{H}_6(g) + \frac{7}{2}\text{O}_2(g) \to 2\text{CO}_2(g) + 3\text{H}_2\text{O}(l) $
- $ \text{H}_2(g) + \frac{1}{2}\text{O}_2(g) \to \text{H}_2\text{O}(l) $
- Apply the formula:
$$
\Delta H_r^\circ = \Delta H_c^\circ (\text{C}_2\text{H}_4) + \Delta H_c^\circ (\text{H}_2) - \Delta H_c^\circ (\text{C}_2\text{H}_6)
$$ - Substitute the values:
$$
\Delta H_r^\circ = (-1411) + (-286) - (-1560)
$$ - Simplify:
$$
\Delta H_r^\circ = -1411 - 286 + 1560 = -137 \, \text{kJ mol}^{-1}
$$ - Thus, the enthalpy change for the reaction is $ \Delta H_r^\circ = -137 \, \text{kJ mol}^{-1} $.
- 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: $$
\Delta H_r^\circ = \sum \Delta H_f^\circ (\text{products}) - \sum \Delta H_f^\circ (\text{reactants})
$$ - 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:
$$
\text{N}_2(g) + 3\text{H}_2(g) \to 2\text{NH}_3(g)
$$
Using the following formation data:
- $ \Delta H_f^\circ (\text{NH}_3) = -46.1 \, \text{kJ mol}^{-1} $
- $ \Delta H_f^\circ (\text{N}_2) = 0 \, \text{kJ mol}^{-1} $
- $ \Delta H_f^\circ (\text{H}_2) = 0 \, \text{kJ mol}^{-1} $
Solution
- Write the formation equation for each species: $$ \text{N}_2(g) + 3\text{H}_2(g) \to 2\text{NH}_3(g) $$
- Apply the formula:
$$
\Delta H_r^\circ = \sum \Delta H_f^\circ (\text{products}) - \sum \Delta H_f^\circ (\text{reactants})
$$ - Substitute the values:
$$
\Delta H_r^\circ = [2(-46.1)] - [1(0) + 3(0)]
$$ - Simplify:
$$
\Delta H_r^\circ = -92.2 - 0 = -92.2 \, \text{kJ mol}^{-1}
$$ - Thus, the enthalpy change for the reaction is $ \Delta H_r^\circ = -92.2 \, \text{kJ mol}^{-1} $.
When using formation data, always ensure your chemical equation is balanced. The coefficients in the balanced equation determine how many times you multiply each $ \Delta H_f^\circ $ value.
Calculating Enthalpy Change Using $ΔH_f$ Values
The overall enthalpy change ($ΔH^\circ$) for a reaction can be calculated as:
$$\Delta H^\circ = \sum \Delta H_f^\circ (\text{products}) - \sum \Delta H_f^\circ (\text{reactants})$$
