Some enthalpy changes cannot be measured directly by experiment. For example, we cannot directly measure the standard enthalpy of formation of carbon monoxide because carbon always reacts with oxygen to form a mixture of carbon monoxide and carbon dioxide. In such cases, we apply Hess's Law to calculate the enthalpy change using alternative, measurable routes.
🔑 Key Principle
Hess's Law is a direct statement of the Law of Conservation of Energy. Since energy cannot be created or destroyed, the total energy change when converting reactants to products must be identical regardless of whether the reaction occurs in a single step or a series of intermediate steps.
The enthalpy change for a chemical reaction is independent of the route taken, provided the initial and final states of the reactants and products are the same.
Constructing Enthalpy Cycles
An enthalpy cycle (or Hess cycle) is a diagrammatic representation of the direct and indirect chemical routes between reactants and products. To solve calculations, we establish two paths:
- Route 1: The direct route from reactants to products (\( \Delta H_{\text{reaction}} \)).
- Route 2: The indirect route via a common set of intermediate substances.
According to Hess's Law:
\[ \text{Route 1} = \text{Route 2} \]
Method 1: Using Enthalpies of Formation (\( \Delta_f H^\theta \))
When constructing a Hess cycle using enthalpies of formation, the common intermediates are the constituent elements in their standard states. Because formation definitions start from elements, the arrows in the cycle must point upwards towards both the reactants and products.
To find the direct route (\( \Delta H_{\text{reaction}} \)), we follow the path of the cycle. Moving against the left arrow (against formation of reactants) means we reverse the sign of the reactants' formation. Moving with the right arrow (with formation of products) means we keep the sign. This gives the standard formula:
- \( \Delta_f H^\theta[\text{Fe}_2\text{O}_3\text{(s)}] = -824\text{ kJ mol}^{-1} \)
- \( \Delta_f H^\theta[\text{Al}_2\text{O}_3\text{(s)}] = -1676\text{ kJ mol}^{-1} \)
Solution:
1. Identify the standard formation values of all species in the reaction. Recall that the standard enthalpy of formation of pure elements in their standard states is exactly zero:
- \( \Delta_f H^\theta[\text{Al(s)}] = 0\text{ kJ mol}^{-1} \)
- \( \Delta_f H^\theta[\text{Fe(s)}] = 0\text{ kJ mol}^{-1} \)
2. Set up the Hess equation:
\[ \Delta H^\theta_{\text{reaction}} = \sum \Delta_f H^\theta\text{(products)} - \sum \Delta_f H^\theta\text{(reactants)} \]
\[ \Delta H^\theta_{\text{reaction}} = [2 \times \Delta_f H^\theta(\text{Fe}) + \Delta_f H^\theta(\text{Al}_2\text{O}_3)] - [2 \times \Delta_f H^\theta(\text{Al}) + \Delta_f H^\theta(\text{Fe}_2\text{O}_3)] \]
3. Substitute the values into the equation:
\[ \Delta H^\theta_{\text{reaction}} = [(2 \times 0) + (-1676)] - [(2 \times 0) + (-824)] \]
\[ \Delta H^\theta_{\text{reaction}} = [-1676] - [-824] = -1676 + 824 \]
\[ \Delta H^\theta_{\text{reaction}} = -852\text{ kJ mol}^{-1} \]
Method 2: Using Enthalpies of Combustion (\( \Delta_c H^\theta \))
When using enthalpies of combustion, the common intermediates are the products of complete combustion (usually \( \text{CO}_2 \) and \( \text{H}_2\text{O} \)). Because combustion definitions represent burning a substance, the arrows in this cycle must point downwards away from both reactants and products towards the combustion products.
To find the direct route (\( \Delta H_{\text{reaction}} \)), we go with the left arrow (reactants to combustion products) and against the right arrow (combustion products to products). This gives the standard formula:
Do not just memorise "products minus reactants" or "reactants minus products". It is very easy to make mistakes if the question asks you to solve for a specific reactant formation value rather than the overall reaction change. Always draw out the triangular cycle and follow the arrows to ensure your signs are correct!
- \( \Delta_c H^\theta[\text{C(s, graphite)}] = -394\text{ kJ mol}^{-1} \)
- \( \Delta_c H^\theta[\text{H}_2\text{(g)}] = -286\text{ kJ mol}^{-1} \)
- \( \Delta_c H^\theta[\text{C}_2\text{H}_5\text{OH(l)}] = -1367\text{ kJ mol}^{-1} \)
Solution:
1. Write out the chemical equation representing the target change (standard enthalpy of formation of ethanol):
\[ 2\text{C(s)} + 3\text{H}_2\text{(g)} + \frac{1}{2}\text{O}_2\text{(g)} \rightarrow \text{C}_2\text{H}_5\text{OH(l)} \]
2. Draw a combustion cycle with the reactants and products at the top and combustion products (\( 2\text{CO}_2\text{(g)} + 3\text{H}_2\text{O(l)} \)) at the bottom. Note that oxygen gas does not combust, so it has a combustion enthalpy of zero.
3. Set up the Hess equation using the combustion data formula:
\[ \Delta H^\theta_{\text{formation}} = \sum \Delta_c H^\theta\text{(reactants)} - \sum \Delta_c H^\theta\text{(products)} \]
\[ \Delta H^\theta_{\text{formation}} = [2 \times \Delta_c H^\theta(\text{C}) + 3 \times \Delta_c H^\theta(\text{H}_2)] - [\Delta_c H^\theta(\text{C}_2\text{H}_5\text{OH})] \]
4. Substitute the combustion values:
\[ \Delta H^\theta_{\text{formation}} = [2 \times (-394) + 3 \times (-286)] - [-1367] \]
\[ \Delta H^\theta_{\text{formation}} = [-788 - 858] + 1367 \]
\[ \Delta H^\theta_{\text{formation}} = -1646 + 1367 = -279\text{ kJ mol}^{-1} \]
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