# Hess' Law ## Hess' Law & Enthalpy 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 - Whether the reaction takes place in one or two steps, the total enthalpy change of the reaction will still be the same

The direct route has the same enthalpy change as the sum of the indirect routes

- Hess' Law is used to calculate enthalpy changes which can't be found experimentally using calorimetry - For example: $3\text{C (s)} + 4\text{H}_2\text{(g)} \rightarrow \text{C}_3\text{H}_8\text{(g)}$ - $\Delta H_f$ (propane) can't be found experimentally as hydrogen and carbon don't react under standard conditions ### Two key rules for calculations - Using enthalpy changes of formation, $\Delta H_f$: - $\Delta H_r = \sum\Delta H_f\text{(products)} - \sum\Delta H_f\text{(reactants)}$ - Using enthalpy changes of combustion, $\Delta H_c$: - $\Delta H_r = \sum\Delta H_c\text{(reactants)} - \sum\Delta H_c\text{(products)}$ ## Calculating $\Delta H_r$ from $\Delta H_f$ using Hess's Law - Knowing the enthalpy change of formation, $\Delta H_f$, allows us to determine the overall enthalpy change of a reaction, $\Delta H_r$ ### Steps for constructing formation cycles 1. Write the equation for the reaction 2. Write the elements with the correct **number of moles** and **state symbols** underneath 3. Draw **upwards pointing arrows** to each compound 4. Write the appropriate **values** on the arrows and **multiply** by the number of moles 5. In a cycle, go from the reactants to the products, changing the sign of the value if the arrow points in the opposite direction ### The formation cycle relationship - The products can be formed: - **Directly** from the elements = $\Delta H_2$ - **OR** - **Indirectly** from the elements = $\Delta H_1 + \Delta H_r$ - Equation: $\Delta H_2 = \Delta H_1 + \Delta H_r$ - Therefore for energy to be conserved: $\Delta H_r = \Delta H_2 - \Delta H_1$ ### Worked Example: Calculating enthalpy change of reaction Use the information in the table to calculate the enthalpy change for this reaction: $$\text{NH}_4\text{NO}_3\text{(s)} + \frac{1}{2}\text{C (s)} \rightarrow \text{N}_2\text{(g)} + 2\text{H}_2\text{O (g)} + \frac{1}{2}\text{CO}_2\text{(g)}$$ | Substance | C (s) | N₂ (g) | H₂O (g) | CO₂ (g) | NH₄NO₃ (s) | |-----------|-------|--------|---------|---------|------------| | $\Delta H_f^{\theta}$ / kJ mol⁻¹ | 0 | 0 | -242 | -394 | -365 | **Step 1:** Write the balanced equation $$\text{NH}_4\text{NO}_3\text{(s)} + \frac{1}{2}\text{C(s)} \longrightarrow \text{N}_2\text{(g)} + 2\text{H}_2\text{O(g)} + \frac{1}{2}\text{CO}_2\text{(g)}$$ **Step 2:** Write the elements with the correct number of moles and state symbols underneath $$\text{N}_2\text{(g)} + 4.5\text{O}_2\text{(g)} + \text{C(s)} + 2\text{H}_2\text{(g)}$$ **Step 3:** Draw upwards pointing arrows to each compound **Step 4:** Write the appropriate values on the arrows and multiply by the number of moles - Arrow from elements to $\text{NH}_4\text{NO}_3\text{(s)}$: $-365$ kJ mol⁻¹ - Arrow from elements to $2\text{H}_2\text{O(g)}$: $2 \times (-242) = -484$ kJ mol⁻¹ - Arrow from elements to $\frac{1}{2}\text{CO}_2\text{(g)}$: $0.5 \times (-394) = -197$ kJ mol⁻¹ **Step 5:** In a cycle, go from the reactants to the products, changing the sign of the value if the arrow points in the opposite direction $$\Delta H_r^{\theta} = +365 - 484 - 197 = -316 \text{ kJ mol}^{-1}$$ - The sign on $-365$ needs reversing as the cycle goes in the opposite direction to the arrow pointing upwards - There is no need to draw arrows from the elements to carbon and nitrogen as $\Delta H_f^{\theta}$ is 0 for elements

Keep your enthalpy values inside their own brackets so that you don't accidentally lose a minus sign

## Calculating $\Delta H_f$ from $\Delta H_c$ using Hess's Law - It can be difficult to find the enthalpy change of formation of compounds experimentally - However, many enthalpy changes of combustion **can** be measured experimentally so these can be used to find the enthalpy of formation ### Steps for constructing combustion cycles 1. Write the equation for the formation of the compound 2. Write the combustion products **below** the equation 3. Draw **downward pointing arrows** from each substance to its combustion products 4. Write **values** on the arrows and **multiply** by the number of moles 5. In a cycle, go from the reactants to the products, changing the sign of the value if the arrow points in the opposite direction ### Worked Example: Calculating enthalpy change of formation Using the data provided, calculate the standard enthalpy change of formation, $\Delta H_f$, of propanone: $$3\text{C (s)} + 3\text{H}_2\text{(g)} + \frac{1}{2}\text{O}_2\text{(g)} \rightarrow \text{CH}_3\text{COCH}_3\text{(l)}$$ | Substance | C (s) | H₂ (g) | CH₃COCH₃ (l) | |-----------|-------|--------|--------------| | $\Delta H_c^{\theta}$ / kJ mol⁻¹ | -394 | -286 | -1821 | **Answer:** **Step 1:** Write the balanced equation $$3\text{C(s)} + 3\text{H}_2\text{(g)} + \frac{1}{2}\text{O}_2\text{(g)} \longrightarrow \text{CH}_3\text{COCH}_3\text{(l)}$$ **Step 2:** Write the combustion products **below** the equation $$3\text{CO}_2 + 3\text{H}_2\text{O}$$ **Step 3:** Draw **downward pointing arrows** from each substance to its combustion product **Step 4:** Write the appropriate values on the arrows and multiply by the number of moles - Arrow from $3\text{C(s)}$ downward: $3 \times (-394) = -1182$ kJ mol⁻¹ - Arrow from $3\text{H}_2\text{(g)}$ downward: $3 \times (-286) = -858$ kJ mol⁻¹ - Arrow from $\text{CH}_3\text{COCH}_3\text{(l)}$ downward: $-1821$ kJ mol⁻¹ **Step 5:** In a cycle, go from the reactants to the products, changing the sign of the value if the arrow points in the opposite direction $$\Delta H_f^{\theta} = -1182 - 858 + 1821 = -219 \text{ kJ mol}^{-1}$$ - The sign on $-1821$ needs reversing as the cycle goes in the opposite direction to the arrow pointing to the combustion products

Don't forget to make sure the number of atoms for each element is balanced when drawing your cycle

## Summary of cycle construction | Cycle Type | Arrow Direction | Intermediate Level | Formula | |------------|----------------|-------------------|---------| | Formation cycles | Upwards from elements | Elements at bottom | $\Delta H_r = \sum\Delta H_f\text{(products)} - \sum\Delta H_f\text{(reactants)}$ | | Combustion cycles | Downwards to combustion products | Combustion products at bottom | $\Delta H_r = \sum\Delta H_c\text{(reactants)} - \sum\Delta H_c\text{(products)}$ |