# 6.4 Hess's Law Hess's Law is a fundamental principle in thermochemistry based on the fact that enthalpy is a *state function*. This means the total enthalpy change for a chemical reaction is independent of the pathway taken from the initial state (reactants) to the final state (products). The change in enthalpy is the same whether the reaction occurs in a single step or in multiple sequential steps. The law, stated by G.H. Hess in 1840, can be summarized as: > **The total enthalpy change in a chemical reaction is the same regardless of whether the reaction is completed in one step or in several steps.** Mathematically, for any cyclic process, the sum of all enthalpy changes is zero: $$ \Sigma \Delta H(\text{Cycle}) = 0 $$ Consider a reaction where reactant **A** transforms into product **B**. * **Direct Path:** $A \rightarrow B$ with an enthalpy change of $\Delta H$. * **Indirect Path:** $A \rightarrow C \rightarrow D \rightarrow B$ with enthalpy changes $\Delta H_1$, $\Delta H_2$, and $\Delta H_3$ for each step. According to Hess's Law, the enthalpy change for the direct path is equal to the sum of the enthalpy changes for the indirect path: $$ \Delta H = \Delta H_1 + \Delta H_2 + \Delta H_3 $$ ## Principle and Application Hess's Law is particularly useful for determining the enthalpy changes of reactions that are difficult or impossible to measure directly in a calorimeter. This includes: - Reactions that are very slow. - Reactions that do not go to completion. - Reactions that produce side products. - Formation of compounds that cannot be synthesized directly from their elements (e.g., $\mathrm{CCl_4}$). - Combustion of elements that form a protective oxide layer (e.g., $\mathrm{Al}$, $\mathrm{B}$). By using Hess's Law, we can calculate the desired enthalpy change by combining the known enthalpy changes of other related reactions. ## Calculating Enthalpies of Reaction from Enthalpies of Formation ($\Delta H_f^{\circ}$) A common application of Hess's Law is to calculate the standard enthalpy change of a reaction ($\Delta H_{\text{reaction}}^{\circ}$) using the standard enthalpies of formation ($\Delta H_f^{\circ}$) of the reactants and products. The general formula is: $$ \Delta H_{\text{reaction}}^{\circ} = \sum n_p \Delta H_f^{\circ}(\text{products}) - \sum n_r \Delta H_f^{\circ}(\text{reactants}) $$ Where: - $n_p$ and $n_r$ are the stoichiometric coefficients of the products and reactants, respectively. - **Important Note:** The standard enthalpy of formation ($\Delta H_f^{\circ}$) of any element in its most stable standard state is defined as **zero**. ## Calculation of Enthalpy of Combustion ($\Delta H_c^{\circ}$) The *enthalpy of combustion*, $\Delta H_c^{\circ}$, is the heat evolved when one mole of a substance burns completely in oxygen under standard conditions. It can be calculated using the same formula as above, with the products typically being $\mathrm{CO_2(g)}$ and $\mathrm{H_2O(l)}$. $$ \Delta H_c^{\circ} = \sum n_p \Delta H_f^{\circ}(\text{products}) - \sum n_r \Delta H_f^{\circ}(\text{reactants}) $$ ## Worked Examples ### Example 6.6: Combustion of Carbon The complete combustion of carbon to carbon dioxide can occur in one step or two steps. **Single Step:** $$ \mathrm{C_{(s)} + O_{2(g)} \rightarrow CO_{2(g)}} \quad \Delta H^{\circ} = -393.5 \,\text{kJ} $$ **Two Steps:** 1. Formation of carbon monoxide: $$ \mathrm{C_{(s)} + \frac{1}{2}O_{2(g)} \rightarrow CO_{(g)}} \quad \Delta H_1^{\circ} = -110.52 \,\text{kJ} $$ 2. Combustion of carbon monoxide: $$ \mathrm{CO_{(g)} + \frac{1}{2}O_{2(g)} \rightarrow CO_{2(g)}} \quad \Delta H_2^{\circ} = -282.98 \,\text{kJ} $$ **Applying Hess's Law:** The total enthalpy change for the two-step process is the sum of the individual steps: $$ \Delta H^{\circ} = \Delta H_1^{\circ} + \Delta H_2^{\circ} = (-110.52) + (-282.98) = -393.50 \,\text{kJ} $$ This matches the enthalpy change for the single-step reaction, demonstrating the path independence. *Figure 6.4: Energy cycle for the reaction between $\mathrm{C(s)}$ and $\mathrm{O_{2(g)}}$ to produce $\mathrm{CO_{2}(g)}$.* ### Example 6.7: Enthalpy of Formation of Methane Calculate the standard enthalpy of formation for methane ($\mathrm{CH_4}$) using the following combustion data. **Target Reaction:** $\mathrm{C_{(s)} + 2H_{2(g)} \rightarrow CH_{4(g)}} \quad \Delta H_f^{\circ} = ?$ **Given Reactions:** (i) $\mathrm{CH_{4(g)} + 2O_{2(g)} \rightarrow CO_{2(g)} + 2H_2O_{(l)}} \quad \Delta H^{\circ} = -890.4 \,\text{kJ mol}^{-1}$ (ii) $\mathrm{H_{2(g)} + \frac{1}{2}O_{2(g)} \rightarrow H_2O_{(l)}} \quad \Delta H^{\circ} = -285.8 \,\text{kJ mol}^{-1}$ (iii) $\mathrm{C_{(s)} + O_{2(g)} \rightarrow CO_{2(g)}} \quad \Delta H^{\circ} = -393.5 \,\text{kJ mol}^{-1}$ **Solution:** 1. **Manipulate the given equations** to match the target reaction: - **Equation (iii):** Use as is, since we need $\mathrm{C_{(s)}}$ as a reactant. - **Equation (ii):** Multiply by 2, since we need $2\mathrm{H_{2(g)}}$ as a reactant. $\Delta H^{\circ}$ is also multiplied by 2. - **Equation (i):** Reverse the reaction, since we need $\mathrm{CH_{4(g)}}$ as a product. The sign of $\Delta H^{\circ}$ is changed. 2. **Combine the manipulated equations:** $$ \begin{align*} \mathrm{C_{(s)} + O_{2(g)}} & \rightarrow \mathrm{CO_{2(g)}} & \Delta H^{\circ} &= -393.5 \,\text{kJ} \\ \mathrm{2H_{2(g)} + O_{2(g)}} & \rightarrow \mathrm{2H_2O_{(l)}} & \Delta H^{\circ} &= 2(-285.8) = -571.6 \,\text{kJ} \\ \mathrm{CO_{2(g)} + 2H_2O_{(l)}} & \rightarrow \mathrm{CH_{4(g)} + 2O_{2(g)}} & \Delta H^{\circ} &= +890.4 \,\text{kJ} \end{align*} $$ 3. **Sum the reactions and cancel common species:** $$ \mathrm{C_{(s)} + \cancel{O_{2(g)}} + 2H_{2(g)} + \cancel{O_{2(g)}} + \cancel{CO_{2(g)}} + \cancel{2H_2O_{(l)}} \rightarrow \cancel{CO_{2(g)}} + \cancel{2H_2O_{(l)}} + CH_{4(g)} + \cancel{2O_{2(g)}}} $$ 4. **Net Reaction and Final Enthalpy:** $$ \mathrm{C_{(s)} + 2H_{2(g)} \rightarrow CH_{4(g)}} $$ $$ \Delta H_f^{\circ} = -393.5 + (-571.6) + 890.4 = \mathbf{-74.7 \,\text{kJ mol}^{-1}} $$ ### Example 6.8: Combustion of Octane (Gasoline) Calculate $\Delta H^{\circ}$ for the combustion of octane, given the standard enthalpies of formation ($\Delta H_f^{\circ}$). **Reaction:** $\mathrm{2C_8H_{18(l)} + 25O_{2(g)} \rightarrow 16CO_{2(g)} + 18H_2O_{(l)}}$ **Given Values:** - $\Delta H_f^{\circ}(\mathrm{C_8H_{18(l)}})$ = -269 kJ/mol - $\Delta H_f^{\circ}(\mathrm{O_{2(g)}})$ = 0 kJ/mol (element in standard state) - $\Delta H_f^{\circ}(\mathrm{CO_{2(g)}})$ = -393.5 kJ/mol - $\Delta H_f^{\circ}(\mathrm{H_2O_{(l)}})$ = -285.8 kJ/mol **Solution:** 1. **Apply the formula:** $$ \Delta H_{\text{reaction}}^{\circ} = \sum n_p \Delta H_f^{\circ}(\text{products}) - \sum n_r \Delta H_f^{\circ}(\text{reactants}) $$ 2. **Substitute values:** $$ \Delta H_{\text{reaction}}^{\circ} = [16 \times \Delta H_f^{\circ}(\mathrm{CO_2}) + 18 \times \Delta H_f^{\circ}(\mathrm{H_2O})] - [2 \times \Delta H_f^{\circ}(\mathrm{C_8H_{18}}) + 25 \times \Delta H_f^{\circ}(\mathrm{O_2})] $$ $$ \Delta H_{\text{reaction}}^{\circ} = [16(-393.5) + 18(-285.8)] - [2(-269) + 25(0)] $$ 3. **Calculate:** $$ \Delta H_{\text{reaction}}^{\circ} = [-6296 - 5144.4] - [-538 + 0] $$ $$ \Delta H_{\text{reaction}}^{\circ} = -11440.4 + 538 $$ $$ \Delta H_{\text{reaction}}^{\circ} = -10902.4 \,\text{kJ} = \mathbf{-1.09 \times 10^4 \,\text{kJ}} $$ ### Example 6.10: Enthalpy of Combustion of Ethyne Calculate the standard enthalpy of combustion ($\Delta H_c^{\circ}$) for ethyne ($\mathrm{C_2H_2}$). **Given Values:** - $\Delta H_f^{\circ}(\mathrm{C_2H_{2(g)}})$ = +226.73 kJ/mol - $\Delta H_f^{\circ}(\mathrm{CO_{2(g)}})$ = -393.5 kJ/mol - $\Delta H_f^{\circ}(\mathrm{H_2O_{(l)}})$ = -285.8 kJ/mol **Solution:** 1. **Write the balanced combustion equation:** $$ \mathrm{C_2H_{2(g)} + \frac{5}{2}O_{2(g)} \rightarrow 2CO_{2(g)} + H_2O_{(l)}} $$ 2. **Balanced Combustion Reaction:** $$ \mathrm{C_2H_{2(g)} + \frac{5}{2}O_{2(g)} \rightarrow 2CO_{2(g)} + H_2O_{(l)}} $$ 3. **Apply the formula:** $$ \Delta H_c^{\circ} = [2 \times \Delta H_f^{\circ}(\mathrm{CO_2}) + 1 \times \Delta H_f^{\circ}(\mathrm{H_2O})] - [1 \times \Delta H_f^{\circ}(\mathrm{C_2H_2}) + \frac{5}{2} \times \Delta H_f^{\circ}(\mathrm{O_2})] $$ $$ \Delta H_c^{\circ} = [2(-393.5) + 1(-285.8)] - [(+226.73) + \frac{5}{2}(0)] $$ 4. **Calculate:** $$ \Delta H_c^{\circ} = [-787.0 - 285.8] - [226.73] $$ $$ \Delta H_c^{\circ} = -1072.8 - 226.73 $$ $$ \Delta H_c^{\circ} = \mathbf{-1299.53 \,\text{kJ/mol}} $$