Hess's law Knowledge-Based Study

Study notes

1. The Principle of Conservation of Energy in Reactions

   In thermochemistry, enthalpy ($H$) is a state function. This means that the change in enthalpy ($\Delta H$) for any chemical transformation depends solely on the initial state of the reactants and the final state of the products, not on the specific pathway or number of steps taken to get from one to the other.

   Formulated by Germain Henri Hess in 1840, Hess's law of constant heat summation states:

   If a reaction can be expressed as the sum of a series of steps, the overall enthalpy change of the reaction is equal to the sum of the enthalpy changes of the individual steps.

   Mathematically, for an overall reaction comprising steps 1, 2, ..., $n$:

   $$\Delta H_{\text{overall}} = \Delta H_1 + \Delta H_2 + \dots + \Delta H_n$$

   This principle allows chemists to determine enthalpy changes for reactions that are difficult, dangerous, or impossible to measure directly using calorimetry.

   Algebraic Manipulation of Equations

   Applying Hess's law involves treating thermochemical equations like algebraic equations. When manipulating a given equation to match a target equation, two core rules apply:

   • Reversing an equation: If a reaction is reversed, the magnitude of $\Delta H$ remains the same, but its sign changes because an exothermic process in the forward direction is endothermic in the reverse direction.
   • Multiplying by a factor: If all stoichiometric coefficients in a reaction are multiplied by an integer or a fraction ($n$), the enthalpy change must be multiplied by that same factor ($n \times \Delta H$), because enthalpy is an extensive property.

   For example, if the combustion of hydrogen is written as:

   $$\text{H}_2\text{(g)} + \frac{1}{2}\text{O}_2\text{(g)} ightarrow \text{H}_2\text{O(l)} \quad \Delta H = -285.8\text{ kJ}$$

   Then decomposing liquid water into its elements requires reversing the reaction:

   $$\text{H}_2\text{O(l)} ightarrow \text{H}_2\text{(g)} + \frac{1}{2}\text{O}_2\text{(g)} \quad \Delta H = +285.8\text{ kJ}$$

   Systematic Problem-Solving Strategy

   To find the enthalpy change of an overall (target) reaction using a set of intermediate reactions, follow these systematic steps:

   1. Identify unique species: Locate substances that appear in only one of the given intermediate equations and compare their position (reactant or product) and stoichiometric coefficient with the target equation.
   2. Adjust intermediate reactions: Reverse equations or multiply them by coefficients so that the unique species align with the target equation.
   3. Sum the adjusted equations: Add the reactants and products together. Cancel out identical chemical species that appear in equal amounts on both sides of the reaction arrow (intermediates).
   4. Calculate overall enthalpy: Add the adjusted $\Delta H$ values algebraically to obtain $\Delta H_{\text{target}}$.

   Worked Example

   Determine the enthalpy change for the synthesis of propane:

   $$3\text{C(s)} + 4\text{H}_2\text{(g)} ightarrow \text{C}_3\text{H}_8\text{(g)}$$

   Given:

   • Reaction 1: $\text{C(s)} + \text{O}_2\text{(g)} ightarrow \text{CO}_2\text{(g)} \quad \Delta H_1 = -393.5\text{ kJ}$
   • Reaction 2: $\text{H}_2\text{(g)} + \frac{1}{2}\text{O}_2\text{(g)} ightarrow \text{H}_2\text{O(l)} \quad \Delta H_2 = -285.8\text{ kJ}$
   • Reaction 3: $\text{C}_3\text{H}_8\text{(g)} + 5\text{O}_2\text{(g)} ightarrow 3\text{CO}_2\text{(g)} + 4\text{H}_2\text{O(l)} \quad \Delta H_3 = -2220.0\text{ kJ}$

   Adjusting the reactions:

   • Multiply Reaction 1 by 3: $\Delta H = 3 \times (-393.5\text{ kJ}) = -1180.5\text{ kJ}$
   • Multiply Reaction 2 by 4: $\Delta H = 4 \times (-285.8\text{ kJ}) = -1143.2\text{ kJ}$
   • Reverse Reaction 3: $\Delta H = -(-2220.0\text{ kJ}) = +2220.0\text{ kJ}$

   Summing the modified steps cancels out $3\text{CO}_2$, $4\text{H}_2\text{O}$, and $5\text{O}_2$, yielding:

   $$\Delta H_{\text{target}} = -1180.5\text{ kJ} + (-1143.2\text{ kJ}) + 2220.0\text{ kJ} = -103.7\text{ kJ}$$

   Enthalpies of Formation as an Alternative Approach

   A direct extension of Hess's law involves using standard enthalpies of formation ($\Delta H^\circ_f$). The standard enthalpy of formation is the enthalpy change when one mole of a substance is formed from its pure elements in their most stable reference states under standard conditions (1 bar, typically 298.15 K).

   By definition, the standard enthalpy of formation of a pure element in its reference state is zero (e.g., $\Delta H^\circ_f[\text{O}_2\text{(g)}] = 0\text{ kJ/mol}$).

   Applying Hess's law to any general reaction allows calculation of the overall reaction enthalpy without writing out separate intermediate steps:

   $$\Delta H^\circ_{\text{rxn}} = \sum n \Delta H^\circ_f(\text{products}) - \sum m \Delta H^\circ_f(\text{reactants})$$

   where $n$ and $m$ represent the stoichiometric coefficients of the products and reactants, respectively.