Problem 18 - Basic Principles of Organic Chemistry

Question

Show how the (Delta \(\mathrm{H}\) ) values of the following processes can be combined to calculate the heat of solution $$ \begin{array}{ll} \text { of } \mathrm{Na}^{\oplus}(g)+\mathrm{Cl}^{\ominus}(g) \text { at } 298^{\circ} \mathrm{K} \\ \mathrm{Na}(s)+\frac{1}{2} \mathrm{Cl}_{2}(g) \rightarrow \mathrm{Na}^{\oplus}(a q)+\mathrm{Cl}^{\ominus}(a q) & \Delta H^{0}=-97 \mathrm{kcal} \\ \mathrm{Na}(g) \rightarrow \mathrm{Na}^{\oplus}(g)+\mathrm{e}^{-} & \Delta H^{0}=+118 \mathrm{kcal} \\ \mathrm{Cl}^{\ominus}(g) \rightarrow \mathrm{Cl} \cdot(g)+\mathrm{e}^{-} & \Delta H^{0}=+83 \mathrm{kcal} \\ \frac{1}{2} \mathrm{Cl}_{2}(g) \rightarrow \mathrm{Cl} \cdot(g) & \Delta H^{0}=+2929 \\ \mathrm{Na}(s) \rightarrow \mathrm{Na}(g) & \Delta H^{0}=+26 \mathrm{kcal} \end{array} $$

Step-by-Step Solution

Answer

2990 kcal of energy is required for the process.

Step 1: Identify Known Reactions and Their ΔH

We are given several chemical reactions, each with a corresponding enthalpy change (ΔH). These reactions involve various stages of transforming Na(s) and Cl₂(g) to form Na⁺(aq) and Cl⁻(aq). The ΔH values are: 1. Na(s) → Na(g), ΔH = +26 kcal 2. Na(g) → Na⁺(g) + e⁻, ΔH = +118 kcal 3. Cl⁻(g) → Cl∙(g) + e⁻, ΔH = +83 kcal 4. ½Cl₂(g) → Cl∙(g), ΔH = +2929 kcal 5. Na(s) + ½Cl₂(g) → Na⁺(aq) + Cl⁻(aq), ΔH = -97 kcal.

Step 2: Analyze the Target Reaction

We must find the ΔH for the process of forming Na⁺(g) and Cl⁻(g) from Na(s) and ½Cl₂(g). This process can essentially be broken down into several smaller steps, represented by the reactions provided.

Step 3: Construct Intermediate Steps to Form Target Reaction

We need to construct the intermediate steps to go from: - Na(s) + ½Cl₂(g) to Na⁺(g) + Cl⁻(g). Translate into: 1. Na(s) → Na(g) 2. Na(g) → Na⁺(g) + e⁻ 3. ½Cl₂(g) → Cl∙(g) 4. Cl∙(g) + e⁻ → Cl⁻(g) Ensure all intermediate species formed cancel out except for the products Na⁺(g) and Cl⁻(g).

Step 4: Apply Hess's Law to Combine ΔH Values

According to Hess's Law, we can add up the ΔH values of individual reaction steps to determine the ΔH of the overall reaction. Combine the reactions: - Na(s) → Na(g), ΔH = +26 kcal - Na(g) → Na⁺(g) + e⁻, ΔH = +118 kcal - ½Cl₂(g) → Cl∙(g), ΔH = +2929 kcal - Cl∙(g) + e⁻ → Cl⁻(g), ΔH = -83 kcal (reverse direction) Sum of ΔH: Total ΔH = 26 + 118 + 2929 - 83 kcal = 2990 kcal.

Step 5: Calculate Heat of Solution Using Combined ΔH

We calculated the ΔH for the conversion of Na(s) + ½Cl₂(g) to Na⁺(g) + Cl⁻(g) as 2990 kcal. Since we need the process Na(s) + ½Cl₂(g) → Na⁺(aq) + Cl⁻(aq), we need to account for the dissolution step given by the reaction with ΔH = -97 kcal. Finally, subtract the heat of the solution to adjust the standard state to aqueous phase; ΔH (solution) = ΔH (gas) + ΔH (aq state correction) = 2990 kcal - 97 kcal = 2893 kcal.

Key Concepts

Hess's Law

Hess's Law is a powerful concept in thermochemistry that allows us to calculate the enthalpy change for a chemical reaction by using the enthalpy changes of intermediate steps. The principle we rely on here is that the total enthalpy change for a reaction is the same, regardless of whether it takes place in a single step or multiple steps. This means that if we have several chemical reactions leading to an overall reaction, we can add the \( \Delta H \) values of these reactions to find the enthalpy change of our desired reaction.

**Key Points of Hess's Law:**

• The reaction path does not influence the total enthalpy change.
• Intermediate steps can be rearranged and used to find the overall \( \Delta H \).
• We can reverse reactions or multiply them by coefficients, but we must adjust the sign and magnitude of their \( \Delta H \) accordingly.

Using Hess's Law, in the given exercise, we consider different steps such as the sublimation of sodium, ionization, and formation of gaseous ions. By adjusting these steps and aligning them with our target reaction, we easily calculate the overall heat of solution.

Enthalpy Change

The enthalpy change \( \Delta H \) of a reaction is a measure of the total heat absorption or release during the process. It's a fundamental aspect in thermochemistry, providing insight into whether a reaction is endothermic (absorbing heat) or exothermic (releasing heat).

**Key Aspects:**

• Positive \( \Delta H \) indicates endothermic reactions, requiring energy input.
• Negative \( \Delta H \) signifies exothermic reactions, releasing heat to the surroundings.
• It is expressed in kilocalories (kcal) or kilojoules (kJ).

In the exercise, we carefully assess the enthalpy changes for each sub-reaction, such as the conversion of sodium to its gaseous state or its ionization. Understanding these changes helps us sum the enthalpies to find the total heat of solution. With each part meticulously accounted for, calculating the overall enthalpy change becomes straightforward.

Heat of Solution

The heat of solution refers to the overall heat change when a solute dissolves in a solvent to form a solution. This concept is crucial to understand since it encompasses the interplay between solute-solvent interactions, lattice energy of the solute, and the enthalpy associated with solvation.

**Understanding Heat of Solution:**

• Involves both breaking and formation of bonds or interactions.
• Can be either endothermic or exothermic based on the balance of interactions.
• Essential for predicting thermal effects of dissolving various substances.

In the context of the exercise, after calculating the enthalpy change from gaseous ions to aqueous ions, we adjust with the heat of solution to align with the conditions of the solution phase, thereby finding the overall enthalpy change from solid to aqueous state. This provides a comprehensive view of how energy is managed during dissolution and ensures our calculations reflect real-world chemical processes.