Problem 45 - Chemistry The Central Science In Si Units

Question

When solutions containing silver ions and chloride ions are mixed, silver chloride precipitates $$ \mathrm{Ag}^{+}(a q)+\mathrm{Cl}^{-}(a q) \longrightarrow \mathrm{AgCl}(s) \quad \Delta H=-65.5 \mathrm{~kJ} $$ (a) Calculate \(\Delta H\) for the production of \(0.450 \mathrm{~mol}\) of \(\mathrm{AgCl}\) by this reaction. (b) Calculate \(\Delta H\) for the production of \(9.00 \mathrm{~g}\) of AgCl. (c) Calculate \(\Delta H\) when \(9.25 \times 10^{-4} \mathrm{~mol}\) of \(\mathrm{AgCl}\) dissolves in water.

Step-by-Step Solution

Answer

(a) -29.475 kJ, (b) -4.11 kJ, (c) +0.0605 kJ.

Step 1: Understanding the given enthalpy change

The reaction \( ext{Ag}^+(aq) + ext{Cl}^-(aq) \rightarrow ext{AgCl}(s)\) has an enthalpy change \(\Delta H = -65.5 \, \text{kJ}\) per mole of \(\text{AgCl}\) formed. This means that for every mole of \(\text{AgCl}\) produced, 65.5 kJ of energy is released.

Step 1: Calculating ΔH for 0.450 mol of AgCl

Since \(\Delta H\) is given per mole, for 0.450 mol of \(\text{AgCl}\), the energy change will be \( \Delta H = -65.5 \, \text{kJ/mol} \times 0.450 \, \text{mol} = -29.475 \, \text{kJ}\). Thus, \(\Delta H\) is \(-29.475 \, \text{kJ}\).

Step 2: Converting grams to moles for 9.00 g of AgCl

First, find the molar mass of \(\text{AgCl}\). The atomic masses are approximately \(\text{Ag} = 107.87 \, \text{g/mol}\) and \(\text{Cl} = 35.45 \, \text{g/mol}\), so the molar mass is \(107.87 + 35.45 = 143.32 \, \text{g/mol}\). Convert grams to moles: \(9.00 \, \text{g} \times \frac{1 \, \text{mol}}{143.32 \, \text{g}} = 0.0628 \, \text{mol}\).

Step 3: Calculating ΔH for 9.00 g of AgCl

Using the mole calculation from Step 2: \( \Delta H = -65.5 \, \text{kJ/mol} \times 0.0628 \, \text{mol} = -4.1144 \, \text{kJ}\). Hence, \(\Delta H\) is \(-4.11 \, \text{kJ}\), to three significant figures.

Step 4: Calculating ΔH for 9.25 x 10^-4 mol of AgCl dissolving

Dissolution is the reverse of precipitation, so the energy change is the negative of the precipitation \(\Delta H\). Therefore: \( \Delta H = +65.5 \, \text{kJ/mol} \times 9.25 \times 10^{-4} \, \text{mol} = +0.0605 \, \text{kJ}\).

Key Concepts

Understanding Chemical Reactions

Chemical reactions are fascinating processes where substances, known as reactants, transform into new substances called products. In our example, when silver ions (\(\mathrm{Ag}^+\)) react with chloride ions (\(\mathrm{Cl}^-\)), they form a new compound, silver chloride (\(\mathrm{AgCl}\)). This is an example of a chemical reaction where ions in solution react to form a solid product. Chemical reactions can be classified into different types: synthesis, decomposition, single replacement, and double replacement, just to name a few. Our case is a simple synthesis, where two ions combine to form a single product. Key characteristics of chemical reactions include:

• The rearrangement of atoms to form new substances.
• Conservation of mass, meaning the total mass of reactants is equal to the total mass of products.
• Energy changes, where energy is either absorbed or released during the reaction.

Understanding these fundamentals helps in predicting the products of a reaction and the energy changes involved. This leads us to our next concept, precipitation reactions.

Exploring Precipitation Reactions

Precipitation reactions occur when two solutions, containing soluble salts, are mixed and form an insoluble solid known as a precipitate. In our scenario, when solutions with silver ions and chloride ions combine, silver chloride solid forms. This reaction can be expressed by the equation: \[\mathrm{Ag}^{+}(aq) + \mathrm{Cl}^{-}(aq) \rightarrow \mathrm{AgCl}(s)\] The key points about precipitation reactions include:

• Formation of a solid from solutions.
• Involves ionic compounds swapping partners.
• Can be predicted using solubility rules that tell us which combinations form precipitates.

In our example, silver chloride is one of those combinations that does not stay dissolved in water and forms a white solid precipitate instead. The beauty of precipitation reactions is that they allow chemists to separate and isolate specific ions from a solution by converting them into an easily removable solid product. Understanding how and why precipitates form can be crucial in many fields, such as environmental science and water purification, where removing certain ions from a solution is necessary.

Mastering Energy Calculations

Energy plays a vital role in chemical reactions. The term enthalpy change (\(\Delta H\)) describes the heat absorbed or released during a reaction at constant pressure. In our reaction, the enthalpy change \(\Delta H = -65.5 \text{ kJ/mol}\) indicates that it is an exothermic reaction, meaning it releases energy. To calculate the energy changes for varying amounts of a substance, you need to understand how \(\Delta H\) is associated per mole:

• For 0.450 mol of \(\mathrm{AgCl}\), energy release equals \(-65.5 \text{ kJ/mol} \times 0.450 \text{ mol} = -29.475 \text{ kJ}\).
• For 9.00 g of \(\mathrm{AgCl}\), convert to moles using the molar mass (\(143.32 \text{ g/mol}\)), giving \(0.0628 \text{ mol}\), resulting in \(-4.11 \text{ kJ}\).
• When \(9.25 \times 10^{-4} \text{ mol}\) of \(\mathrm{AgCl}\) dissolves, it's an endothermic process, reversing the reaction: \(\Delta H = +0.0605 \text{ kJ}\).

Each calculation illustrates that precise energy considerations are crucial in reactions, allowing scientists to predict and understand the behavior and feasibility of chemical processes. This insight not only applies in the lab but also in industrial and environmental applications, optimizing processes for energy efficiency and impact.