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 mol of 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 AgCl dissolves in water.

Short answer

\( \Delta H_{0.450} = -29.475\,\mathrm{kJ} \) \( \Delta H_{9.00} = -4.110\,\mathrm{kJ} \) \( \Delta H_{9.25 \times 10^{-4}} = +0.06058\,\mathrm{kJ} \)

Step-by-step solution

To calculate ΔH for 0.450 mol of AgCl production, we start with ΔH per mole (given in the exercise) and multiply it by the number of moles (0.450 mol): $$ \Delta H_{0.450} = \Delta H \times n $$ $$ \Delta H_{0.450} = -65.5\,\mathrm{kJ/mol} \times 0.450\,\mathrm{mol} $$tag_title# Calculate the ΔH value for 0.450 mol of AgCl formed.

Now, we can calculate the ΔH value for 0.450 mol of AgCl formed: $$ \Delta H_{0.450} = -29.475\,\mathrm{kJ} $$ #b) ΔH for 9.00 g of AgCl production:

Calculate the moles of AgCl from the mass given.

First, we need to convert the mass of AgCl (9.00 g) into moles. To do this, we use the molar mass of silver chloride (AgCl): $$ \mathrm{Molar\,mass\,of\, AgCl} = M_{\mathrm{Ag}} + M_{\mathrm{Cl}} = 107.87\,\mathrm{g/mol} + 35.45\,\mathrm{g/mol} = 143.32\,\mathrm{g/mol} $$ $$ n = \frac{m}{M} = \frac{9.00\,\mathrm{g}}{143.32\,\mathrm{g/mol}} $$ Now we calculate the moles of AgCl: $$ n = 0.06277\,\mathrm{mol} $$tag_title# Calculate the ΔH value for 9.00 g of AgCl formed.

Now, we can calculate the ΔH value for 9.00 g (0.06277 mol) of Ag Cl formed: $$ \Delta H_{9.00} = \Delta H \times n $$ $$ \Delta H_{9.00} = -65.5\,\mathrm{kJ/mol} \times 0.06277\,\mathrm{mol} $$ Now we calculate the ΔH value for 9.00 g (0.06277 mol) of AgCl formed: $$ \Delta H_{9.00} = -4.110\,\mathrm{kJ} $$ #c) ΔH for 9.25 x 10⁻⁴ mol of AgCl dissolution:

For the reverse reaction, change the sign of ΔH.

Since dissolution is the reverse reaction, we need to change the sign of ΔH: $$ \Delta H_{dissolution} = +65.5\,\mathrm{kJ/mol} $$tag_title# Calculate the ΔH value for 9.25 x 10⁻⁴ mol of AgCl dissolved.

Now, we calculate the ΔH value for 9.25 x 10⁻⁴ mol of AgCl dissolved: $$ \Delta H_{9.25 \times 10^{-4}} = \Delta H_{dissolution} \times n $$ $$ \Delta H_{9.25 \times 10^{-4}} = +65.5\,\mathrm{kJ/mol} \times 9.25 \times 10^{-4}\,\mathrm{mol} $$ Now we calculate the ΔH value for 9.25 x 10⁻⁴ mol of AgCl dissolved: $$ \Delta H_{9.25 \times 10^{-4}} = +0.06058\,\mathrm{kJ} $$

Key concepts

Chemical Thermodynamics

Chemical thermodynamics is a fascinating study that combines the principles of physics and chemistry to analyze the energy changes associated with chemical reactions and processes. In the context of our problem, we focus on a particular aspect of thermodynamics called enthalpy (represented as \( \Delta H \)), which expresses the heat change at constant pressure. \( \Delta H \) is essential for predicting whether reactions will occur spontaneously and for understanding how much energy is released or absorbed during a reaction.

Enthalpy change is measured in kilojoules per mole (kJ/mol) and can be positive (endothermic reaction) or negative (exothermic reaction). Here, our textbook problem indicates that the formation of silver chloride (AgCl) from its ions is exothermic, as \( \Delta H \) is a negative value (-65.5 kJ/mol). This negative sign signifies that the reaction releases energy to the surroundings, and is thermodynamically favorable under standard conditions.

Precipitation Reactions

Precipitation reactions are a type of chemical reaction that occur when two soluble salts in aqueous solutions are mixed together to form an insoluble solid, known as a precipitate. In our textbook exercise, silver ions \( (\mathrm{Ag}^{+}) \) and chloride ions \( (\mathrm{Cl}^{-}) \) in solution combine to form silver chloride \( (\mathrm{AgCl}) \) as a solid precipitate.

These reactions are central to various fields such as analytical chemistry, environmental science, and materials science. Understanding how to calculate the enthalpy change, \( \Delta H \), for precipitation reactions allows students to predict the energy required or released when forming a certain amount of precipitate. This can aid in the design and optimization of chemical processes, including the purification of substances or the manufacture of advanced materials.

Mole Concept

The mole concept is a foundational principle in chemistry that enables chemists to count particles like atoms, molecules, or ions by weighing them. One mole corresponds to Avogadro's number (approximately \(6.022 \times 10^{23}\) entities) and is defined as the quantity of a substance that contains as many particles as there are atoms in 12 grams of carbon-12.

Calculations involving moles are crucial when dealing with chemical equations and reactions. In the exercise, for instance, to calculate the enthalpy changes for different amounts of AgCl, we convert grams to moles using the molar mass of AgCl, then multiply by the enthalpy change per mole. This application of the mole concept enables precise predictions about the energy changes that come with the formation or dissolution of compounds, highlighting its importance in chemical thermodynamics and quantitative analysis.