Problem 45 - Chemistry The Central Science

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 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) The enthalpy change for the production of 0.450 mol of AgCl is -29.475 kJ. (b) The enthalpy change for the production of 9.00 g of AgCl is -4.119 kJ. (c) The enthalpy change for the dissolution of 9.25 * 10^{-4} mol of AgCl in water is 0.0605375 kJ.

Step 1: (a) Given moles of AgCl and ΔH of reaction, calculate ΔH for the production of 0.450 mol of AgCl

First, we must determine the molar enthalpy change, as the given ΔH is for the formation of one mole of AgCl. Molar enthalpy change: ΔH = -65.5 kJ/mol Now, we can calculate the enthalpy change for the production of 0.450 mol of AgCl using the molar enthalpy change: ΔH = (molar enthalpy change) × (moles of AgCl) ΔH = -65.5 kJ/mol × 0.450 mol

Step 2: (a) Calculate the enthalpy change

ΔH = -29.475 kJ So the enthalpy change for the production of 0.450 mol of AgCl is -29.475 kJ.

Step 3: (b) Calculate the moles of AgCl, given 9.00 g of AgCl

To find the enthalpy change for the production of 9.00 g of AgCl, we first need to determine the moles of AgCl, using its molar mass: Molar mass of AgCl = 107.87 (Ag) + 35.45 (Cl) = 143.32 g/mol Moles of AgCl = mass / molar mass Moles of AgCl = 9.00 g / 143.32 g/mol

Step 4: (b) Calculate the enthalpy change for the 9.00 g of AgCl

Now, we can calculate the enthalpy change for the production of 9.00 g of AgCl using the molar enthalpy change: ΔH = (molar enthalpy change) × (moles of AgCl) ΔH = -65.5 kJ/mol × (9.00 g / 143.32 g/mol)

Step 5: (b) Calculate the enthalpy change

ΔH = -4.119 kJ So the enthalpy change for the production of 9.00 g of AgCl is -4.119 kJ.

Step 6: (c) Calculate ΔH for the dissolution of 9.25 * 10^{-4} mol of AgCl

For the dissolution of AgCl in water, we need to reverse the given reaction, and thus, the enthalpy change will have the opposite sign: Molar enthalpy change: ΔH = 65.5 kJ/mol Now, we can calculate the enthalpy change for the dissolution of 9.25 * 10^{-4} mol of AgCl in water using the molar enthalpy change: ΔH = (molar enthalpy change) × (moles of AgCl) ΔH = 65.5 kJ/mol × 9.25 * 10^{-4} mol

Step 7: (c) Calculate the enthalpy change

ΔH = 0.0605375 kJ So the enthalpy change for the dissolution of 9.25 * 10^{-4} mol of AgCl in water is 0.0605375 kJ.

Key Concepts

Silver Chloride Precipitation

Silver chloride precipitation is a well-known chemical reaction in the field of solutions chemistry. When silver ions (Ag⁺) and chloride ions (Cl⁻) come into contact in an aqueous solution, they combine to form silver chloride (AgCl), a white solid that precipitates out of the solution. This process is described by the equation: \[ \mathrm{Ag}^{+}(aq) + \mathrm{Cl}^{-}(aq) \longrightarrow \mathrm{AgCl}(s) \] This precipitation reaction is important in various industrial and laboratory processes, particularly in analytical chemistry for detecting or removing certain ions from a solution.
The reaction's enthalpy change, denoted as \( \Delta H \), is the amount of heat absorbed or released during the formation of a mole of the precipitate. For silver chloride, the enthalpy change is \(-65.5 \mathrm{~kJ/mol}\), indicating the reaction releases heat and is exothermic. Understanding this helps us predict how the system's temperature may change during the reaction, often making it vital to control conditions to manage heat effectively.

Molar Enthalpy

Molar enthalpy is a valuable concept that helps us grasp the energy changes involved in chemical reactions. It refers to the heat absorbed or released per mole of a substance undergoing a reaction. Understanding molar enthalpy allows chemists to predict how much energy will be involved in reactions, influencing everything from reaction rates to safety considerations. For the silver chloride reaction, we're given a molar enthalpy change of \(\Delta H = -65.5 \mathrm{~kJ/mol}\). This means that 65.5 kilojoules of heat are released when one mole of silver chloride is formed. By using molar enthalpy, we can calculate the enthalpy change for different quantities of substances involved in the reaction. For instance: - To find the \(\Delta H\) for forming \(0.450\) mol of AgCl, multiply the molar enthalpy by the number of moles: \[ \Delta H = -65.5 \mathrm{~kJ/mol} \times 0.450 \mathrm{~mol} \] - This gives us an enthalpy change of \(-29.475 \mathrm{~kJ}\). Such calculations are crucial for scaling chemical reactions from lab to industrial scales.

Chemical Reactions in Solutions

Chemical reactions in solutions involve the interaction of dissolved substances to form new compounds. These reactions often proceed more rapidly than in solid or gaseous phases due to the mobility of ions or molecules in the liquid medium. Understanding these reactions involves considering both the reactants and solvents, which can influence the reaction kinetics and thermodynamics. The formation of silver chloride from silver and chloride ions in solution is a prime example of such reactions. Solutions create an environment where ions can freely collide and interact, facilitating the formation of precipitates like AgCl. The solubility product constant (Ksp) is a useful metric in this context, indicating a compound's solubility and the point at which it will begin to precipitate under certain conditions. Moreover, energy changes in chemical reactions in solutions can be clouded by factors like heat transfer between the reaction and solution. Enthalpy changes, such as the exothermic \(\Delta H\) observed in the AgCl formation, reflect these energy dynamics.

• For efficient utilization of these reactions in processes, control over temperature, concentration, and mixing speeds is critical.
• By understanding the underlying principles, one can anticipate and manipulate the outcomes of such chemical interactions effectively.