Question

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

Short answer

(a) The enthalpy change for the production of 0.450 moles of silver chloride is -29.475 kJ. (b) The enthalpy change for the production of 9.00 g of silver chloride is approximately -4.113 kJ. (c) The enthalpy change for the dissolution of 9.25 x 10^-4 mol of silver chloride is approximately 0.0606 kJ.

Step-by-step solution

(a) Enthalpy change for 0.450 mol AgCl production

To find the enthalpy change for this specific situation, we can use the direct proportionality between enthalpy change and moles of silver chloride produced: \(\Delta H = -65.5\ \mathrm{kJ/mol} \times 0.450\ \mathrm{mol}\) \(\Delta H = -29.475\ \mathrm{kJ}\) The enthalpy change for the production of 0.450 moles of silver chloride is -29.475 kJ.

(b) Enthalpy change for 9.00 g AgCl production

Firstly, let's convert the mass of AgCl to moles, given that the molar mass of AgCl = 143.32 g/mol. Moles of AgCl: \(n = \frac{9.00\ \mathrm{g}}{143.32\ \mathrm{g/mol}}\) n ≈ 0.0628 mol Now, we can calculate the enthalpy change for the production of 9.00 g (0.0628 mol) of silver chloride: \(\Delta H = -65.5\ \mathrm{kJ/mol} \times 0.0628\ \mathrm{mol}\) \(\Delta H ≈ -4.113\ \mathrm{kJ}\) The enthalpy change for the production of 9.00 g of silver chloride is approximately -4.113 kJ.

(c) Enthalpy change for 9.25 x 10^-4 mol AgCl dissolution

As the enthalpy change given corresponds to the formation of solid AgCl from aqueous Ag+ and Cl-, we need to reverse the process for dissolution of AgCl back into its soluble ions. So, the sign of enthalpy change will be reversed. Now, we can find the enthalpy change for the dissolution of 9.25 x 10^-4 mol of AgCl: \(\Delta H = 65.5\ \mathrm{kJ/mol} \times 9.25 \times 10^{-4}\ \mathrm{mol}\) \(\Delta H ≈ 0.0606\ \mathrm{kJ}\) The enthalpy change for the dissolution of 9.25 x 10^-4 mol of silver chloride is approximately 0.0606 kJ.

Key concepts

Silver Chloride Precipitation

Silver chloride precipitation is a fascinating chemical process in which silver ions (\(\text{Ag}^{+}\)) react with chloride ions (\(\text{Cl}^{-}\)) to form solid silver chloride (\(\text{AgCl}\)). This process can be represented by the following reaction equation:
\[\text{Ag}^{+}(aq) + \text{Cl}^{-}(aq) \rightarrow \text{AgCl}(s)\]
The formation of an insoluble solid from soluble ions is what makes this a precipitation reaction. These solid particles settle out of the solution, and you can often see this as a cloudy formation in the mixture. \(\text{AgCl}\) is particularly famous for its low solubility in water, which makes the precipitation almost instantaneous.
Precipitation reactions are useful in various applications. They can help in the purification of water by removing unwanted ions, determining unknown concentrations, or synthesizing new compounds. In this exercise, we are focusing on how much heat (\(\Delta H\)) is either absorbed or released when the reaction occurs.

Stoichiometry

Stoichiometry in chemistry involves the calculation of reactants and products in a chemical reaction. It allows us to predict quantities based on the ratios established by the balanced chemical equation. For the reaction between silver ions and chloride ions: \[\text{Ag}^{+}(aq) + \text{Cl}^{-}(aq) \rightarrow \text{AgCl}(s)\]
The stoichiometric coefficients (which are all 1 in this equation) tell us that one mole of silver ions reacts with one mole of chloride ions to produce one mole of silver chloride. This process is straightforward, which is evident in our calculations.

• When you need to find out how much heat is released or absorbed, first determine the number of moles involved.
• Next, multiply the moles by the enthalpy change per mole (\(\Delta H\)), which is negative in this case because the reaction releases heat.

Using stoichiometry helps ensure accurate calculations, even when the amount of product or reactant is not in such neat mole quantities. Remember that these proportions must always be respected to maintain the integrity of any calculations.

Enthalpy Calculations

Enthalpy is a measure of heat content in a chemical reaction. In our case, the change in enthalpy (\(\Delta H\)) tells us whether the reaction absorbs heat (endothermic) or releases it (exothermic). For silver chloride precipitation, \(\Delta H = -65.5\ \text{kJ/mol}\) indicates the process is exothermic.
The formula used for calculating \(\Delta H\) is:
\[\Delta H = \Delta H_{\text{per mole}} \times \text{number of moles}\]

• For example, when producing 0.450 mol of \(\text{AgCl}\), the enthalpy change would be \(-65.5\ \text{kJ/mol} \times 0.450\ \text{mol} = -29.475\ \text{kJ}\).
• Similarly, for 9.00 g of \(\text{AgCl}\), it's essential first to convert this mass into moles: \(\text{Moles} = \frac{9.00\ \text{g}}{143.32\ \text{g/mol}}\), which is then used to calculate \(\Delta H\).

Enthalpy calculations provide insight into the energy changes in chemical reactions, helping us understand and quantify them for practical applications.

Dissolution Reaction

A dissolution reaction occurs when a solid compound breaks down into its ions in a solution. For \(\text{AgCl}\), the dissolution can be described as the reverse of precipitation:
\[\text{AgCl}(s) \rightarrow \text{Ag}^{+}(aq) + \text{Cl}^{-}(aq)\]
Here, the solid \(\text{AgCl}\) dissociates into its ionic components in water. This process requires an uptake of energy, since \(\text{AgCl}\) is insoluble in water under normal circumstances.
This is reflected in the positive \(\Delta H\) when calculating the enthalpy change for dissolution: reversing the original \(\Delta H\) value given for precipitation. For 9.25 x 10^-4 mol of \(\text{AgCl}\) dissociation, the calculation is:
\[\Delta H = 65.5\ \text{kJ/mol} \times 9.25 \times 10^{-4}\ \text{mol}\]
Dissolution reactions are key in many chemical fields, as they are fundamental to processes like solubility studies and reactions in aqueous environments.