Problem 95 - Chemistry and Chemical Reactivity

Question

Insoluble AgCl(s) precipitates when solutions of \(\mathrm{AgNO}_{3}(\mathrm{aq})\) and \(\mathrm{NaCl}(\mathrm{aq})\) are mixed. \(\operatorname{AgNO}_{3}(\mathrm{aq})+\mathrm{NaCl}(\mathrm{aq}) \rightarrow \mathrm{AgCl}(\mathrm{s})+\mathrm{NaNO}_{3}(\mathrm{aq})\) $$ \Delta_{i} H^{0}=? $$ To measure the energy evolved in this reaction, 250\. mL. of 0.16 M AgNO \(_{3}\) (aq) and 125 mL. of \(0.32 \mathrm{M} \mathrm{NaCl}(\mathrm{aq})\) are mixed in a coffee-cup calorimeter. The temperature of the mixture rises from \(21.15^{\circ} \mathrm{C}\) to \(22.90^{\circ} \mathrm{C} .\) Calculate the enthalpy change for the precipitation of \(\mathrm{AgCl}(\mathrm{s}),\) in \(\mathrm{kJ} / \mathrm{mol}\). (Assume the density of the solution is \(1.0 \mathrm{g} / \mathrm{mL}\) and its specific heat capacity is \(4.2 \mathrm{J} / \mathrm{g} \cdot \mathrm{K} .)\)

Step-by-Step Solution

Answer

The enthalpy change for the reaction is \(-68.91 \text{ kJ/mol}\).

Step 1: Calculate the Moles of Reactants

First, calculate the moles of \( \mathrm{AgNO}_3 \) using the formula: \( \text{moles} = \text{Volume (L)} \times \text{Concentration (M)} \). For \( \mathrm{AgNO}_3 \), the volume is 0.250 L and the concentration is 0.16 M, so: \[ \text{moles of } \mathrm{AgNO}_3 = 0.250 \times 0.16 = 0.040 \text{ mol} \].\Similarly, calculate the moles of \( \, \mathrm{NaCl} \, \) using 0.125 L and 0.32 M: \[ \text{moles of } \mathrm{NaCl} = 0.125 \times 0.32 = 0.040 \text{ mol} \].

Step 2: Determine the Limiting Reactant

Since both \( \mathrm{AgNO}_3 \) and \( \mathrm{NaCl} \) have equal moles (0.040 mol), either reactant can be used as the limiting reactant for this reaction. This means both react completely to form \( \mathrm{AgCl} \).

Step 3: Calculate the Heat Absorbed by the Solution

Next, use \( q = mc\Delta T \) to find the heat absorbed by the solution. The total mass \( m \) of the solution is the sum of the volumes of the solutions (375 mL) times the density (1.0 g/mL), which equals 375 grams. \( c \) is the specific heat capacity (4.2 J/g·K). \( \Delta T \) is the change in temperature: \( 22.90 - 21.15 = 1.75 \) °C. Thus, the heat \( q \) is: \[ q = 375 \times 4.2 \times 1.75 = 2756.25 \text{ J} \].

Step 4: Convert Heat Absorbed to kJ

Since we are working with energies in kJ, convert 2756.25 J to kJ: \[ q = \frac{2756.25}{1000} = 2.75625 \text{ kJ} \].

Step 5: Calculate the Enthalpy Change per Mole of \( \mathrm{AgCl} \) Formed

The enthalpy change for the reaction is based on the formation of \( \mathrm{AgCl} \). Since 0.040 mol of AgCl formed, the enthalpy change \( \Delta H^0 \) is: \[ \Delta H^0 = \frac{2.75625}{0.040} = 68.90625 \text{ kJ/mol} \].Note: As the temperature increased, the reaction releases heat, making \( \Delta H^0 \) negative, so \( \Delta H^0 = -68.90625 \text{ kJ/mol} \).

Key Concepts

Precipitation Reaction

A precipitation reaction is a fascinating chemical reaction that occurs when two soluble salts are combined in a solution, resulting in the formation of an insoluble solid known as the precipitate. In the context of our exercise, when solutions of silver nitrate (AgNO extsubscript{3}) and sodium chloride (NaCl) are mixed, they undergo a precipitation reaction to produce solid silver chloride (AgCl).
This results in the equation: \[\operatorname{AgNO}_{3}(\mathrm{aq}) + \mathrm{NaCl}(\mathrm{aq}) \rightarrow \mathrm{AgCl}(\mathrm{s}) + \mathrm{NaNO}_{3}(\mathrm{aq})\] Here, the precipitate AgCl falls out of the solution as a solid because it is not soluble in water.
Precipitation reactions are commonly used in chemistry to isolate specific compounds from a solution, making them incredibly useful for purification processes. This reaction is also a perfect example of how different ions in solution can interact to form an insoluble product.

Limiting Reactant

The limiting reactant in a chemical reaction is the substance that is completely consumed first, thus determining the maximum amount of product that can be formed. In our experiment involving AgNO extsubscript{3} and NaCl, both reactants have equivalent moles of 0.040 mol, meaning that they are present in the stoichiometric ratio.
In such a case, either reactant can be considered as the limiting reactant as both will be completely consumed to form the precipitate AgCl.
Understanding the concept of a limiting reactant allows chemists to predict the amount of product formed and to effectively use reactants without wastage. It is crucial in yield optimization of chemical processes.

Calorimetry

Calorimetry is the process of measuring the amount of heat released or absorbed during a chemical reaction. This exercise makes use of a coffee-cup calorimeter, a simple yet effective tool for measuring enthalpy changes in solution-based reactions. By mixing two solutions and measuring the temperature change, the heat exchanged with the surroundings is calculated.
The equation used here is:\[q = mc\Delta T\]where:

• \(q\) is the heat exchanged.
• \(m\) is the mass of the solution.
• \(c\) is the specific heat capacity.
• \(\Delta T\) is the temperature change.

In this experiment, a rise in temperature signifies an exothermic reaction, where heat is released into the surroundings.

AgCl Formation

Silver chloride (AgCl) is a notable compound not just because of its interesting chemical properties, but also due to its applications such as its role in photography and electrochemistry. In our precipitation reaction, AgCl is formed when the nitrate and chloride ions dissociate from their salts and then recombine to form the stable and insoluble compound AgCl.
The chemical process of AgCl formation is represented as:\[\text{Ag}^+ (\text{aq}) + \text{Cl}^- (\text{aq}) \rightarrow \text{AgCl} (\text{s})\]This solid's lack of solubility in water is key, leading to its use in many filtering and separation processes as well. Because AgCl is sensitive to light, it is used in photographic film as a light-sensitive material. Its unique properties emphasize its significance beyond mere laboratory exercises.