Question

In a coffee-cup calorimeter, \(50.0 \mathrm{~mL}\) of \(0.100 \mathrm{M} \mathrm{AgNO}_{3}\) and \(50.0 \mathrm{~mL}\) of \(0.100 \mathrm{M} \mathrm{HCl}\) are mixed to yield the following reaction: $$ \mathrm{Ag}^{+}(a q)+\mathrm{Cl}^{-}(a q) \longrightarrow \operatorname{AgCl}(s) $$ The two solutions were initially at \(22.60^{\circ} \mathrm{C}\), and the final temperature is \(23.40^{\circ} \mathrm{C}\). Calculate the heat that accompanies this reaction in \(\mathrm{kJ} / \mathrm{mol}\) of \(\mathrm{AgCl}\) formed. Assume that the combined solution has a mass of \(100.0 \mathrm{~g}\) and a specific heat capacity of \(4.18 \mathrm{~J} /{ }^{\circ} \mathrm{C} \cdot \mathrm{g} .\)

Short answer

The heat that accompanies the reaction of \(\mathrm{Ag}^{+}(a q)+\mathrm{Cl}^{-}(a q) \longrightarrow \operatorname{AgCl}(s)\) is 66.9 \(\mathrm{kJ} / \mathrm{mol}\) of \(\mathrm{AgCl}\) formed.

Step-by-step solution

Calculate the moles of AgCl formed

We are given that 50.0 mL of 0.100 M \(\mathrm{AgNO}_{3}\) solution is mixed with 50.0 mL of 0.100 M \(\mathrm{HCl}\) solution. The reaction is: $$ \mathrm{Ag}^{+}(a q)+\mathrm{Cl}^{-}(a q) \longrightarrow \operatorname{AgCl}(s) $$ Since the reaction has a 1:1 stoichiometry, the amounts of \(\mathrm{Ag}^{+}\) and \(\mathrm{Cl}^{-}\) that react are equal. So we can find the moles of \(\mathrm{AgCl}\) formed by finding the moles of either \(\mathrm{Ag}^{+}\) or \(\mathrm{Cl}^{-}\) in the initial solutions and taking the lowest amount as the limiting reagent. Moles of \(\mathrm{Ag}^{+}\) = Molarity × Volume = 0.100 mol/L × 0.0500 L = 0.00500 mol Moles of \(\mathrm{Cl}^{-}\) = Molarity × Volume = 0.100 mol/L × 0.0500 L = 0.00500 mol Both have equal moles, so either can be considered the limiting reagent. Thus, 0.00500 mol of \(\mathrm{AgCl}\) is formed.

Calculate the heat transferred

The two solutions are initially at 22.60 \(^\circ\mathrm{C}\) and finally reach a temperature of 23.40 \(^\circ\mathrm{C}\). The mass of the combined solution is 100.0 g, and its specific heat capacity is 4.18 \(\mathrm{~J} / { }^{\circ} \mathrm{C} \cdot \mathrm{g}\). To calculate the heat transferred, we will use the formula: q = mcΔT where: - q is the heat transferred - m is the mass of the combined solution (100.0 g) - c is the specific heat capacity (4.18 \(\mathrm{~J} /{ }^{\circ} \mathrm{C} \cdot \mathrm{g}\)) - ΔT is the change in temperature (23.40 \(^\circ\mathrm{C}\) - 22.60 \(^\circ\mathrm{C}\)) q = (100.0 g) × (4.18 \(\mathrm{~J} /{ }^{\circ} \mathrm{C} \cdot \mathrm{g}\)) × (23.40 \(^\circ\mathrm{C}\) - 22.60 \(^\circ\mathrm{C}\)) q = (100.0 g) × (4.18 \(\mathrm{~J} /{ }^{\circ} \mathrm{C} \cdot \mathrm{g}\)) × (0.80 \(^\circ\mathrm{C}\)) q = 334.4 J

Calculate the heat per mole of AgCl

Now we have the total heat transferred (q) and the moles of \(\mathrm{AgCl}\) formed. To calculate the heat per mole of \(\mathrm{AgCl}\) formed, we will use the formula: Heat per mole of AgCl = \(\frac{q}{n}\) where: - q is the heat transferred (334.4 J) - n is the moles of \(\mathrm{AgCl}\) formed (0.00500 mol) Heat per mole of AgCl = \(\frac{334.4 \mathrm{~J}}{0.00500 \mathrm{~mol}}\) = 66880 J/mol Now we will convert the heat per mole of AgCl from J/mol to kJ/mol: 66880 J/mol × \(\frac{1 \mathrm{~kJ}}{1000 \mathrm{~J}}\) = 66.9 \(\mathrm{kJ} / \mathrm{mol}\) The heat that accompanies this reaction is 66.9 \(\mathrm{kJ} / \mathrm{mol}\) of \(\mathrm{AgCl}\) formed.

Key concepts

Enthalpy Change

In a chemical reaction, the enthalpy change is the heat absorbed or released under constant pressure. In the context of calorimetry, we measure this heat to find how much energy is involved in a particular chemical reaction. When determining the enthalpy change for the formation of a compound such as AgCl in our reaction, the measured heat (q) of the system is considered equal to the enthalpy change per mole when recorded under constant pressure.

The heat change is calculated using the formula:

• \( q = mc\Delta T \)
• where \( m \) is the mass of the solution, \( c \) is the specific heat capacity, and \( \Delta T \) is the temperature change.

It's essential to convert this value into a per-mole basis to understand how much energy is released or absorbed per mole of the product formed, which in our example was calculated to be 66.9 \( \mathrm{kJ} / \mathrm{mol} \) of AgCl.

Stoichiometry

Stoichiometry is fundamental for understanding the quantitative aspects of chemical reactions. It involves the calculation of reactants and products in a chemical reaction. In our problem, stoichiometry helps us determine the amount of AgCl produced and understand the ratio in which reactants react.

The reaction equation provided:

• \( \mathrm{Ag}^{+}(a q) + \mathrm{Cl}^{-}(a q) \longrightarrow \operatorname{AgCl}(s) \)

is a straightforward example of a 1:1 stoichiometry. This means that one mole of \( \mathrm{Ag}^{+} \) ions reacts with one mole of \( \mathrm{Cl}^{-} \) ions to produce one mole of \( \mathrm{AgCl} \).

By calculating the moles from the molarity and volume of each reactant, and recognizing the 1:1 ratio, we find that 0.00500 moles of \( \mathrm{AgCl} \) are formed in this process.

Specific Heat Capacity

Specific heat capacity is the amount of heat required to change the temperature of one gram of a substance by one degree Celsius. It's a crucial property in calorimetry because it determines how much energy is needed to heat up or cool down a given quantity of the material.

For our solution, we used a specific heat capacity of:

• 4.18 \( \mathrm{~J} /{ }^{\circ} \mathrm{C} \cdot \mathrm{g} \)

This value represents the average specific heat of water, which in this case is a good approximation for the dilute aqueous solutions of the reactants. By knowing the mass of the solution is 100.0 g and the change in temperature is 0.8 \( ^\circ \mathrm{C} \), we use the specific heat capacity to calculate the heat absorbed by the solution.

Understanding this concept helps in determining how energy flows in the system during the reaction.

Limiting Reagent

In chemical reactions, the limiting reagent is the reactant that gets completely consumed first, determining the maximum amount of product that can be formed. Identifying the limiting reagent is essential in stoichiometry calculations.

In our calorimetry problem, both solutions of \( \mathrm{AgNO}_{3} \) and \( \mathrm{HCl} \) were mixed in equal molar concentrations and volumes: 0.100 M solutions at 50.0 mL each. Therefore, both \( \mathrm{Ag}^{+} \) ions and \( \mathrm{Cl}^{-} \) ions were present in equal amounts (0.00500 mol each). Because of the 1:1 stoichiometry of the reaction, either of these could be the limiting reagent, leading to the formation of precisely 0.00500 mol of \( \mathrm{AgCl} \).

With both reactants limiting in this particular example, it simplifies calculations since the amount of product formed depends directly on the initial concentration and volume of the limiting reactants.