Problem 39 - Chemistry

Question

A quantity of \(2.00 \times 10^{2} \mathrm{~mL}\) of \(0.862 \mathrm{M} \mathrm{HCl}\) is mixed with \(2.00 \times 10^{2} \mathrm{~mL}\) of \(0.431 \mathrm{M} \mathrm{Ba}(\mathrm{OH})_{2}\) in a constant-pressure calorimeter of negligible heat capacity. The initial temperature of the \(\mathrm{HCl}\) and \(\mathrm{Ba}(\mathrm{OH})_{2}\) solutions is the same at \(20.48^{\circ} \mathrm{C}\). For the process $$\mathrm{H}^{+}(a q)+\mathrm{OH}^{-}(a q) \longrightarrow \mathrm{H}_{2} \mathrm{O}(l)$$ the heat of neutralization is \(-56.2 \mathrm{~kJ} / \mathrm{mol}\). What is the final temperature of the mixed solution? Assume the specific heat of the solution is the same as that for pure water.

Step-by-Step Solution

Answer

The final temperature of the solution is 14.69°C.

Step 1: Calculate Moles of Reactants

First, find the moles of HCl and Ba(OH)₂ in each solution. The volume of each solution is 200 mL, which is equivalent to 0.2 L. For HCl: \( moles = 0.862 \, M \times 0.2 \, L = 0.1724 \, moles \). For Ba(OH)₂, since it provides two OH⁻ ions per molecule: \( moles = 0.431 \, M \times 0.2 \, L = 0.0862 \, moles \times 2 = 0.1724 \, moles \).

Step 2: Determine Limiting Reagent and Reaction Extent

The balanced reaction for neutralization is:\[ H^+(aq) + OH^-(aq) \rightarrow H_2O(l) \]Both HCl and Ba(OH)₂ provide quantities of ions sufficient for complete neutralization, so they react in a 1:1 ratio. Since both provide equal moles of H⁺ and OH⁻ ions, all are used in the reaction.

Step 3: Calculate Heat Evolved in Reaction

Since the reaction involves \(0.1724 \, mol\) of \(H^+\) and \(OH^-\), and the enthalpy change (heat of neutralization) is \(-56.2 \, kJ/mol\), the total heat evolved \( q \) is:\[ q = 0.1724 \, mol \times (-56.2 \, \frac{kJ}{mol}) = -9.69 \, kJ \]

Step 4: Calculate Temperature Change

The total mass of the solution is calculated by assuming the density of water (\(1 \,g/mL\)), thus 400 mL corresponds to 400 g. Using the formula for heat transfer \( q = mc\Delta T \), where \( m = 400 \, g \) and the specific heat \( c = 4.18 \, J/g \cdot ^\circ C \):\[ -9690 \, J = 400 \, g \times 4.18 \, \frac{J}{g^\circ C} \times \Delta T \]Solving for \( \Delta T \), we find:\[ \Delta T = \frac{-9690}{400 \times 4.18} = -5.79 ^\circ C \]

Step 5: Find Final Temperature

Subtract the temperature change from the initial temperature to find the final temperature:\[ T_{final} = 20.48^\circ C - 5.79^\circ C = 14.69^\circ C \]

Key Concepts

Neutralization Reaction

When an acid and a base react together, a neutralization reaction occurs. In these reactions, typically a hydrogen ion, \( \text{H}^+ \), from the acid reacts with a hydroxide ion, \( \text{OH}^- \), from the base to form water. In the given exercise, hydrochloric acid \( (\text{HCl}) \) reacts with barium hydroxide \( (\text{Ba(OH)}_2) \) to achieve this neutralization. The overall process can be simplified to:

• \( \text{H}^+(aq) + \text{OH}^-(aq) \rightarrow \text{H}_2\text{O}(l) \)

The reaction is highly exothermic, meaning it releases heat. This release of heat is an essential aspect to consider when calculating changes in temperature during the reaction. Neutralization reactions are particularly significant in calorimetry, as they enable scientists to determine the heat exchange between the chemical system and its surroundings.

Enthalpy Change

Enthalpy change \( (\Delta H) \) measures the heat change at constant pressure during a chemical reaction. For neutralization reactions, a known enthalpy change allows us to calculate the heat evolved or absorbed. In this exercise, the enthalpy change of the neutralization reaction is given as \(-56.2 \text{ kJ/mol}\). This negative sign indicates heat is released, confirming the reaction is exothermic.To find the total heat evolved during the reaction, multiply the moles of the reacting substances by the enthalpy change. Using the exercise's values:

• Total heat \( q = 0.1724 \text{ mol} \times (-56.2 \text{ kJ/mol}) = -9.69 \text{ kJ} \)

This calculation derives the heat change as \(-9.69 \text{ kJ}\), indicating that the reaction releases this amount of energy as heat into the surroundings. Understanding enthalpy change is crucial for grasping the influence of reactions on temperature changes in calorimetric experiments.

Limiting Reagent

In chemical reactions, the limiting reagent is the reactant that is completely consumed first, limiting the extent of the reaction. To identify it, compare the moles of each reactant based on the reaction stoichiometry. In this instance:

• Hydrochloric acid \( \text{(HCl)} \) provides \(0.1724 \text{ mol} \)
• Barium hydroxide \( \text{(Ba(OH)}_2) \) provides \(0.0862 \text{ mol} \), but as it offers 2 \( \text{OH}^- \) ions per molecule, it effectively provides \(0.1724 \text{ mol} \) of \( \text{OH}^- \)

Both reactants provide equal moles of \( \text{H}^+ \) and \( \text{OH}^- \) ions. Hence, neither reactant limits the reaction since all ions react completely with each other, ensuring a complete neutralization without excess. Identifying limiting reagents is pivotal in determining how much product can form and calculating the heat released accurately.

Specific Heat Capacity

Specific heat capacity \( (c) \) is the amount of heat required to raise the temperature of a unit mass of a substance by one degree Celsius. For this exercise, it's assumed to be the same as water's, which is \(4.18 \text{ J/g} \cdot ^\circ \text{C} \).The equation for heat transfer \( q = mc\Delta T \) involves calculating the mass \( m \) of the solution, its specific heat \( c \), and the temperature change \( \Delta T \). Given:

• Volume = \( 400 \text{ mL} \) translates to mass \( 400 \text{ g} \)
• Temperature change \( \Delta T = \frac{-9690}{400 \times 4.18} \)

Using the values in this calculation:

• \( \Delta T = -5.79 ^\circ \text{C} \)

Incorporating specific heat capacity is critical to determine how much temperature change occurs as a result of the heat released in the reaction, completing our understanding of the reaction in calorimetry.