Step-by-Step Solution

The problem asks for the enthalpy of dissociation of acetic acid, which can be determined by comparing the heat evolved from its neutralization with a strong base (Experiment 2) to the heat evolved from the neutralization of a strong acid with a strong base (Experiment 1). First, we use Experiment 1 to find the heat capacity of the calorimeter (calorimeter constant, C).

Part 1: Determine the Calorimeter Constant (C) from Experiment 1

1. Reaction: The neutralization reaction between a strong acid (HCl) and a strong base (NaOH) is: $\text{H}^+(aq) + \text{OH}^-(aq) \rightarrow \text{H}_2\text{O}(l)$ The standard enthalpy of this reaction is given as $\Delta H_{neut} = -57.0 \text{ kJ/mol}$.

2. Calculate moles of reactants:

   • Moles of HCl = Molarity $\times$ Volume = $1.0 \text{ mol/L} \times 0.100 \text{ L} = 0.1 \text{ mol}$
   • Moles of NaOH = Molarity $\times$ Volume = $1.0 \text{ mol/L} \times 0.100 \text{ L} = 0.1 \text{ mol}$ Since the moles are equal, 0.1 mol of water is formed.

3. Calculate the heat released by the reaction ($q_{rxn}$): $q_{rxn} = \text{moles} \times (-\Delta H_{neut}) = 0.1 \text{ mol} \times (57.0 \text{ kJ/mol}) = 5.7 \text{ kJ} = 5700 \text{ J}$

4. Calculate the heat absorbed by the solution and calorimeter ($q_{abs}$):

   • Total volume of the solution = 100 mL + 100 mL = 200 mL.
   • Density of the solution = 1.0 g/mL, so mass of the solution ($m_s$) = 200 mL $\times$ 1.0 g/mL = 200 g.
   • Specific heat capacity of the solution ($c_s$) = 4.2 J/gK.
   • Temperature rise, $\Delta T_1 = 5.7^\circ\text{C} = 5.7 \text{ K}$.
   • The heat absorbed is given by $q_{abs} = (m_s c_s + C) \Delta T_1$, where C is the calorimeter constant. $q_{abs} = (200 \text{ g} \times 4.2 \text{ J/gK} + C) \times 5.7 \text{ K} = (840 + C) \times 5.7 \text{ J}$

5. Equate heat released and absorbed to find C: For an insulated system, $q_{rxn} = q_{abs}$. $5700 = (840 + C) \times 5.7$ $\frac{5700}{5.7} = 840 + C$ $1000 = 840 + C$ $C = 1000 - 840 = 160 \text{ J/K}$

Part 2: Determine the Enthalpy of Dissociation from Experiment 2

1. Reaction: The neutralization of a weak acid (CH₃COOH) with a strong base (NaOH) is: $\text{CH}_3\text{COOH}(aq) + \text{OH}^-(aq) \rightarrow \text{CH}_3\text{COO}^-(aq) + \text{H}_2\text{O}(l)$

2. Calculate moles of reactants:

   • Moles of CH₃COOH = $2.0 \text{ mol/L} \times 0.100 \text{ L} = 0.2 \text{ mol}$
   • Moles of NaOH = $1.0 \text{ mol/L} \times 0.100 \text{ L} = 0.1 \text{ mol}$ NaOH is the limiting reactant, so the reaction proceeds based on 0.1 mol.

3. Calculate the heat released in Experiment 2 ($q_2$): The experimental conditions are identical, so the total mass and specific heat capacity are the same. The total heat capacity of the system (solution + calorimeter) is $m_s c_s + C = 840 \text{ J/K} + 160 \text{ J/K} = 1000 \text{ J/K}$.

   • The temperature rise is $\Delta T_2 = 5.6^\circ\text{C} = 5.6 \text{ K}$.
   • The heat released, which is absorbed by the system, is: $q_2 = (\text{Total heat capacity}) \times \Delta T_2 = 1000 \text{ J/K} \times 5.6 \text{ K} = 5600 \text{ J}$

4. Calculate the molar enthalpy of neutralization for acetic acid ($\\\Delta H_{neut, weak}$): This heat ($q_2$) was released from the neutralization of 0.1 mol of acetic acid. $\Delta H_{neut, weak} = -\frac{q_2}{\text{moles}} = -\frac{5600 \text{ J}}{0.1 \text{ mol}} = -56000 \text{ J/mol} = -56.0 \text{ kJ/mol}$

5. Calculate the enthalpy of dissociation ($\\\Delta H_{diss}$): The neutralization of a weak acid involves two processes: (i) Dissociation of the weak acid: $\text{CH}_3\text{COOH}(aq) \rightarrow \text{H}^+(aq) + \text{CH}_3\text{COO}^-(aq)$ with enthalpy change $\Delta H_{diss}$. (ii) Neutralization of H⁺ by OH⁻: $\text{H}^+(aq) + \text{OH}^-(aq) \rightarrow \text{H}_2\text{O}(l)$ with enthalpy change $\Delta H_{neut} = -57.0 \text{ kJ/mol}$.

   By Hess's Law, the overall enthalpy change is the sum of the enthalpy changes of the steps: $\Delta H_{neut, weak} = \Delta H_{diss} + \Delta H_{neut}$ Substituting the known values: $-56.0 \text{ kJ/mol} = \Delta H_{diss} + (-57.0 \text{ kJ/mol})$ Solving for $\Delta H_{diss}$: $\Delta H_{diss} = -56.0 + 57.0 = 1.0 \text{ kJ/mol}$

Thus, the enthalpy of dissociation of acetic acid is 1.0 kJ/mol.