When `100mL` of `1.0M HCl` was mixed with `100 mL` of `1.0 M NaOH` in an insulated beaker at constant pressure, a temperature increase of `5.7^(@)C` was measured for the beaker and its contents (Expt.1). Because the enthalpy of neutralisation of a strong acid with a strong base is constant `(-57.0kJmol^(-1))` this experiment couldbe used to measure the calorimeter constant. In a second experiment (Expt.2), `100mL` of `2.0M` acetic acid `K_(a)=2.0xx10^(-5))` was mixed with `100mL` of `1.0M NaOH` (under identical conditions to Expt. 1) where a temperature rise of `5.6^(@)C` was measured.
Enthalpy of dissociation (in `kJ mol^(-1)`) of acetic acid obtained from the Expt. 2 is

Options

A. 1
B. 10
C. 24.5
D. 51.4

Explanation

To solve the problem, we need to calculate the enthalpy of dissociation of acetic acid based on the data provided from the two experiments. Here’s a step-by-step solution:

### Step 1: Calculate the heat evolved in Experiment 1
In Experiment 1, we mixed 100 mL of 1.0 M HCl with 100 mL of 1.0 M NaOH.

1. **Calculate moles of HCl and NaOH:**
\[ \text{Moles of HCl} = \text{Molarity} \times \text{Volume} = 1.0 \, \text{mol/L} \times 0.1 \, \text{L} = 0.1 \, \text{mol} \]
Since the reaction is 1:1, moles of NaOH = 0.1 mol.

2. **Calculate heat evolved:**
\[ \text{Heat evolved} = \text{Moles} \times \text{Enthalpy of neutralization} = 0.1 \, \text{mol} \times (-57.0 \, \text{kJ/mol}) = -5.7 \, \text{kJ} = -5700 \, \text{J} \]

### Step 2: Calculate the heat absorbed by the solution
Using the formula \( Q = m \cdot s \cdot \Delta T \):

1. **Calculate the mass of the solution:**
\[ \text{Mass} = \text{Density} \times \text{Volume} = 1 \, \text{g/mL} \times 200 \, \text{mL} = 200 \, \text{g} \]

2. **Calculate the heat absorbed by the solution:**
\[ Q_{\text{solution}} = m \cdot s \cdot \Delta T = 200 \, \text{g} \times 4.2 \, \text{J/g°C} \times 5.7 \, \text{°C} = 4788 \, \text{J} \]

### Step 3: Calculate the calorimeter constant
The heat absorbed by the calorimeter can be calculated as:
\[ Q_{\text{calorimeter}} = Q_{\text{total}} - Q_{\text{solution}} = -5700 \, \text{J} - 4788 \, \text{J} = -912 \, \text{J} \]

Using the calorimeter constant \( C \):
\[ C \cdot \Delta T = 912 \, \text{J} \quad \Rightarrow \quad C = \frac{912 \, \text{J}}{5.7 \, \text{°C}} \approx 160 \, \text{J/°C} \]

### Step 4: Calculate the heat evolved in Experiment 2
In Experiment 2, we mixed 100 mL of 2.0 M acetic acid with 100 mL of 1.0 M NaOH.

1. **Calculate moles of acetic acid:**
\[ \text{Moles of acetic acid} = 2.0 \, \text{mol/L} \times 0.1 \, \text{L} = 0.2 \, \text{mol} \]
However, only 0.1 mol of NaOH will react, so effectively 0.1 mol of acetic acid will react.

2. **Calculate heat absorbed by the solution:**
\[ Q_{\text{solution}} = 200 \, \text{g} \times 4.2 \, \text{J/g°C} \times 5.6 \, \text{°C} = 4704 \, \text{J} \]

3. **Calculate total heat evolved:**
\[ Q_{\text{total}} = Q_{\text{solution}} + Q_{\text{calorimeter}} = 4704 \, \text{J} + (160 \, \text{J/°C} \times 5.6 \, \text{°C}) = 4704 \, \text{J} + 896 \, \text{J} = 5600 \, \text{J} \]

### Step 5: Calculate the enthalpy of dissociation of acetic acid
The heat of dissociation for acetic acid can be calculated as:
\[ \Delta H_{\text{dissociation}} = Q_{\text{total}} - Q_{\text{neutralization (strong)}} = 5600 \, \text{J} - 5700 \, \text{J} = -100 \, \text{J} \]

Since this is for 0.1 mol, to find per mole:
\[ \Delta H_{\text{dissociation}} = \frac{-100 \, \text{J}}{0.1 \, \text{mol}} = -1000 \, \text{J/mol} = -1 \, \text{kJ/mol} \]

### Final Answer
The enthalpy of dissociation of acetic acid obtained from Experiment 2 is **1 kJ/mol**.

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