Calculate the heat of neutralisation from the following data:
`200mL` of `1M HCI` is mixed with `400mL` of `0.5M NaOH`. The temperature rise in calorimeter was found to be `4.4^(@)C`. Water equivalent of calorimeter is `12g` and specific heat is `1cal mL^(-1) degree^(-1)` for solution.

Options

A. `-13.464 kcal`
B. `-12.02 kcal`
C. `-15.262 kcal`
D. `-20 . 92 kcal`

Explanation

To calculate the heat of neutralization from the given data, we can follow these steps:

### Step 1: Calculate the total volume of the solution
We have:
- Volume of HCl = 200 mL
- Volume of NaOH = 400 mL

Total volume of the solution = Volume of HCl + Volume of NaOH
\[ \text{Total Volume} = 200 \, \text{mL} + 400 \, \text{mL} = 600 \, \text{mL} \]

### Step 2: Calculate the milliequivalents of HCl and NaOH
Milliequivalents (mEq) can be calculated using the formula:
\[ \text{mEq} = \text{Normality} \times \text{Volume (mL)} \]

For HCl:
- Normality of HCl = 1 N (since HCl is a strong acid and fully dissociates)
- Volume of HCl = 200 mL

\[ \text{mEq of HCl} = 1 \, \text{N} \times 200 \, \text{mL} = 200 \, \text{mEq} \]

For NaOH:
- Normality of NaOH = 0.5 N (since NaOH is a strong base and fully dissociates)
- Volume of NaOH = 400 mL

\[ \text{mEq of NaOH} = 0.5 \, \text{N} \times 400 \, \text{mL} = 200 \, \text{mEq} \]

### Step 3: Determine the heat evolved in the reaction
The heat evolved can be calculated using the formula:
\[ q = m \cdot C \cdot \Delta T \]
where:
- \( m \) = mass (in grams) = volume (in mL) since the density of the solution is approximately 1 g/mL
- \( C \) = specific heat capacity (given as 1 cal/mL/°C)
- \( \Delta T \) = temperature change (given as 4.4 °C)

**Heat from the calorimeter:**
- Water equivalent of calorimeter = 12 g
- Specific heat = 1 cal/g/°C
- Temperature rise = 4.4 °C

\[ q_{\text{calorimeter}} = 12 \, \text{g} \times 1 \, \text{cal/g/°C} \times 4.4 \, \text{°C} = 52.8 \, \text{cal} \]

**Heat from the solution:**
- Mass of solution = 600 g (since 600 mL of solution)
- Specific heat = 1 cal/mL/°C
- Temperature rise = 4.4 °C

\[ q_{\text{solution}} = 600 \, \text{g} \times 1 \, \text{cal/g/°C} \times 4.4 \, \text{°C} = 2640 \, \text{cal} \]

### Step 4: Total heat evolved
\[ q_{\text{total}} = q_{\text{calorimeter}} + q_{\text{solution}} = 52.8 \, \text{cal} + 2640 \, \text{cal} = 2692.8 \, \text{cal} \]

### Step 5: Calculate the heat of neutralization per 1000 mEq
Since we have 200 mEq of HCl and NaOH, we can find the heat of neutralization for 1000 mEq:
\[ \text{Heat of neutralization for 200 mEq} = 2692.8 \, \text{cal} \]
\[ \text{Heat of neutralization for 1000 mEq} = \frac{1000}{200} \times 2692.8 \, \text{cal} = 5 \times 2692.8 \, \text{cal} = 13464 \, \text{cal} \]

### Step 6: Convert to kilocalories
\[ \text{Heat of neutralization} = \frac{13464 \, \text{cal}}{1000} = 13.464 \, \text{kcal} \]

### Step 7: Final answer
Since the reaction is exothermic, we express it as:
\[ \Delta H = -13.464 \, \text{kcal} \]

### Conclusion
The heat of neutralization is **-13.464 kcal**.

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