The enthalpy change when 1 L of 1 M `H _(2)SO_(4)` is completely neutralized by 1L of1 M `Ca(OH)_(2)` will be

Options

A. `-13.7 kcal`
B. `-27.4 kcal`
C. `-1.37 kcal`
D. `- 2.74 kcal`

Explanation

To find the enthalpy change when 1 L of 1 M \( H_2SO_4 \) is completely neutralized by 1 L of 1 M \( Ca(OH)_2 \), we can follow these steps:

### Step 1: Understand the Reaction
The neutralization reaction between sulfuric acid (\( H_2SO_4 \)) and calcium hydroxide (\( Ca(OH)_2 \)) can be represented as:
\[ H_2SO_4 + Ca(OH)_2 \rightarrow CaSO_4 + 2H_2O \]
This shows that one mole of \( H_2SO_4 \) reacts with one mole of \( Ca(OH)_2 \) to produce calcium sulfate and water.

### Step 2: Determine the Molarity and Volume
We have:
- 1 L of 1 M \( H_2SO_4 \) which means we have 1 mole of \( H_2SO_4 \).
- 1 L of 1 M \( Ca(OH)_2 \) which means we have 1 mole of \( Ca(OH)_2 \).

### Step 3: Calculate the Normality
For \( H_2SO_4 \):
- The valency factor (n) is 2 because it can donate 2 protons (H\(^+\)).
- Therefore, the normality of \( H_2SO_4 \) is \( 1 \, M \times 2 = 2 \, N \).

For \( Ca(OH)_2 \):
- The valency factor (n) is also 2 because it can donate 2 hydroxide ions (OH\(^-\)).
- Therefore, the normality of \( Ca(OH)_2 \) is \( 1 \, M \times 2 = 2 \, N \).

### Step 4: Calculate the Enthalpy Change
The enthalpy change for the neutralization of 1 mole of \( H_2SO_4 \) is given as -13.7 kcal. Since we have 1 mole of \( H_2SO_4 \) and it reacts completely with 1 mole of \( Ca(OH)_2 \), we can calculate the total enthalpy change for the reaction:
\[ \Delta H = -13.7 \, \text{kcal} \times 2 = -27.4 \, \text{kcal} \]

### Final Answer
The enthalpy change when 1 L of 1 M \( H_2SO_4 \) is completely neutralized by 1 L of 1 M \( Ca(OH)_2 \) is:
\[ \Delta H = -27.4 \, \text{kcal} \]

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