When 1 mole of strong diprotic acid `H_(2)A` (completely ionised) is completely neutralized by `NaOH` then amount of heat is released is `"E" KJ mol^(-1)` . Calculate amount of heat relased when one mole of weak monoprotic acid `HB` completely neutralized by `NaOH`.

Options

A. `E`
B. less than `E//2`
C. greater than `E//2`
D. None of these

Explanation

To solve the problem, we need to analyze the heat released during the neutralization of a strong diprotic acid and a weak monoprotic acid.

### Step-by-Step Solution:

1. **Understanding the Strong Diprotic Acid Reaction**:
- The strong diprotic acid \( H_2A \) completely ionizes and reacts with sodium hydroxide (\( NaOH \)).
- The reaction can be written as:
\[ H_2A + 2 NaOH \rightarrow 2 NaA + 2 H_2O \]
- Given that the heat released during this reaction is \( E \) kJ/mol for 1 mole of \( H_2A \).

2. **Understanding the Weak Monoprotic Acid Reaction**:
- The weak monoprotic acid \( HB \) only has one acidic hydrogen and reacts with sodium hydroxide (\( NaOH \)).
- The reaction can be written as:
\[ HB + NaOH \rightarrow NaB + H_2O \]
- Since \( HB \) is a weak acid, it does not completely ionize like the strong acid \( H_2A \).

3. **Calculating Heat Released for Weak Monoprotic Acid**:
- For the diprotic acid \( H_2A \), the heat released is \( E \) kJ for 2 moles of acidic hydrogen.
- Therefore, for 1 mole of \( HB \) (which has only 1 acidic hydrogen), the heat released will be half of that of the diprotic acid.
- Thus, the heat released when 1 mole of weak monoprotic acid \( HB \) is neutralized is:
\[ \Delta H = \frac{E}{2} \text{ kJ/mol} \]

4. **Conclusion**:
- The amount of heat released when one mole of weak monoprotic acid \( HB \) is completely neutralized by \( NaOH \) is \( \frac{E}{2} \) kJ/mol.