It is known that entropy of neutralisation of a strong acid and strong base is -13.68kcal/mol.If enthalpy of neutralisation of `N_2H_4`(hydrazine) with strong acid is equal to -11.68 kcal/mol, then enthalpy of the reaction
`N_2H_4+H_2OtoN_2H_5^+(aq)+OH^(-)(aq)` in Kcal/mol is :

Explanation

To solve the problem, we need to determine the enthalpy of the reaction:

\[ N_2H_4 + H_2O \rightarrow N_2H_5^+ (aq) + OH^− (aq) \]

We are given the following information:

1. The entropy of neutralization of a strong acid and strong base is \(-13.68 \, \text{kcal/mol}\).
2. The enthalpy of neutralization of hydrazine (\(N_2H_4\)) with a strong acid is \(-11.68 \, \text{kcal/mol}\).

### Step-by-Step Solution:

**Step 1: Understand the given values.**
- The enthalpy of neutralization for a strong acid and strong base is a reference value, indicating the heat released during the neutralization process.
- The enthalpy of neutralization of \(N_2H_4\) with a strong acid is less than that of the strong acid-strong base neutralization.

**Step 2: Calculate the difference in enthalpy.**
- We can find how much less the enthalpy of neutralization of \(N_2H_4\) is compared to the strong acid-strong base neutralization.

\[ \Delta H_{\text{difference}} = \Delta H_{\text{strong acid-strong base}} - \Delta H_{\text{N2H4 with strong acid}} \]

Substituting the values:

\[ \Delta H_{\text{difference}} = -13.68 \, \text{kcal/mol} - (-11.68 \, \text{kcal/mol}) = -13.68 + 11.68 = -2 \, \text{kcal/mol} \]

**Step 3: Interpret the result.**
- The negative value indicates that the enthalpy of neutralization of \(N_2H_4\) is less by \(2 \, \text{kcal/mol}\) compared to the strong acid-strong base neutralization.

**Step 4: Determine the enthalpy of the reaction.**
- Since the reaction involves the ionization of hydrazine, we can conclude that the energy consumed for ionization is equal to the difference calculated.

Thus, the enthalpy of the reaction \(N_2H_4 + H_2O \rightarrow N_2H_5^+ (aq) + OH^− (aq)\) is:

\[ \Delta H = 2 \, \text{kcal/mol} \]

### Final Answer:
The enthalpy of the reaction is \(2 \, \text{kcal/mol}\).