Heat of neutralization of HCl by NaOH is 13.7 kcal per equivalent
and by `NH_4OH` is 12.27 kcal. The heat of dissociation of
`NH_4OH` is

Options

A. `-25.97` kcal
B. 25.97 kcal
C. `-1.43` kcal
D. 1.43 kcal

Explanation

To find the heat of dissociation of \( NH_4OH \), we can break down the problem into steps as follows:

### Step 1: Understand the Neutralization Reactions
The neutralization of \( HCl \) by \( NaOH \) and \( NH_4OH \) can be represented as:
1. \( HCl + NaOH \rightarrow NaCl + H_2O \) with a heat of neutralization \( \Delta H_2 = -13.7 \, \text{kcal} \).
2. \( HCl + NH_4OH \rightarrow NH_4Cl + H_2O \) with a heat of neutralization \( \Delta H_1 = -12.27 \, \text{kcal} \).

### Step 2: Write the Dissociation Reaction of \( NH_4OH \)
The dissociation of \( NH_4OH \) can be represented as:
\[ NH_4OH \rightleftharpoons NH_4^+ + OH^- \]
Let the heat of dissociation of \( NH_4OH \) be \( \Delta H_d \).

### Step 3: Relate the Heats of Reaction
The total heat of the neutralization reaction involving \( NH_4OH \) can be expressed as the sum of the heat of dissociation and the heat of neutralization:
\[ \Delta H_{total} = \Delta H_d + \Delta H_2 \]
Where \( \Delta H_2 \) is the heat released when \( H^+ \) combines with \( OH^- \) to form water, which is \( -13.7 \, \text{kcal} \).

### Step 4: Set Up the Equation
From the previous steps, we can write:
\[ -12.27 = \Delta H_d + (-13.7) \]

### Step 5: Solve for \( \Delta H_d \)
Rearranging the equation gives:
\[ \Delta H_d = -12.27 + 13.7 \]
\[ \Delta H_d = 1.43 \, \text{kcal} \]

### Conclusion
Thus, the heat of dissociation of \( NH_4OH \) is \( 1.43 \, \text{kcal} \).