If heat of dissocitation of `CHCl_(2)COOH` is 0.7 kcal/mole then `Delta H` for the reaction:
`CHCl_(2)COOH+KOH rarrCHCl_(2)COOK+H_(2)O`

Options

A. `-13` kcal
B. `+13` kcal
C. `-14.4` kcal
D. `-13.7` kcal

Answer

To solve the problem, we need to determine the change in enthalpy (ΔH) for the reaction:

\[ \text{CHCl}_2\text{COOH} + \text{KOH} \rightarrow \text{CHCl}_2\text{COOK} + \text{H}_2\text{O} \]

### Step 1: Identify the given values
- Heat of dissociation of CHCl₂COOH = 0.7 kcal/mole
- Heat of neutralization (ΔH_neutralization) = -13.7 kcal/mole (this is a known value for the neutralization of a strong base with a weak acid)

### Step 2: Write the formula for ΔH of the reaction
The change in enthalpy for the reaction can be calculated using the following formula:

\[ \Delta H_{\text{reaction}} = \Delta H_{\text{neutralization}} + \Delta H_{\text{dissociation}} \]

### Step 3: Substitute the known values into the formula
Now we can substitute the known values into the equation:

\[ \Delta H_{\text{reaction}} = (-13.7 \text{ kcal/mole}) + (0.7 \text{ kcal/mole}) \]

### Step 4: Perform the calculation
Now, we perform the calculation:

\[ \Delta H_{\text{reaction}} = -13.7 + 0.7 = -13.0 \text{ kcal/mole} \]

### Final Answer
Thus, the change in enthalpy (ΔH) for the reaction is:

\[ \Delta H = -13.0 \text{ kcal/mole} \]