Heat of neutralisation of strong acid and strong base under `1 atm` and `25^@C` is `-13.7 kcal`. If standard Gibbs energy change for dissociation of water to `H^+` and `OH^-` is `-19.14 kcal`, the change in standard entropy for dissociation of water is:

Options

A. 18.25
B. 110.2
C. `-18.25`
D. None of these

Answer

To solve the problem, we will use the relationship between Gibbs free energy change (ΔG), enthalpy change (ΔH), and entropy change (ΔS) given by the equation:

\[ \Delta G = \Delta H - T \Delta S \]

### Step 1: Identify the given values
- Heat of neutralization (ΔH) = -13.7 kcal
- Gibbs energy change (ΔG) = -19.14 kcal
- Temperature (T) = 25°C = 298 K (since we need to use Kelvin for calculations)

### Step 2: Convert the values to consistent units
Since ΔG and ΔH are given in kilocalories, we will convert them to calories for consistency:
- ΔH = -13.7 kcal = -13.7 × 10³ cal = -13700 cal
- ΔG = -19.14 kcal = -19.14 × 10³ cal = -19140 cal

### Step 3: Substitute the values into the Gibbs free energy equation
Using the equation:

\[ \Delta G = \Delta H - T \Delta S \]

Substituting the known values:

\[ -19140 = -13700 - 298 \Delta S \]

### Step 4: Rearranging the equation to solve for ΔS
Rearranging the equation gives:

\[ -19140 + 13700 = -298 \Delta S \]

\[ -5430 = -298 \Delta S \]

### Step 5: Solve for ΔS
Dividing both sides by -298:

\[ \Delta S = \frac{5430}{298} \]

Calculating this gives:

\[ \Delta S \approx 18.255 cal/K \]

### Conclusion
The change in standard entropy for the dissociation of water is approximately **18.255 cal/K**.