The standard enthalpy of neutralization of `KOH` with `HCN` and HCl in dilute solution is `-2480 cal.mol^(-1)` and `-13.68 kcalmol^(-1)` respectively. Find the enthalpy of dissocitation of `HCN` at the same temperature.

Answer

To find the enthalpy of dissociation of HCN given the standard enthalpy of neutralization of KOH with HCN and HCl, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Given Data:**
- Enthalpy of neutralization of KOH with HCl: \( \Delta H_{neutralization, HCl} = -13.68 \, \text{kcal/mol} \)
- Enthalpy of neutralization of KOH with HCN: \( \Delta H_{neutralization, HCN} = -2480 \, \text{cal/mol} \)

2. **Convert Units:**
- Convert the enthalpy of neutralization with HCl from kcal to cal:
\[ -13.68 \, \text{kcal/mol} = -13.68 \times 1000 \, \text{cal/mol} = -13680 \, \text{cal/mol} \]

3. **Write the Reaction for Neutralization:**
- For KOH with HCl:
\[ \text{KOH} + \text{HCl} \rightarrow \text{KCl} + \text{H}_2\text{O} \]
- For KOH with HCN:
\[ \text{KOH} + \text{HCN} \rightarrow \text{KCN} + \text{H}_2\text{O} \]

4. **Set Up the Relation for Enthalpy of Dissociation:**
- The dissociation of HCN can be represented as:
\[ \text{HCN} \rightarrow \text{H}^+ + \text{CN}^- \]
- The enthalpy change for the dissociation of HCN can be expressed as:
\[ \Delta H_{dissociation} = \Delta H_{neutralization, HCN} - \Delta H_{neutralization, HCl} \]

5. **Substitute the Values:**
- Substitute the values we have:
\[ \Delta H_{dissociation} = -2480 \, \text{cal/mol} - (-13680 \, \text{cal/mol}) \]
\[ \Delta H_{dissociation} = -2480 \, \text{cal/mol} + 13680 \, \text{cal/mol} \]
\[ \Delta H_{dissociation} = 11100 \, \text{cal/mol} \]

6. **Final Answer:**
- The enthalpy of dissociation of HCN is:
\[ \Delta H_{dissociation} = 11100 \, \text{cal/mol} \]