The enthalpy of neutralisation of HCl by NaOH IS `-55.9 kJ ` and that of HCN by NaOH is `-12.1 kJ mol^(-1)`. The enthalpy of ionisation of HCN is

Options

A. `-43.8 KJ`
B. `43.8 KJ`
C. 68 KJ
D. `-68` KJ

Explanation

To find the enthalpy of ionization of HCN, we can use the enthalpy of neutralization values provided for HCl and HCN. Here’s the step-by-step solution:

### Step 1: Understand the Enthalpy of Neutralization
The enthalpy of neutralization is the heat change that occurs when an acid reacts with a base to form water. For strong acids like HCl, the reaction is complete, and the enthalpy change is a fixed value.

Given:
- Enthalpy of neutralization of HCl by NaOH = \(-55.9 \, \text{kJ/mol}\)
- Enthalpy of neutralization of HCN by NaOH = \(-12.1 \, \text{kJ/mol}\)

### Step 2: Write the Reactions
1. For HCl:
\[ \text{HCl} + \text{NaOH} \rightarrow \text{NaCl} + \text{H}_2\text{O} \]
The enthalpy change for this reaction is \(-55.9 \, \text{kJ/mol}\).

2. For HCN:
\[ \text{HCN} + \text{NaOH} \rightarrow \text{NaCN} + \text{H}_2\text{O} \]
The enthalpy change for this reaction is \(-12.1 \, \text{kJ/mol}\).

### Step 3: Apply Hess's Law
Hess's Law states that the total enthalpy change for a reaction is the sum of the enthalpy changes for the individual steps. We can express the enthalpy of ionization of HCN as follows:

\[ \Delta H_{\text{ionization}} + \Delta H_{\text{neutralization of HCN}} = \Delta H_{\text{neutralization of HCl}} \]

### Step 4: Rearranging the Equation
Rearranging the equation gives us:
\[ \Delta H_{\text{ionization}} = \Delta H_{\text{neutralization of HCl}} - \Delta H_{\text{neutralization of HCN}} \]

### Step 5: Substitute the Values
Substituting in the values we have:
\[ \Delta H_{\text{ionization}} = (-55.9 \, \text{kJ/mol}) - (-12.1 \, \text{kJ/mol}) \]
\[ \Delta H_{\text{ionization}} = -55.9 + 12.1 \]
\[ \Delta H_{\text{ionization}} = -43.8 \, \text{kJ/mol} \]

### Conclusion
The enthalpy of ionization of HCN is \(+43.8 \, \text{kJ/mol}\).

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