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. `-68.0kJ`
B. `-43.8 kJ`
C. `68.0kJ`
D. `43.8kJ`.

Explanation

To find the enthalpy of ionization of HCN using the given enthalpies of neutralization, we can follow these steps:

### Step 1: Write the reactions and their enthalpy changes

1. **Neutralization of HCl by NaOH:**
\[ \text{HCl} + \text{NaOH} \rightarrow \text{NaCl} + \text{H}_2\text{O} \]
Enthalpy change, \(\Delta H_1 = -55.9 \, \text{kJ/mol}\)

2. **Neutralization of HCN by NaOH:**
\[ \text{HCN} + \text{NaOH} \rightarrow \text{NaCN} + \text{H}_2\text{O} \]
Enthalpy change, \(\Delta H_2 = -12.1 \, \text{kJ/mol}\)

### Step 2: Relate the enthalpy of ionization of HCN

The ionization of HCN can be represented as:
\[ \text{HCN} \rightleftharpoons \text{H}^+ + \text{CN}^- \]
Let the enthalpy of ionization of HCN be \(\Delta H_{\text{ionization}}\).

### Step 3: Use Hess's Law

We can use Hess's law to relate the enthalpy changes. The overall reaction for the ionization of HCN can be derived from the two neutralization reactions:

1. The enthalpy of neutralization of HCl gives us:
\[ \Delta H_1 = -55.9 \, \text{kJ/mol} \]

2. The enthalpy of neutralization of HCN gives us:
\[ \Delta H_2 = -12.1 \, \text{kJ/mol} \]

We can express the enthalpy of ionization of HCN in terms of these two reactions:
\[ \Delta H_{\text{ionization}} = \Delta H_2 - \Delta H_1 \]

### Step 4: Substitute the values

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

### Conclusion

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

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