`CsOH +HCI rarr CsCI +H_(2)O, DeltaH =- 13.4 kcal mol^(-1)……….(i)`
`CsOH +HF rarr CsF +H_(2)O, DeltaH =- 16.4 kcal mol^(-1) ………(ii)`
Calculate `DeltaH` for the ionisation of `HF` in `H_(2)O`.

Options

A. `3.0 kcal`
B. `-3.0 kcal`
C. `6.0 kcal`
D. `0.3 kcal`

Explanation

To calculate the enthalpy change (\( \Delta H \)) for the ionization of HF in water, we can use the given reactions and their enthalpy changes. Let's break down the solution step by step:

### Step 1: Write down the given reactions and their enthalpy changes.
1. **Reaction (i)**:
\[ \text{CsOH} + \text{HCl} \rightarrow \text{CsCl} + \text{H}_2\text{O}, \quad \Delta H_1 = -13.4 \, \text{kcal/mol} \]

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