`Delta_(f)H(H_(2)O) =- 68 kcal mol^(-1)` and `DeltaH` of neutralisation is `-13.7 kcal mol^(-1)`, then the heat of formation of `overset(Theta)OH` is

Options

A. `-68 kcal mol^(-1)`
B. `-54.3 kcal mol^(-1)`
C. `54.3 kcal mol^(-1)`
D. `-71.7 kcal mol^(-1)`

Explanation

To find the heat of formation of the hydroxide ion (OH⁻), we can use the given enthalpy values and the concept of enthalpy of neutralization. Here’s a step-by-step solution:

### Step 1: Understand the Reaction
In a neutralization reaction, an acid reacts with a base to produce water (H₂O) and a salt. The general reaction can be represented as:
\[ \text{Acid} + \text{Base} \rightarrow \text{H}_2\text{O} + \text{Salt} \]

### Step 2: Write the Enthalpy Change for the Reaction
The enthalpy change (ΔH) for the neutralization reaction is given as:
...