On the basis of the following thermochemical data `:` `(Delta_(f)G^(@)H_((aq.))^(+)=0)`
`H_(2)O_((l))rarr H_((aq.))^(+)+OH_((aq.))^(-),DeltaH=57.32kJ`
`H_(2(g))+(1)/(2)O_(2(g))rarrH_(2)O_((l)),DeltaH=-286.20kJ`
The value of enthalpy of formation of `OH^(-)` ion at `25^(@)C` is `:`

Options

A. `22.88 KJ`
B. `-228.88KJ`
C. `+228.88kJ`
D. `-343. 52KJ`

Explanation

To find the enthalpy of formation of the hydroxide ion (OH⁻) at 25°C, we can use the given thermochemical data and apply Hess's law. Here's the step-by-step solution:

### Step 1: Write down the given reactions and their enthalpy changes.

1. **Reaction 1**:
\[ H_2O_{(l)} \rightarrow H^+_{(aq)} + OH^-_{(aq)}, \quad \Delta H = 57.32 \, \text{kJ} \]

2. **Reaction 2**:
\[ H_2(g) + \frac{1}{2} O_2(g) \rightarrow H_2O_{(l)}, \quad \Delta H = -286.20 \, \text{kJ} \]

### Step 2: Combine the reactions to find the overall reaction.

To find the enthalpy of formation of OH⁻, we can combine the two reactions. We need to reverse Reaction 2 to get H₂O on the reactant side:

**Reversed Reaction 2**:
\[ H_2O_{(l)} \rightarrow H_2(g) + \frac{1}{2} O_2(g), \quad \Delta H = +286.20 \, \text{kJ} \]

Now, we can add Reaction 1 and the reversed Reaction 2:

\[ H_2O_{(l)} \rightarrow H^+_{(aq)} + OH^-_{(aq)} \quad (\Delta H = 57.32 \, \text{kJ}) \]
\[ H_2O_{(l)} \rightarrow H_2(g) + \frac{1}{2} O_2(g) \quad (\Delta H = +286.20 \, \text{kJ}) \]

### Step 3: Write the overall reaction and calculate the total enthalpy change.

Adding these reactions gives:
\[ H_2(g) + \frac{1}{2} O_2(g) + H^+_{(aq)} + OH^-_{(aq)} \rightarrow H_2O_{(l)} + H^+_{(aq)} + OH^-_{(aq)} \]

The overall reaction simplifies to:
\[ H_2(g) + \frac{1}{2} O_2(g) \rightarrow H^+_{(aq)} + OH^-_{(aq)} \]

### Step 4: Calculate the total enthalpy change.

The total enthalpy change for the overall reaction is:
\[ \Delta H = 57.32 \, \text{kJ} + 286.20 \, \text{kJ} = -228.88 \, \text{kJ} \]

### Step 5: Use the enthalpy change to find the enthalpy of formation of OH⁻.

Using the formula:
\[ \Delta H = \sum \Delta H_f \text{(products)} - \sum \Delta H_f \text{(reactants)} \]

For the products:
- \( \Delta H_f(H^+) = 0 \)
- \( \Delta H_f(OH^-) = x \) (unknown)

For the reactants:
- \( \Delta H_f(H_2) = 0 \)
- \( \Delta H_f(O_2) = 0 \)

Thus, we have:
\[ -228.88 \, \text{kJ} = 0 + x - 0 \]
\[ x = -228.88 \, \text{kJ} \]

### Final Answer:
The enthalpy of formation of the hydroxide ion (OH⁻) at 25°C is:
\[ \Delta H_f(OH^-) = -228.88 \, \text{kJ} \]

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