Given that ,
`H_(2) O(l) rarr H^(+) (aq) +OH^(-) (aq) , DeltaH = 57.32 kJ`
`H_(2)(g) +1/2 O_(2) (g) rarr H_2 O(l), DeltaH = -286.02 kJ`
Then calculate the enthalpy of formation of `OH^(-) ` at `25^@C`

Options

A. `-22.8 kj`
B. `-343.52 kJ`
C. `+228.8 kJ`
D. `+343.52 kJ`

Explanation

To calculate the enthalpy of formation of the hydroxide ion (OH⁻) at 25°C, we can use the given reactions and their enthalpy changes. Let's break down the solution step by step.

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

1. **Ionization of Water:**
\[ H_2O(l) \rightarrow H^+(aq) + OH^-(aq), \quad \Delta H = 57.32 \, \text{kJ} \]

2. **Formation of Water:**
\[ H_2(g) + \frac{1}{2}O_2(g) \rightarrow H_2O(l), \quad \Delta H = -286.02 \, \text{kJ} \]

### Step 2: Identify the enthalpy of formation of water.

From the second reaction, we know that the enthalpy of formation of water (\(H_2O\)) is:
\[ \Delta H_f(H_2O) = -286.02 \, \text{kJ} \]

### Step 3: Write the enthalpy change for the ionization of water.

The enthalpy change for the ionization of water can be expressed in terms of the enthalpy of formation of the ions:
\[ \Delta H = \Delta H_f(H^+) + \Delta H_f(OH^-) - \Delta H_f(H_2O) \]

### Step 4: Substitute known values into the equation.

We know that the enthalpy of formation of \(H^+\) is zero because it is a standard state:
\[ \Delta H_f(H^+) = 0 \]
Substituting this into the equation gives:
\[ 57.32 \, \text{kJ} = 0 + \Delta H_f(OH^-) - (-286.02 \, \text{kJ}) \]

### Step 5: Simplify the equation.

This simplifies to:
\[ 57.32 \, \text{kJ} = \Delta H_f(OH^-) + 286.02 \, \text{kJ} \]

### Step 6: Solve for \(\Delta H_f(OH^-)\).

Rearranging the equation to isolate \(\Delta H_f(OH^-)\):
\[ \Delta H_f(OH^-) = 57.32 \, \text{kJ} - 286.02 \, \text{kJ} \]
\[ \Delta H_f(OH^-) = -228.70 \, \text{kJ} \]

### Final Answer:
The enthalpy of formation of \(OH^-\) at 25°C is:
\[ \Delta H_f(OH^-) = -228.70 \, \text{kJ} \]

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