The enthalpy of formation of `H_(2)O(l)` is -280.70 kJ/mol and enthalpy of neutralisation of a strong acid and strong base is -56.70 kJ/mol. What is the enthalpy of formation of `OH^(-)` ions?

Options

A. `-22.9` kJ/mol
B. `-224` kJ/mol
C. `58.7` kJ/mol
D. `214` kJ/mol

Answer

To find the enthalpy of formation of OH⁻ ions, we can use the given data about the enthalpy of formation of water (H₂O) and the enthalpy of neutralization of a strong acid with a strong base. Here’s a step-by-step solution:

### Step 1: Write the formation reaction of water
The formation reaction of water can be represented as:
\[ \text{H}_2(g) + \frac{1}{2} \text{O}_2(g) \rightarrow \text{H}_2O(l) \]
The enthalpy change for this reaction (ΔH_f) is given as:
\[ \Delta H_f = -280.70 \text{ kJ/mol} \]

### Step 2: Write the neutralization reaction
The neutralization of a strong acid (like HCl) with a strong base (like NaOH) can be represented as:
\[ \text{H}^+(aq) + \text{OH}^-(aq) \rightarrow \text{H}_2O(l) \]
The enthalpy change for this reaction (ΔH_neutralization) is given as:
\[ \Delta H_{neutralization} = -56.70 \text{ kJ/mol} \]

### Step 3: Reverse the neutralization reaction
If we reverse the neutralization reaction, we get:
\[ \text{H}_2O(l) \rightarrow \text{H}^+(aq) + \text{OH}^-(aq) \]
The enthalpy change for this reversed reaction will be the opposite of the neutralization enthalpy:
\[ \Delta H_{reverse} = +56.70 \text{ kJ/mol} \]

### Step 4: Combine the reactions
Now, we can combine the formation reaction of water and the reversed neutralization reaction:
1. Formation of water:
\[ \text{H}_2(g) + \frac{1}{2} \text{O}_2(g) \rightarrow \text{H}_2O(l) \quad (\Delta H = -280.70 \text{ kJ/mol}) \]
2. Reversed neutralization:
\[ \text{H}_2O(l) \rightarrow \text{H}^+(aq) + \text{OH}^-(aq) \quad (\Delta H = +56.70 \text{ kJ/mol}) \]

When we add these two reactions, the water on the product side cancels out:
\[ \text{H}_2(g) + \frac{1}{2} \text{O}_2(g) \rightarrow \text{H}^+(aq) + \text{OH}^-(aq) \]

### Step 5: Calculate the overall enthalpy change
The overall enthalpy change for the combined reaction is:
\[ \Delta H_{overall} = \Delta H_f + \Delta H_{reverse} \]
Substituting the values:
\[ \Delta H_{overall} = -280.70 \text{ kJ/mol} + 56.70 \text{ kJ/mol} \]
\[ \Delta H_{overall} = -224.00 \text{ kJ/mol} \]

### Conclusion
The enthalpy of formation of OH⁻ ions is:
\[ \Delta H_f(\text{OH}^-) = -224.00 \text{ kJ/mol} \]

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