Enthalpy of neutralisation of acetic acid by `NaOH` is `-50.6KJ//mol` and the heat of neutralisation of a storng acid with a strong bases is `-55..9KJ//mol` . What is the value of `DeltaH` for the ionisation of `CH_(3)COOH` ?

Options

A. `+5.3` kJ/mol
B. `+6.2` kJ/mol
C. `+8.2` kJ/mol
D. `+9.3` kJ/mol

Answer

To find the value of ΔH for the ionization of acetic acid (CH₃COOH), we can use the enthalpy changes provided for the neutralization reactions. Here’s a step-by-step solution:

### Step 1: Understand the Reactions
We have two reactions to consider:
1. The neutralization of a strong acid (H⁺) with a strong base (OH⁻):
\[ \text{H}^+_{\text{(aq)}} + \text{OH}^-_{\text{(aq)}} \rightarrow \text{H}_2\text{O}_{\text{(l)}} \]
The enthalpy change for this reaction (ΔH₁) is given as -55.9 kJ/mol.

2. The neutralization of acetic acid (CH₃COOH) with NaOH:
\[ \text{CH}_3\text{COOH}_{\text{(aq)}} + \text{OH}^-_{\text{(aq)}} \rightarrow \text{CH}_3\text{COO}^-_{\text{(aq)}} + \text{H}_2\text{O}_{\text{(l)}} \]
The enthalpy change for this reaction (ΔH₂) is given as -50.6 kJ/mol.

### Step 2: Set Up the Equation
To find the enthalpy change for the ionization of acetic acid (ΔH for the reaction):
\[ \text{CH}_3\text{COOH}_{\text{(aq)}} \rightarrow \text{CH}_3\text{COO}^-_{\text{(aq)}} + \text{H}^+_{\text{(aq)}} \]
we can relate it to the two neutralization reactions.

### Step 3: Use Hess's Law
According to Hess's law, the total enthalpy change for a reaction is the sum of the enthalpy changes for the individual steps. Therefore, we can express the enthalpy change for the ionization of acetic acid as:
\[ \Delta H = \Delta H_2 - \Delta H_1 \]

### Step 4: Substitute the Values
Now, substituting the values we have:
\[ \Delta H = (-50.6 \text{ kJ/mol}) - (-55.9 \text{ kJ/mol}) \]
\[ \Delta H = -50.6 + 55.9 \]
\[ \Delta H = 5.3 \text{ kJ/mol} \]

### Conclusion
Thus, the value of ΔH for the ionization of acetic acid is **5.3 kJ/mol**.

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