Given, `CH_(3)COOH(aq)toCH_(3)COO^(-)(aq)+H^(+)(aq)`
`DeltaH_(rxn)^(@)=0.004" kcal "gm^(-1)`
Enthalpy change when 1 mole of `Ba(OH)_(2)`, a strong base, is completely neutralized by `CH_(3)COOH(aq)` is (`DeltaH^(@)` of neutralization of strong acid with strong base is=`-13.7" kcal "mol^(-1)`)

Options

A. `-27.46kcal`/mol
B. `27.46kcal`/mol
C. `-26.92kcal`/mol
D. `-13.46kcal`/mol

Answer

To solve the problem, we need to calculate the enthalpy change when 1 mole of barium hydroxide (Ba(OH)₂) is completely neutralized by acetic acid (CH₃COOH). We will follow these steps:

### Step 1: Understand the dissociation of acetic acid
The dissociation of acetic acid is given by the equation:
\[ \text{CH}_3\text{COOH} (aq) \rightarrow \text{CH}_3\text{COO}^- (aq) + \text{H}^+ (aq) \]
The enthalpy change for this reaction is given as:
\[ \Delta H_{\text{rxn}} = 0.004 \, \text{kcal/g} \]

### Step 2: Calculate the enthalpy change for 1 mole of acetic acid
The molecular mass of acetic acid (CH₃COOH) is approximately 60 g/mol. Therefore, the enthalpy change for 1 mole of acetic acid can be calculated as:
\[ \Delta H_{\text{1 mole}} = 0.004 \, \text{kcal/g} \times 60 \, \text{g} = 0.24 \, \text{kcal} \]

### Step 3: Write the neutralization reaction
The neutralization reaction between barium hydroxide and acetic acid can be represented as:
\[ \text{Ba(OH)}_2 + 2 \text{CH}_3\text{COOH} \rightarrow \text{Ba(CH}_3\text{COO)}_2 + 2 \text{H}_2\text{O} \]

### Step 4: Use the enthalpy of neutralization for strong acid and strong base
The enthalpy change of neutralization for a strong acid with a strong base is given as:
\[ \Delta H_{\text{neutralization}} = -13.7 \, \text{kcal/mol} \]
Since 2 moles of acetic acid are used in the reaction, we need to consider the enthalpy change for both moles.

### Step 5: Calculate the total enthalpy change for the neutralization
The total enthalpy change for the neutralization reaction can be calculated as follows:
\[ \Delta H_{\text{total}} = \Delta H_{\text{neutralization}} + 2 \times \Delta H_{\text{dissociation}} \]
Where:
- \(\Delta H_{\text{dissociation}} = 0.24 \, \text{kcal}\) (for 1 mole of acetic acid)

Thus, for 2 moles:
\[ \Delta H_{\text{dissociation (2 moles)}} = 2 \times 0.24 \, \text{kcal} = 0.48 \, \text{kcal} \]

Now substituting the values:
\[ \Delta H_{\text{total}} = -13.7 \, \text{kcal} + 0.48 \, \text{kcal} = -13.22 \, \text{kcal} \]

### Final Answer
The enthalpy change when 1 mole of Ba(OH)₂ is completely neutralized by CH₃COOH is approximately:
\[ \Delta H_{\text{total}} \approx -13.22 \, \text{kcal/mol} \]

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