The enthalpy of neutralization of `40.0 g` of NaOH by 60.0 g of `CH_(3)COOH` will be:

Options

A. `57.1 kJ "equiv"^(-1)`
B. less than `57.1 kJ "equiv"^(-1)`
C. more than `57.1 kJ "equiv"^(-1)`
D. `13.7 kJ"equiv"^(-1)`

Answer

To solve the problem of finding the enthalpy of neutralization of 40.0 g of NaOH by 60.0 g of CH₃COOH, we can follow these steps:

### Step 1: Determine the moles of NaOH and CH₃COOH
To find the moles, we need to use the formula:

\[ \text{Moles} = \frac{\text{mass (g)}}{\text{molar mass (g/mol)}} \]

- **Molar mass of NaOH**:
- Na: 23 g/mol
- O: 16 g/mol
- H: 1 g/mol
- Total = 23 + 16 + 1 = 40 g/mol

- **Molar mass of CH₃COOH** (acetic acid):
- C: 12 g/mol (2 Carbons)
- H: 1 g/mol (4 Hydrogens)
- O: 16 g/mol (2 Oxygens)
- Total = (2 × 12) + (4 × 1) + (2 × 16) = 24 + 4 + 32 = 60 g/mol

Now, calculate the moles:

\[ \text{Moles of NaOH} = \frac{40.0 \, \text{g}}{40 \, \text{g/mol}} = 1 \, \text{mol} \]

\[ \text{Moles of CH₃COOH} = \frac{60.0 \, \text{g}}{60 \, \text{g/mol}} = 1 \, \text{mol} \]

### Step 2: Write the neutralization reaction
The neutralization reaction between NaOH and CH₃COOH can be represented as:

\[ \text{NaOH} + \text{CH}_3\text{COOH} \rightarrow \text{CH}_3\text{COONa} + \text{H}_2\text{O} \]

### Step 3: Identify the enthalpy of neutralization
The enthalpy of neutralization for a strong base reacting with a weak acid is typically less than 57.1 kJ/mol because some energy is used to ionize the weak acid.

### Step 4: Conclusion
Since we have 1 mole of NaOH and 1 mole of CH₃COOH reacting, the enthalpy of neutralization will be less than 57.1 kJ/mol.

Thus, the final answer is that the enthalpy of neutralization of 40.0 g of NaOH by 60.0 g of CH₃COOH is less than 57.1 kJ/mol.