The `K_(a)` for `CH_(3)COOH` at `300` and `310 K` are `1.8 xx 10^(-5)` and `1.805 xx 10^(-5)`, respectively. The enthalpy of deprotonation for acetic acid is `51.6 cal`.

Explanation

To solve the problem, we will use the Van't Hoff equation, which relates the change in the equilibrium constant (K) with temperature and the enthalpy change (ΔH) of the reaction.

### Step-by-Step Solution:

1. **Identify the Given Values:**
- \( K_a \) at \( 300 \, K \) (T1) = \( 1.8 \times 10^{-5} \)
- \( K_a \) at \( 310 \, K \) (T2) = \( 1.805 \times 10^{-5} \)
- Enthalpy of deprotonation (ΔH) = \( 51.6 \, \text{cal} \)
- Universal gas constant (R) = \( 1.987 \, \text{cal/(mol·K)} \)

2. **Write the Van't Hoff Equation:**
\[ \ln \left( \frac{K_2}{K_1} \right) = -\frac{\Delta H}{R} \left( \frac{1}{T_2} - \frac{1}{T_1} \right) \]
Here, \( K_1 = K_a \) at \( T_1 \) and \( K_2 = K_a \) at \( T_2 \).

3. **Substitute the Known Values:**
\[ \ln \left( \frac{1.805 \times 10^{-5}}{1.8 \times 10^{-5}} \right) = -\frac{\Delta H}{1.987} \left( \frac{1}{310} - \frac{1}{300} \right) \]

4. **Calculate the Left Side:**
- First, calculate the fraction:
\[ \frac{1.805 \times 10^{-5}}{1.8 \times 10^{-5}} \approx 1.00278 \]
- Now, take the natural logarithm:
\[ \ln(1.00278) \approx 0.00277 \]

5. **Calculate the Right Side:**
- Calculate \( \frac{1}{310} - \frac{1}{300} \):
\[ \frac{1}{310} \approx 0.0032258, \quad \frac{1}{300} \approx 0.0033333 \]
\[ \frac{1}{310} - \frac{1}{300} \approx -0.0001075 \]
- Substitute back into the equation:
\[ 0.00277 = -\frac{\Delta H}{1.987} \times (-0.0001075) \]

6. **Rearrange to Solve for ΔH:**
\[ \Delta H = \frac{0.00277 \times 1.987}{0.0001075} \]
- Calculate:
\[ \Delta H \approx \frac{0.0055}{0.0001075} \approx 51.2 \, \text{cal} \]

7. **Conclusion:**
- The calculated ΔH (approximately \( 51.2 \, \text{cal} \)) is not equal to the given ΔH (51.6 cal). Therefore, the statement regarding the enthalpy of deprotonation is **false**.