1. **State the problem:** We need to find the half-life of a second-order reaction given the rate constant $K=0.498$ L/mol·s and the initial concentration $[A]_0=0.0050$ mol/L. 2. **Formula for half-life of second-order reaction:** The half-life $t_{1/2}$ for a second-order reaction is given by: $$ t_{1/2} = \frac{1}{K [A]_0} $$ where $K$ is the rate constant and $[A]_0$ is the initial concentration. 3. **Substitute the given values:** $$ t_{1/2} = \frac{1}{0.498 \times 0.0050} $$ 4. **Calculate the denominator:** $$ 0.498 \times 0.0050 = 0.00249 $$ 5. **Calculate the half-life:** $$ t_{1/2} = \frac{1}{0.00249} \approx 401.61 \text{ seconds} $$ **Final answer:** The half-life of the second-order reaction is approximately $401.61$ seconds.