To determine the total bond energy in the products of the reaction \( \text{H}_2 + \text{I}_2 \rightarrow 2 \text{HI} \), we need to calculate the bond energy of the products, which consist of 2 molecules of hydroiodic acid (HI).
From the table provided, the bond energy for the H-I bond is given as 297 kJ/mol. Since there are 2 molecules of HI produced in the reaction, we will multiply the bond energy of H-I by 2:
\[ \text{Total bond energy in products} = 2 \times \text{Bond Energy of H-I} \]
\[ \text{Total bond energy in products} = 2 \times 297 \text{ kJ/mol} = 594 \text{ kJ/mol} \]
Therefore, the correct answer is 594 kJ/mol. The reasoning is that for each molecule of HI formed, there is one H-I bond, and since 2 molecules of HI are produced, we multiply the bond energy by 2.