To calculate the total bond energy in the products of the reaction \( H_2 + I_2 \rightarrow 2HI \), we first need to look at the bonds in the products, which are 2 moles of HI.
The bond energy for H-I is given as 297 kJ/mol. Since there are 2 moles of HI produced, we multiply the bond energy by 2:
\[ \text{Total bond energy in products} = 2 \times \text{Bond energy of HI} \] \[ = 2 \times 297 , \text{kJ/mol} = 594 , \text{kJ/mol} \]
Thus, the total bond energy in the products (hydroiodic acid) is 594 kJ/mol.
The correct answer is 594.