To calculate the total bond energy in the products of the reaction \( H_2 + I_2 \rightarrow 2 HI \), we will consider the bond energies of the hydroiodic acid (HI) produced.
Since the reaction produces 2 moles of HI, we need to take into account the bond energy of 1 mole of HI and then multiply it by 2.
Given the bond energy of HI is 297 kJ/mol, the total bond energy for the products is:
\[ \text{Total bond energy} = 2 \times \text{Bond energy of HI} \] \[ = 2 \times 297 \text{ kJ/mol} \] \[ = 594 \text{ kJ/mol} \]
Therefore, the total bond energy in the products of the reaction is 594 kJ/mol.