The bond energy of H-Br bond can be calculated using the following equation:
bond energy = ∆H(products) - ∆H(reactants)
where ∆H is the change in enthalpy for the reaction.
From the given data, we can calculate the bond energy of H-Br bond as follows:
bond energy = (2 HBr + 150 kJ) - (H₂ + Br₂)
= (2 HBr - H₂ - Br₂) + 150 kJ
= (2 HBr - H₂ - Br₂ - 430 kJ) + (150 kJ - 180 kJ)
= (-H₂ - Br₂ + 2 HBr - 430 kJ + 150 kJ - 180 kJ)
= (-H₂ - Br₂ + 2 HBr - 480 kJ)
= (-2H + 2 HBr - 480 kJ)
= (2 HBr - 2H - 480 kJ)
= (2 HBr - 2H + 2H₂ - 2H - 480 kJ)
= (2 HBr - 2H₂ - 480 kJ)
= (-2H₂ + 2 HBr - 480 kJ)
= (-2H₂ + 2 HBr + 430 kJ)
= (-2H₂ + 2 HBr + 2H₂ - 2H - 180 kJ)
= (-2H₂ + 2 HBr + 2H₂ - 2H + 430 kJ - 180 kJ)
= (2H₂ + 2 HBr - 2H + 250 kJ)
= (2H₂ + 2 HBr - 2H - 430 kJ + 680 kJ)
= (2H₂ + 2 HBr - 2H - 430 kJ + 2H₂ - 2H + 680 kJ)
= (4H₂ + 2 HBr - 430 kJ + 680 kJ)
= (4H₂ + 2 HBr + 250 kJ)
= (4H₂ + 2 HBr - 180 kJ + 430 kJ)
= (4H₂ + 2 HBr + 250 kJ - 180 kJ)
= (4H₂ + 2 HBr + 70 kJ)
Therefore, the bond energy of H-Br bond is 70 kJ.
The correct answer is therefore [d] 240 kJ.