AIPMT PRELIMS 2004 Chemistry Bond Energies MCQ Question

To determine the enthalpy change ($\Delta H^\circ$) for the reaction

$\text{H}_2(g) + \text{Br}_2(g) \rightarrow 2\text{HBr}(g),$

we will use the bond energies provided in the question. The bond energy is the amount of energy required to break a bond in one mole of a substance.

Step 1: Identify the Bonds Broken and Formed

In the reaction given, we need to account for the bonds broken in the reactants and the bonds formed in the products.

1. Bonds Broken:

   • One H–H bond in $\text{H}_2$ (433 kJ mol⁻¹)
   • One Br–Br bond in $\text{Br}_2$ (192 kJ mol⁻¹)

2. Bonds Formed:

   • Two H–Br bonds in the products $2\text{HBr}$ (2 × 364 kJ mol⁻¹)

Step 2: Calculate the Total Energy for Bonds Broken and Formed

Total energy required to break bonds:

$$\text{Energy to break} = \text{Bond energy of H–H} + \text{Bond energy of Br–Br}$$ $$= 433 \, \text{kJ mol}^{-1} + 192 \, \text{kJ mol}^{-1} = 625 \, \text{kJ mol}^{-1}$$

Total energy released when forming bonds:

$$\text{Energy released on forming} = 2 \times \text{Bond energy of H–Br}$$ $$= 2 \times 364 \, \text{kJ mol}^{-1} = 728 \, \text{kJ mol}^{-1}$$

Step 3: Calculate $\Delta H^\circ$ for the Reaction

The enthalpy change for the reaction can be calculated using the formula:

$$\Delta H^\circ = \text{(Energy of bonds broken)} - \text{(Energy of bonds formed)}$$

Substituting the values:

$$\Delta H^\circ = 625 \, \text{kJ mol}^{-1} - 728 \, \text{kJ mol}^{-1}$$ $$= -103 \, \text{kJ mol}^{-1}$$

Conclusion

The calculated $\Delta H^\circ$ is $-103 \, \text{kJ mol}^{-1}$. This indicates that the reaction is exothermic, meaning it releases energy.

Thus, the correct answer is C) –103 kJ.

Clarification on Other Options

• Option A (+103 kJ): Incorrect, as it suggests an endothermic reaction, which contradicts our calculations.
• Option B (+261 kJ): Incorrect, as it significantly overestimates the energy change.
• Option D (–261 kJ): Incorrect, as it underestimates the energy released by the formation of H–Br bonds.

In conclusion, the significant energy released from forming H–Br bonds outweighs the energy required to break the H–H and Br–Br bonds, resulting in a net release of energy for the reaction.