Question

If the bond energies of H–H, Br – Br and H –Br are 433, 192 and 364 kJ mol^–1 respectively the ∆Hº for the reaction H₂(g) + Br₂(g) \(\rightarrow\)2HBr(g)

• A. + 103 kJ
• B. 261kJ
• C. –103 kJ
• D. –261 kJ

Answer 1

The Correct Option is C

 To solve this problem, we need to calculate the change in enthalpy (\(\Delta H^\circ\)) for the given reaction using bond energies. The reaction provided is:

\(H_2(g) + Br_2(g) \rightarrow 2 HBr(g)\)

Step-by-Step Solution:

1. Bond energies indicate the energy required to break a bond in one mole of a compound. The given bond energies are:
   • H–H = 433 kJ mol^–1
   • Br–Br = 192 kJ mol^–1
   • H–Br = 364 kJ mol^–1
2. Calculate the total energy required to break the bonds of the reactants:
   • Energy to break H–H bond: 433 kJ
   • Energy to break Br–Br bond: 192 kJ
3. Calculate the total energy released when new bonds are formed:
   • Energy released in forming 2 moles of H–Br = 2 × 364 = 728 kJ
4. Calculate the change in enthalpy (\(\Delta H^\circ\)) using the formula:

Hence, the enthalpy change for the reaction is \(-103 \text{ kJ mol}^{-1}\), which corresponds to the correct answer: –103 kJ.

Therefore, the correct option is –103 kJ.