Question

Use bond energies to estimate the enthalpy of formation of HBr(g).
BE(H-H) = 436 kJ/mol
BE(Br-Br) = 192 kJ/mol
BE(H-Br) = 366 kJ/mol

• A. +152 kJ/mol
• B. −104 kJ/mol
• C. +262 kJ/mol
• D. −52 kJ/mol

Answer 1

Answer: (D)

Step 1:

Enthalpy of reaction is the difference in energy absorbed in breaking bonds in reactants and energy released when new bonds are formed in products.

$\Delta H^o={\sum }_{bondenergy\left(reactants\right)}-{\sum }_{bondenergy\left(products\right)}$

Step 2:

Since two moles of HBr are formed in the reaction, for 1 mole 52 kJ/mol energy is required. Since energy is released during bond formation the value would be negative.

$\Delta H^o=$ 436 + 192 - 2$\times$ 366 = -104 KJ/mol

Hence, option (B) is correct.