NEET2020ChemistryThermodynamics (C)Actual
At standard conditions, if the change in the enthalpy for the following reaction is - 109   kJmol - 1 H 2 ( g ) + Br 2 ( g ) → 2 HBr ( g ) Given that bond energy of H 2 and Br 2 is 435   kJmol - 1 and 192   kJmol - 1 , respectively, what is the bond energy (in kJ   mol - 1 ) of HBr ?
Options
- A368
- B736
- C518
- D259
Correct answer
A. 368
Step-by-step solution
Δ H = Σ ( B · E ) Reactants  - Σ ( B . E ) Products  - 109 = B · E ( H - H ) + B · E ( Br - Br ) - 2 × B · E ( H - Br ) - 109 = 435 + 192 - 2 × B · E ( H - Br ) B . E ( H - Br ) = 435 + 192 + 109 2 = 368   KJ / mol