Question

Bond dissociation enthalpy of H2, Br2 and HBr are 435, 192, 368 kJ mol–1 respectively. Enthalpy of formation of HBr is


–109 kJ/mol


–54.5 kJ/mol


–86.4 kJ/mol


–75 kJ/mol

Answer 1

-109

FIRST OPTION

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Answer 2

Given:

Bond dissociation enthalpy of $H_2, Br_2$ and HBr are 435, 192, 368 kJ $mol^{{-} 1 }$respectively.

To Find:

What is the enthalpy of formation of HBr?

Solution:

$\Delta H=\sum$the bond energy of reactants  $- \sum$the bond energy of products

$=435+192-2\times 368$

$=627 - 736$

$=-109$kJ/mol

Therefore, the enthalpy of formation of HBr is -109kJ/mol