1. Tardigrade
  2. Question
  3. Chemistry
  4. At standard conditions, if the change in the enthalpy for the following reaction is -109 kJ mol-1 H2(g)+Br2(g) → 2HBr(g) Given that bond energy of H2 and Br2 is 435 KJ mol-1 and 192 KJ mol-1 , respectively, what is the bond energy (in KJ mol-1) of HBr ?

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Q. At standard conditions, if the change in the enthalpy for the following reaction is $-109 \,kJ\, mol^{-1}$ $H_{2(g)}+Br_{2(g)} \to 2HBr_{(g)}$ Given that bond energy of $H_{2}$ and $Br_{2}$ is $435\,KJ\,mol^{-1}$ and $192\,KJ\,mol^{-1}$ , respectively, what is the bond energy (in $KJ \,mol^{-1})$ of $HBr$ ?

NEETNEET 2020Thermodynamics

Solution:

$\Delta H=\sum\left(B.E\right)_{\text{Reactants}}-\sum \left(B.E\right)_{\text{proudcts}}$
$-109=\left[B.E._{\left(H-H\right)}+B.E_{\left(Br-Br\right)}\right]-\left[2\times B.E_{\left(H-Br\right)}\right]$
$-109=435+192-2\times B.E_{\left(H-Br\right)}$
$B.E_{\left(H-Br\right)}=\frac{435+192+109}{2}=368\,KJ/mol$