Question

Given that bond energies of $H - H$ and $Cl - Cl$ are $430 \,kJ\, mol ^{-1}$ and $240\, kJ\, mol ^{-1}$ respectively and $\Delta_{ f } H$ for $HCl$ is $-90\, kJ\, mol ^{-1}$. Bond enthalpy of $HCl$ is :

• A. $245\, kJ\, mol ^{-1}$
• B. $290\, kJ \,mol ^{-1}$
• C. $380 \,kJ\, mol ^{-1}$
• D. $425 \,kJ \,mol ^{-1}$

Answer 1

Answer: (D)

Given :
Bond energies of $H - H$ and $Cl - Cl$ bonds which can be represented as follows :
$H _{2}( g ) \rightarrow 2 H ( g ) \quad \Delta H _{( H )}=+430 kJmol ^{-1}$
$Cl _{2}( g ) \rightarrow 2 Cl ( g ) \quad \Delta H _{( Cl )}=+240 kJ mol ^{-1}$
$HCl ( g ) \rightarrow H ( g )+ Cl ( g ) \quad \Delta H _{( Hcl )}=?$
$\frac{1}{2} H _{2}+\frac{1}{2} Cl _{2} \rightarrow HCl$
$\Delta_{ f } H _{ HCl }=\Delta H _{ BE }$ Reactant $+\Delta H _{ BE }$ Product $- BE$ of $HCl$
$\Rightarrow -90=\frac{1}{2} \times 430+\frac{1}{2} \times 240- BE$ of $HCl$
$\Rightarrow BE$ of $HCl =215+120+90=305+120=425 kJ mol ^{-1}$
Where, BE = Bond enthalpy