Question

Given that bond energies of $H-H$ and $Cl-Cl$ are $430kJmo{l}^{-1}$ and $240kJmo{l}^{-1}$ respectively and ${\Delta }_{f}H$ for $HCl$ is $-90kJmo{l}^{-1}$. Bond enthalpy of $HCl$ is :

• A. $245kJmo{l}^{-1}$
• B. $290kJmo{l}^{-1}$
• C. $380kJmo{l}^{-1}$
• D. $425kJmo{l}^{-1}$

Answer 1

Answer: (D)

Given :
Bond energies of $H-H$ and $Cl-Cl$ bonds which can be represented as follows :
$H_2(g)\to 2H(g)\Delta H_{(H)}=+430kJmo{l}^{-1}$
$C{l}_2(g)\to 2Cl(g)\Delta H_{(Cl)}=+240kJmo{l}^{-1}$
$HCl(g)\to H(g)+Cl(g)\Delta H_{(Hcl)}=?$
$\frac{1}{2}{H}_2+\frac{1}{2}C{l}_2\to HCl$
${\Delta }_f{H}_{HCl}=\Delta H_{BE}$ 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 $HCl=215+120+90=305+120=425kJmo{l}^{-1}$
Where, BE = Bond enthalpy