Question

Given that, for the reaction,
$\frac{1}{2} H _{2}+\frac{1}{2} Cl _{2} \longrightarrow HCl$
$\Delta H_{f}( HCl )=-93\, kJ / mol$
$BE \left( H _{2}\right)=434\, kJ / mol$
$BE \left( Cl _{2}\right)=242\, kJ / mol$
The bond dissociation energy of $HCl$ is

• A. $232\, kJ / mol$
• B. $331\, kJ / mol$
• C. $431\, kJ / mol$
• D. $530\, kJ / mol$

Answer 1

Answer: (C)

Given,
$\frac{1}{2} H _{2}+\frac{1}{2} Cl _{2} \longrightarrow HCl , \Delta H _{f}=-93\, kJ / mol$
$\because$ In this reaction, one mole of $HCl$ is formed.
$\therefore \Delta H_{f}$ of the reaction will remain same, i.e. $-93\, kJ / mol$.
Let the bond dissociation energy of $H - Cl$ bond be $x$.
For the given reaction,
$\Delta H_{f} =\Sigma B\cdot E _{(\text {reactants })}-\Sigma B \cdot E _{\text {(products) }}$
$-93 =\frac{1}{2}(434)+\frac{1}{2}(242)-x$
$-93 =+338-x$
$\Rightarrow x=431\, kJ / mol$