Question

The $H - H$ bond energy is $430\, kJ\, mol ^{-1}$ and $Cl - Cl$ bond energy is $240\, kJ\, mol ^{-1} . \Delta H$ for $HCl$ is $-90\, kJ$. The $H$-Cl bond energy is about

• A. $180\, kJ\, mol^{-1}$
• B. $360\, kJ\, mol^{-1}$
• C. $213\, kJ\, mol^{-1}$
• D. $425\, kJ\, mol^{-1}$

Answer 1

Answer: (D)

$\Delta H_{\text {reaction }}= \Sigma$ bond energy of reactant
$-\Sigma$ bond energy of product The formation of one mole of $HCl$ can be represented as
$\frac{1}{2} H _{2}+\frac{1}{2} Cl _{2} \rightarrow HCl$
$\Delta H_{\text {reaction }}=\frac{1}{2} \Delta H_{ H - H }+\frac{1}{2} \Delta H_{ Cl - Cl } -\Delta H_{ H - Cl }$
Here, $\Delta H$ for $HCl =-90\, kJ$
$\Delta H_{ H - H } =430\, kJ\, mol ^{-1}$
$\Delta H_{ Cl - Cl } =240\, kJ\, mol ^{-1}$
therefore,
$-90 =\frac{1}{2} \times 430+\frac{1}{2} \times 240-\Delta H_{ H - Cl }$
$-90 =215+120-\Delta H_{ H - Cl }$
$\Delta H_{ H - Cl } =215+120+90$
$=425\, kJ\, mol ^{-1}$