Question

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

• A. $180kJmo{l}^{-1}$
• B. $360kJmo{l}^{-1}$
• C. $213kJmo{l}^{-1}$
• D. $425kJmo{l}^{-1}$

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
$\frac{1}{2}{H}_2+\frac{1}{2}C{l}_2\to HCl$
$\Delta H_{\text{reaction }}=\frac{1}{2}\Delta H_{H-H}+\frac{1}{2}\Delta H_{Cl-Cl}-\Delta H_{H-Cl}$
Here, $\Delta H$ for $HCl=-90kJ$
$\Delta H_{H-H}=430kJmo{l}^{-1}$
$\Delta H_{Cl-Cl}=240kJmo{l}^{-1}$
therefore,
$-90=\frac{1}{2}\times 430+\frac{1}{2}\times 240-\Delta H_{H-Cl}$
$-90=215+120-\Delta H_{H-Cl}$
$\Delta H_{H-Cl}=215+120+90$
$=425kJmo{l}^{-1}$