$ \mathrm{~A} \cdot \mathrm{e}^{-\mathrm{E}_{\mathrm{a}} / \mathrm{RT}}=\frac{\mathrm{A}_1 \mathrm{~A}_2}{\mathrm{~A}_3} \cdot \mathrm{e}^{-\frac{\left(\mathrm{E}_{\mathrm{a}_1}+\mathrm{E}_{\mathrm{a}_2}-\mathrm{E}_{\mathrm{a}_3}\right)}{R T}}$
$\mathrm{E}_{\mathrm{a}}=\mathrm{E}_{\mathrm{a}_1}+\mathrm{E}_{\mathrm{a}_2}-\mathrm{E}_{\mathrm{a}_3}=40+50-60=30 \mathrm{~kJ} / \mathrm{mole}$
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$A.$ $SO _{4}^{2-}$ and $CrO _{4}^{2-}$
$B.$ $SiCl _{4}$ and $TiCl _{4}$
$C.$ $NH _{3}$ and $NO _{3}^{-}$
$D.$ $BCl _{3}$ and $BrCl _{3}$
$BCl _{3}$ and $BrCl _{3}$
| $S \,No.$ | Time $(s)$ | Total pressure $(atm)$ |
| $1$ | $0$ | $0.5$ |
| $2$ | $100$ | $0.6$ |
Calculate the rate constant
| $H - H$ bond energy | $:\, 431.37 \,kJ\, mol^{-1}$ |
| $C= C$ bond energy | $:\, 606.10\, kJ \,mol^{-1}$ |
| $C - C$ bond energy | $:\, 336.49\, kJ\, mol^{-1}$ |
| $C - H$ bond energy | $:\, 410.50\, kJ\, mol^{-1}$ |
Enthalpy for the reaction,
$\begin{array}{*{20}{c}}
{H\,\,\,\,H} \\
{|\,\,\,\,\,\,\,\,|} \\
{C = C} \\
{|\,\,\,\,\,\,\,\,\,|} \\
{H\,\,\,\,H}
\end{array}\, + \,H - H\, \to \,\begin{array}{*{20}{c}}
{H\,\,\,\,H} \\
{|\,\,\,\,\,\,\,\,|} \\
{H - C - C - H} \\
{|\,\,\,\,\,\,\,\,\,|} \\
{H\,\,\,\,H}
\end{array}\,$
will be .............. $\mathrm{kJ \,mol}^{-1}$