$H - H$ બંધઊર્જા | $:\, 431.37 \,kJ\, mol^{-1}$ |
$C= C$ બંધઊર્જા | $:\, 606.10\, kJ \,mol^{-1}$ |
$C - C$ બંધઊર્જા | $:\, 336.49\, kJ\, mol^{-1}$ |
$C - H$ બંધઊર્જા | $:\, 410.50\, kJ\, mol^{-1}$ |
પ્રક્રિયા : $\begin{array}{*{20}{c}}
{H\,\,\,\,H} \\
{|\,\,\,\,\,\,\,\,|} \\
{C = C} \\
{|\,\,\,\,\,\,\,\,\,|} \\
{H\,\,\,\,H}
\end{array}\, + \,H - H\, \to \,\begin{array}{*{20}{c}}
{H\,\,\,\,H} \\
{|\,\,\,\,\,\,\,\,|} \\
{H - C - C - H} \\
{|\,\,\,\,\,\,\,\,\,|} \\
{H\,\,\,\,H}
\end{array}\,$
$\Delta H=(606.10+4 \times 410.5+431.37)$$-(6 \times 410.50+336.49)$
$=-120.2 \mathrm{kJ} / \mathrm{mol}$
$(ii)$ $SO_2$$_{(g)} +$ $\frac{1}{2} O_2$$_{(g)}$ $\rightarrow$ $SO_3$ $_(g) + y\, Kcal,$ તો $SO_2$ ની નિર્માણ ઉષ્મા શોધો.