Question

If, $H_2\left(g\right)+C{l}_2\left(g\right)\to 2HCl\left(g\right)$; $\Delta H=-44Kcal$
$2Na\left(s\right)+2HCl\left(g\right)\to 2NaCl\left(s\right)+H_2\left(g\right)$; $\Delta H=-152Kcal$
then, $Na\left(s\right)+\frac{1}{2}C{l}_2\left(g\right)\to NaCl\left(s\right)$; $\Delta H=?$

• A. +108 Kcal
• B. -196 Kcal
• C. -98 Kcal
• D. 50 Kcal

Answer 1

The correct option is C -98 Kcal

$H_2\left(g\right)+C{l}_2\left(g\right)\to 2HCl\left(g\right)$....(i)
$2Na\left(s\right)+2HCl\left(g\right)\to 2NaCl\left(s\right)+H_2\left(g\right)$....(ii)
$\frac{\left(ii\right)}{2}+\frac{1}{2}\times \left(i\right)$ gives
$Na\left(s\right)+\frac{1}{2}C{l}_2\left(g\right)\to NaCl\left(s\right)$
$\therefore \Delta H=\frac{-44}{2}-\frac{152}{2}$
$=\frac{-196}{2}$
$=-98$ Kcal

Option C is correct.