Question

If, $H_2\left(g\right)+C{l}_2\left(g\right)\to 2HCl\left(g\right);\Delta H^o=-44Kcal$
$2Na\left(s\right)+2HCl\left(g\right)\to 2NaCl\left(s\right)+H_2\left(g\right);$
$\Delta H=-152Kcal$
Then, $Na\left(s\right)+0.5C{l}_2\left(g\right)\to NaCl\left(s\right);\Delta H^o=?$

• A. $108kcal$
• B. $196kcal$
• C. $-98kcal$
• D. $54kcal$

Answer 1

The correct option is C $-98kcal$

The thermochemical reactions are as given below.
$H_2\left(g\right)+C{l}_2\left(g\right)\to 2HCl\left(g\right);$

$\Delta H^o=-44Kcal$
$2Na\left(s\right)+2HCl\left(g\right)\to 2NaCl\left(s\right)+H_2\left(g\right);$
$\Delta H=-152Kcal$
Both reactions are multiplied by $0.5$ and then added.
The reaction for the formation of $NaCl$ is $Na\left(s\right)+0.5C{l}_2\left(g\right)\to NaCl\left(s\right)$.

Its heat of formation is $0.5(-44)+0.5(-152)$
$=-98$kcal