Question

The bond dissociation energy of gaseous $H_2,C{l}_2$ and $HCl$ are 104 , 58 and 103 kcal / mole respectively . Calculate the enthalpy of formation of HCl gas.

• A. -75 Kcal
• B. -22 Kcal
• C. -45 Kcal
• D. none of these

Answer 1

The correct option is B -22 Kcal

$\frac{1}{2}+H_2+\frac{1}{2}C{l}_2\to HCl$

$\Delta h+=$(Bond energy of reactant)- (Bond energy of Product)

$=\left(\frac{1}{2}\Delta H_B\right(H_2)+\frac{1}{2}\Delta H_B(C{l}_2\left)\right)-\left(\Delta H_B\right(HCl\left)\right)$

$=\left(\frac{1}{2}\right(104)+\frac{1}{2}(58\left)\right)-\left(103\right)$

$=52+29-103$

$=-22$