Bond dissociation energies of H2, Cl2 and HCl(g) are 104, 58 and 103 kcal mol-1 respectively. Calculate the enthalpy of formation of HCl gas.

Question

Bond dissociation energies of H2, Cl2 and HCl(g) are 104, 58 and 103 kcal mol-1 respectively. Calculate the enthalpy of formation of HCl gas. (A) -22 kcal (B) + 22 kcal (C) +184 kcal (D) -184 kcal

Tardigrade Admin · Accepted Answer

(1/2)H2+(1/2)Cl2→ HCl Δ H = (1/2)×104 + (1/2)× 58 - 103 = -22 kcal

Q. Bond dissociation energies of $H_{2}$ , $C l_{2}$ and $H C l_{(g)}$ are $104$ , $58$ and $103 k c a l m o l^{- 1}$ respectively. Calculate the enthalpy of formation of $H Cl$ gas.

$- 22 k c a l$

$+ 22 k c a l$

$+ 184 k c a l$

$- 184 k c a l$

$\frac{1}{2} H_{2} + \frac{1}{2} C l_{2} \to H Cl$
$Δ H = \frac{1}{2} \times 104 + \frac{1}{2} \times 58 - 103 = - 22 k c a l$

Q. Bond dissociation energies of H2​, Cl2​ and HCl(g)​ are 104, 58 and 103kcalmol−1 respectively. Calculate the enthalpy of formation of HCl gas.

−22kcal

+22kcal

+184kcal

−184kcal