Question

Bond energies of few bonds are given below :
$Cl — Cl = 242.8 \,kJ\, mol^{-1}$, $H — Cl = 431.8\, kJ\, mol^{-1,}$
$O — H = 464\, kJ\, mol^{-1}$, $O = O = 442\, kJ\, mol^{-1}$
Using the $B.E.$, calculate $\Delta H$ for the following reaction, $2Cl_2 + 2H_2O \to 4HCl + O_2$

• A. $906 \,kJ\, mol^{-1}$
• B. $172.4 \,kJ\, mol^{-1}$
• C. $198.8 \,kJ\, mol^{-1}$
• D. $442 \,kJ\, mol^{-1}$

Answer 1

Answer: (B)

$2Cl_2 + 2H_2O \to 4HCl + O_2$
$\Delta H = B.E.$ of $(2 \times Cl — Cl) + (2 \times 2 \times O — H)$
$- (4 \times H — Cl) + (O = O)$
$= 2 \times 242.8 + 4 \times 464 - 4 \times 431.8 - 442$
$= 172.4 \,kj\, mol^{-1}$