Given that, for the reaction, (1/2) H 2+(1/2) Cl 2 longrightarrow HCl Δ Hf( HCl )=-93 kJ / mol BE ( H 2)=434 kJ / mol BE ( Cl 2)=242 kJ / mol The bond dissociation energy of HCl is

Question

Given that, for the reaction, (1/2) H 2+(1/2) Cl 2 longrightarrow HCl Δ Hf( HCl )=-93 kJ / mol BE ( H 2)=434 kJ / mol BE ( Cl 2)=242 kJ / mol The bond dissociation energy of HCl is (A) 232 kJ / mol (B) 331 kJ / mol (C) 431 kJ / mol (D) 530 kJ / mol

Tardigrade Admin · Accepted Answer

Given, (1/2) H 2+(1/2) Cl 2 longrightarrow HCl , Δ H f=-93 kJ / mol ∵ In this reaction, one mole of HCl is formed. ∴ Δ Hf of the reaction will remain same, i.e. -93 kJ / mol. Let the bond dissociation energy of H - Cl bond be x. For the given reaction, Δ Hf = Sigma B⋅ E ( text reactants )- Sigma B ⋅ E text (products) -93 =(1/2)(434)+(1/2)(242)-x -93 =+338-x ⇒ x=431 kJ / mol

Q. Given that, for the reaction,
$\frac{1}{2} H_{2} + \frac{1}{2} C l_{2} ⟶ H Cl$
$Δ H_{f} (H Cl) = - 93 k J / m o l$
$BE (H_{2}) = 434 k J / m o l$
$BE (C l_{2}) = 242 k J / m o l$
The bond dissociation energy of $H Cl$ is

$232 k J / m o l$

$331 k J / m o l$

$431 k J / m o l$

$530 k J / m o l$

Q. Given that, for the reaction, 21​H2​+21​Cl2​⟶HCl ΔHf​(HCl)=−93kJ/mol BE(H2​)=434kJ/mol BE(Cl2​)=242kJ/mol The bond dissociation energy of HCl is

232kJ/mol

331kJ/mol

431kJ/mol

530kJ/mol

Q. Given that, for the reaction,
$\frac{1}{2} H_{2} + \frac{1}{2} C l_{2} ⟶ H Cl$
$Δ H_{f} (H Cl) = - 93 k J / m o l$
$BE (H_{2}) = 434 k J / m o l$
$BE (C l_{2}) = 242 k J / m o l$
The bond dissociation energy of $H Cl$ is

$232 k J / m o l$

$331 k J / m o l$

$431 k J / m o l$

$530 k J / m o l$