Bond dissociation enthalpy of H2, Cl2 and HCl are 434, 242 and 431 kJ mol- 1 respectively. Enthalpy of formation of HCl is

Question

Bond dissociation enthalpy of H2, Cl2 and HCl are 434, 242 and 431 kJ mol- 1 respectively. Enthalpy of formation of HCl is (A) 245 kJ mol- 1 (B) 93 kJ mol- 1 (C) -245 kJ mol- 1 (D) -93 kJ mol- 1

Tardigrade Admin · Accepted Answer

The reaction between hydrogen and chlorine molecules to form hydrogen chloride. This reaction can be broken down into these simple steps:

H_{2} (g) + C l_{2} (g) \to 2 H Cl (g)

Step 1: Dissociation of Hydrogen and Chlorine molecules into their respective atoms

H_{2} (g) \to 2 H (g)

C l_{2} (g) \to 2 Cl (g)

Step 2: Combination of these atoms to form hydrogen chloride

2 H (g) + 2 Cl (g) \to 2 H Cl (g)

Since in the first step, one-mole of each

H - H

and

Cl - Cl

bonds are broken, it is an endothermic process

(Δ H_{1} > 0)

while in the second step two moles of

H Cl

are being formed which releases energy making

Δ H_{2} < 0

Now,

Δ H_{1} = Δ H_{Cl - Cl} + Δ H_{H - H}

= 434 + 242

= 676 k J m o l^{- 1}

Now,

Δ H_{2} = - 2Δ H_{H - Cl}

= 862 k J m o l^{- 1}

Applying Hess Law, i.e,

Δ H^{'} = Δ H_{1} + Δ H_{2}

, we get

Δ H^{'} = - 186 k J m o l^{- 1}

This is the enthalpy of formation of two moles of

H Cl

. Hence the enthalpy of formation of one mole of

H Cl

will be

Δ H = - 93 k J m o l^{- 1}

Q. Bond dissociation enthalpy of $H_{2}, C l_{2}$ and $H Cl$ are $434, 242$ and $431 k J m o l^{- 1}$ respectively. Enthalpy of formation of $H Cl$ is

$245 k J m o l^{- 1}$

$93 k J m o l^{- 1}$

$- 245 k J m o l^{- 1}$

$- 93 k J m o l^{- 1}$

Q. Bond dissociation enthalpy of H2​,Cl2​ and HCl are 434,242 and 431kJmol−1 respectively. Enthalpy of formation of HCl is

245kJmol−1

93kJmol−1

−245kJmol−1

−93kJmol−1

Q. Bond dissociation enthalpy of $H_{2}, C l_{2}$ and $H Cl$ are $434, 242$ and $431 k J m o l^{- 1}$ respectively. Enthalpy of formation of $H Cl$ is

$245 k J m o l^{- 1}$

$93 k J m o l^{- 1}$

$- 245 k J m o l^{- 1}$

$- 93 k J m o l^{- 1}$