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Heat of Formation

Bond dissocia...

Question

Bond dissociation enthalpy of $H_{2}, C l_{2}$ and $H C l$ are 434, 242 and 431 kJ $m o l^{- 1}$ respectively.

Enthalpy of formation of $H C l$ is:

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

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245 k J m o l^{- 1}

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93 k J m o l^{- 1}

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

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Solution

The correct option is D $- 93 k J m o l^{- 1}$
The reaction between hydrogen and chlorine molecules to form hydrogen chloride. This reaction can be broken down into these simple steps:

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

Step 1: Dissociation of Hydrogen and Chlorine molecules into their respective atoms
$H_{2} (g) \to 2 H (g)$
$C l_{2} (g) \to 2 C l (g)$

Step 2: Combination of these atoms to form hydrogen chloride
$2 H (g) + 2 C l (g) \to 2 H C l (g)$

Since in the first step, one-mole of each $H - H$ and $C l - C l$ bonds are broken, it is an endothermic process ( $Δ H_{1} > 0$ ) while in the second step two moles of $H C l$ are being formed which releases energy making $Δ H_{2} < 0$

Now, $Δ H_{1} = Δ H_{C l - C l} + Δ H_{H - H}$
$= 434 + 242$
$= 676 k J m o l^{- 1}$

Now, $Δ H_{2} = - 2 Δ H_{H - C l}$
$= - 862 k J m o l^{- 1}$

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 $H C l$ . Hence the enthalpy of formation of one mole of $H C l$ will be

$Δ H = - 93 k J m o l^{- 1}$

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Similar questions

Q. Bond dissociation enthalpy of

H_{2}, C l_{2}

and

H C l

are

434, 242

and

431 {kJ mol}^{- 1}

respectively.
Enthalpy of formation of

H C l

{kJ mol}^{- 1}

X

, then value of

X

is:

Given that bond energies of
$H - H$ and $C l - C l$ are $430 k J m o l^{- 1}$ and $240 k J m o l^{- 1}$ respectively and $Δ_{f} H$ for $H C l$ is -90 kJ $m o l^{- 1} .$ The bond enthalpy of $H C l$ is:

Q. Bond dissociation enthalpies of

H_{2} (g) a n d N_{2} (g)

are

436.0 k J m o l^{- 1}

and

941.3 k J m o l^{- 1}

respectively. Enthalpy of formation of

N H_{3}

(g) is

- 46.0 k J m o l^{- 1}

. What is the enthalpy of atomization of

N H_{3}

Q. Bond dissociation enthalphies of

H_{2} (g)

and

N_{2} (g)

are 436.0 kJ

m o l^{- 1}

and 941.8 kJ

m o l^{- 1}

respectively and enthalpy of formation of

N H_{3} (g)

is -46 kJ

m o l^{- 1}

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Q. Bond dissociation enthalpies of

H_{2_{(} g)}

and

N_{2_{(} g)}

are 436.0

K J

m o l^{- 1}

and 941.8

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m o l^{- 1}

, respectively and enthalpy of formation of

N H_{3_{(} g)}

- 46 K J

m o l^{- 1}

What is the enthalpy of atomization of

N H_{3_{(} g)}

and the average bond enthalpy of

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

Bond dissociation enthalpy of $H_{2}, C l_{2}$ and $H C l$ are 434, 242 and 431 kJ $m o l^{- 1}$ respectively.

Enthalpy of formation of $H C l$ is: