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Enthalpy

The bond enth...

Question

The bond enthalpies of $H - H, C l - C l$ and $H - C l$ are $435, 243$ and $431$ kJ $m o l^{- 1}$ , respectively. The enthalpy of formation of $H C l (g)$ will be:

92

kJ mol

^{- 1}

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- 92

kJ mol

^{- 1}

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247

kJ mol

^{- 1}

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770

kJ mol

^{- 1}

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Solution

The correct option is C $- 92$ kJ mol $^{- 1}$
Enthalpy of formation $H C l$ would be

$\frac{1}{2} H_{2} + \frac{1}{2} C l_{2} \to H C l$

So fromation enthalphy $= \frac{1}{2} \times (435 + 235) - 431 = - 92 k J m o l^{- 1}$

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Similar questions

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. 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

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

. Bond enthalpy of

H C l

is :

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:

Q. Calculate

△ H_{r}^{o}

for the reaction

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

The average bond enthalpies of C - H and C - Cl bonds are

414 k J m o l^{- 1}

and

330 k J m o l^{- 1} .

Q. The enthalpies of formation of

O H (g)

H (g)

and

O (g)

are 21

k J m o l^{- 1}

, 109

k J m o l^{- 1}

and 124

k J m o l^{- 1} .

The value of bond enthalpy

O - H

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The bond enthalpies of H−H, Cl−Cl and H−Cl are 435,243 and 431 kJ mol−1, respectively. The enthalpy of formation of HCl(g) will be:

The correct option is C −92 kJ mol−1Enthalpy of formation HCl would be 12H2+12Cl2→HClSo fromation enthalphy =12×(435+235)−431=−92 kJmol−1

The bond enthalpies of $H - H, C l - C l$ and $H - C l$ are $435, 243$ and $431$ kJ $m o l^{- 1}$ , respectively. The enthalpy of formation of $H C l (g)$ will be:

The correct option is C $- 92$ kJ mol $^{- 1}$
Enthalpy of formation $H C l$ would be

$\frac{1}{2} H_{2} + \frac{1}{2} C l_{2} \to H C l$

So fromation enthalphy $= \frac{1}{2} \times (435 + 235) - 431 = - 92 k J m o l^{- 1}$