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Standard XII

Chemistry

Bond Order and Enthalpy

Enthalpy of f...

Question

Enthalpy of formation of $C_{2} H_{6} = - 83 k J / m o l$
Enthalpy of sublimation of graphite = $719 k J / m o l$
Enthalpy of bond dissociation of $H_{2} = 435 k J / m o l$
C - H bond enthalpy = $414 k J / m o l$
Calculate the bond enthalpy of C - C

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Solution

(i) Enthalpy of formation of $C_{2} H_{6} = - 83 K J m o l^{- 1}$
$2 C (s) + 3 H_{2} (g) ⟶ C_{2} H_{6} (g) . . . . - 83 K J m o l^{- 1}$
(ii) Enthalpy of sublimation of $C (s) ⟶ C (g) = 719 K J m o l^{- 1}$
Multiply by $2$
$∴ 2 C (s) ⟶ 2 C (g) = 1438 K J m o l^{- 1}$
(iii) Enthalpy of bond dissociation of $H_{2} (g) = 435 K J m o l^{- 1}$
Multiply by $3$
$3 H_{2} (g) ⟶ 6 H (g) = 1305 K J m o l^{- 1}$
Adding equation (ii) & (iii) and subtract from equation (i)
$2 C (g) + 6 H (g) ⟶ C_{2} H_{6} (g) ⟶ - 2826 K J m o l^{- 1}$
This means that enthalpy of formation of $6$ moles of $C - H$ bonds and one mole of $C - C$ bonds is $2826 K J m o l^{- 1}$
Enthalpy of formation of $6$ moles of $C - H$ bonds is $6 \times 414 K J m o l^{- 1} = 2484 K J m o l^{- 1}$
$∴$ Enthalpy of formation of $C - C$ bond is
$⟶ 2826 - 2484 = 342 K J m o l^{- 1}$

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Similar questions

Q. From the following data :
Enthalpy of formation of

C H_{3} C N = 87.86

kJ / mol
Enthalpy of formation of

C_{2} H_{6} = - 83.68

kJ / mol
Enthalpy of sublimation of graphite

= 719.65

kJ / mol
Enthalpy of dissociation of nitrogen

= 945.58

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Enthalpy of dissociation of

H_{2} = 435.14

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Calculate the bond enthalpy of ( i )

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Q. If enthalpy of a reaction is

1482 k J / m o l

and bond dissociation enthalpy of

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415 k J / m o l

than bond dissociation enthalpy of

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Q. Bond dissociation enthalpies of

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and enthalpy of formation of

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Q. Bond dissociation enthalpies of

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C = O = 728.0 k J {m o l}^{- 1}

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Δ H_{s u b} C (s) = 718.4 k J {m o l}^{- 1}

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Enthalpy of formation of C2H6=−83kJ/molEnthalpy of sublimation of graphite = 719kJ/molEnthalpy of bond dissociation of H2=435kJ/molC - H bond enthalpy = 414kJ/molCalculate the bond enthalpy of C - C

Enthalpy of formation of $C_{2} H_{6} = - 83 k J / m o l$
Enthalpy of sublimation of graphite = $719 k J / m o l$
Enthalpy of bond dissociation of $H_{2} = 435 k J / m o l$
C - H bond enthalpy = $414 k J / m o l$
Calculate the bond enthalpy of C - C