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Bond Order and Enthalpy

If Δ f H ⊖ ...

Question

If $Δ_{f} H^{⊖} o f I C l (g), C l (g), a n d I (g) i s 17.57, 121.34 a n d 106.96 J^{- 1}$ , respectively. Then bond dissociation energy of I-Cl bond is:

A

35.15 J^{- 1}

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B

106.69 J^{- 1}

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C

210.73 J^{- 1}

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D

420.9 J^{- 1}

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Solution

The correct option is A $35.15 J^{- 1}$
Bond dissociation energy $= 2 \times Δ H_{f}$ of $I C l (g)$

Hence $D E = 2 \times 17.57$ $J^{- 1}$

$D E = 35.15$ $J^{- 1}$

This is because $I_{2} + C l_{2} ⇌ 2 I - C l$

$2$ moles of $I C l$ are formed

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Similar questions

Q. If values of

Δ H_{f}^{o}

I C l (g), C l (g), I (g)

are respectively

17.57, 121.34, 106.96

^{- 1}

I - C l

(bond energy) in J mol

^{- 1}

Q. Calculate free energy change for the reaction at

27^{o} C

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

by using the bond energy and energy data.
Bond energies of

H - H, C l - C l

H - C l

435 k J m o l^{- 1}, 240 k J m o l^{- 1}

430 k J m o l^{- 1}

respectively. Standard entropies of

H_{2}, C l_{2}

H C l

131, 223

187 J K^{- 1} m o l^{- 1}

Q. Calculate the bond energy of

C l - C l

bond from the following data:

C H_{4 (g)} + C l_{(g)} \to C H_{3} C l_{(g)} + H C l_{(g)}; Δ H = - 100.3 k J

. Also the bond enthalpies of

C - H, C - C l, H - C l

413, 326

431

k J m o l^{- 1}

Q. Bond dissociation energy of

C H_{4} is 360 k J m o l^{- 1}

C_{2} H_{6} is 620 k J m o l^{- 1}

. Then bond dissociation energy of

C - C

Q. Bond energies of H - H and CI - CI are

430 k J m o l^{- 1}

242 k J m o l^{- 1}

Δ H_{f}

91 k J m o l^{- 1}

. What will be the bond energy of H - Cl bond (per mole value)?

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