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Factors Affected by Hydrogen Bonding

In solid NH...

Question

In solid $N H_{3}$ , each $N H_{3}$ molecule has six other $N H_{3}$ molecules as nearest neighbours. $Δ H$ of sublimation of $N H_{3}$ at the melting point is $30.8$ kJ/mol and the estimated $Δ H$ of sublimation in absence of $H -$ bonding is $14.3$ kJ/mol. The strength of $H -$ bond in $N H_{3}$ ion kJ/mol is:

A

5.5

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B

11.0

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C

16.5

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D

2.75

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Solution

The correct option is A $5.5$
Total strength of $H -$ bonds $= 30.8 - 14.4 = 16.4$ kJ/mol.

There are $6$ nearest neighbors but each $-$ bond involves $2$ molecules.

Thus, total $H -$ bond $= \frac{6}{2} = 3$

$∴$ $H -$ bond energy $= \frac{16.4}{3} = 5.5$ kJ/mol

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Similar questions

Q. In solid ammonia, each

{N H}_{3}

molecule has six other

{N H}_{3}

molecules as nearest neighbours.

Δ H

of sublimation of

{N H}_{3}

at the melting point is

30.8 k J

{m o l}^{- 1}

and the estimated

Δ H

of sublimation in the absence of hydrogen bonding is

14.4 k J

{m o l}^{- 1}

. What is the strength of hydrogen bond in solid ammonia in

k J

{m o l}^{- 1}

? Give your answer in nearest integer number.

Q. The standard enthalpy of formation of

N H_{3}

46.0

kJ/mol. If the enthalpy of formation of

H_{2}

from its atoms is

- 436

kJ/mol and that of

N_{2}

- 712

kJ/mol, the average bond enthalpy of

N - H

N H_{3}

The standard enthalpy of formation of $N H_{3}$ is $46.0 k J m o l^{- 1}$ . If the enthalpy of formation of $H_{2}$ from its atom is $- 436 k J m o l^{- 1}$ and that of $N_{2}$ is $- 712 k J m o l^{- 1}$ , the average bond enthalpy of $N - H$ bond in $N H_{3}$ is

N_{2} (g) + 3 H_{2} (g) ⟶ 2 N H_{3} : Δ H = - 92 k J

is Haber`s process for manufacture of

N H_{3}

. What is the heat of formation of

N H_{3}

Q. Consider the following data:

Δ_{f} H^{\circ} (N_{2} H_{4}, l) = 50 kJ/mol, Δ_{f} H^{\circ} (N H_{3}, g) = - 46 kJ/mol

B . E (N - H) = 393 kJ/mol

B . E (H - H) = 436 kJ/mol

Δ_{v a p} H (N_{2} H_{4} l) = 18 kJ/mol

. Calculate the N-N bond energy in kJ/mol for

N_{2} H_{4}

.

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