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Factors Affected by Hydrogen Bonding

Each NH3 mole...

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Each NH3 molecule has six other NH3 molecules as its nearest neighbours in solid state; △ H of sublimation of NH3 molecule at the m.p. is 30.8 kJ/mol, and in the absence of H-bonding estimated △ H of sublimation is 14.4 kJ/mol. Hence strength of H-bond in solid NH3 is-1)5.47 KJ/mol, 2)10.93 kJ/mol, 3)16.40 KJ/mol.

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N H_{3}

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 -

14.3

kJ/mol. The strength of

H -

N H_{3}

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

Q. White phosphorous is a tetra atomic solid

[P_{4} (s)]

at room temperature and on strong heating in absence of oxygen, it polymerizes into red phosphorus as:

Δ H = - 104 kJ/mol of P_{4}

The enthalpy of sublimation

[P_{4} (s) \to P_{4} (g)]

59 kJ/mol

and enthalpy of atomization is

316.25 kJ/mol

P (g)

.
Now find the average

P - P

bond enthalpy in

P_{4}

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Factors Affected by Hydrogen Bonding

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