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Standard XII

Chemistry

Enthalpy

Heat of forma...

Question

Heat of formation of $2$ moles of ${N H}_{3} (g)$ is $- 90 k J$ ; bond energies of $H - H$ and $N - N$ bonds are $435 k J$ and $390 k J$ ${m o l}^{- 1}$ respectively. The value of the bond energy of $N \equiv N$ will be:

- 472.5 k J

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- 945 k J

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472.5 k J

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945 k J

{m o l}^{- 1}

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Solution

The correct option is D $945 k J$ ${m o l}^{- 1}$
$Δ H_{r e a c t i o n} = \sum {(B E)}_{r e a c t a n t s} - \sum {(B E)}_{p r o d u c t s}$
$- 90 = x + 3 \times 435 - 6 \times 390$
$x = 945 k J$ ${m o l}^{- 1}$

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Similar questions

Q. Calculate the enthalpy of formation of ammonia from the following bond energy data:

(N - N)

bond

= 398 k J {m o l}^{- 1}

;

(H - H)

bond

= 435 k J {m o l}^{- 1}

and

(N \equiv N)

bond

= 945.36 k J {m o l}^{- 1}

Q. The heat of formation of NO from its elements is

+ 90 k j {m o l}^{- 1}

. What is the approximate bond dissociation energy of the bond in NO?
Bond Bond energy Bond Bond energy
N=N 941kj

{m o l}^{- 1}

O=O 499kJ

{m o l}^{- 1}

Q. Bond dissociation enthalpies of

H_{2_{(} g)}

and

N_{2_{(} g)}

are 436.0

K J

m o l^{- 1}

and 941.8

K J

m o l^{- 1}

, respectively and enthalpy of formation of

N H_{3_{(} g)}

- 46 K J

m o l^{- 1}

What is the enthalpy of atomization of

N H_{3_{(} g)}

and the average bond enthalpy of

N - H

bond respectively (in

K J

m o l^{- 1}

Q. The enthalpy of formation of ammonia when calculated from the following bond energy data is:

[B.E. of

N - H, H - H, N \equiv N

is 389 kJ

m o l^{- 1}

, 435 kJ

m o l^{- 1}

, 945.36 kJ

m o l^{- 1}

respectively]

Q. Given that the bond energies of

: N \equiv N

946

m o l^{- 1}

H - H

435

m o l^{- 1}

N - N

159

m o l^{- 1}

, and

N - H

389

m o l^{- 1}

, the heat of formation of hydrazine in the gas phase in kJ

m o l^{- 1}

is:

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Heat of formation of 2 moles of NH3(g) is −90kJ; bond energies of H−H andN−N bonds are 435kJ and 390kJ mol−1 respectively. The value of the bond energy of N≡N will be:

The correct option is D 945 kJ mol−1ΔHreaction=∑(BE)reactants−∑(BE)products−90=x+3×435−6×390x=945kJ mol−1

Heat of formation of $2$ moles of ${N H}_{3} (g)$ is $- 90 k J$ ; bond energies of $H - H$ and $N - N$ bonds are $435 k J$ and $390 k J$ ${m o l}^{- 1}$ respectively. The value of the bond energy of $N \equiv N$ will be:

The correct option is D $945 k J$ ${m o l}^{- 1}$
$Δ H_{r e a c t i o n} = \sum {(B E)}_{r e a c t a n t s} - \sum {(B E)}_{p r o d u c t s}$
$- 90 = x + 3 \times 435 - 6 \times 390$
$x = 945 k J$ ${m o l}^{- 1}$