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Standard XII

Chemistry

Nitrogen

Enthalpy of f...

Question

Enthalpy of formation of NH3 is Xkj and Bond Energy H-H,Bond Energy N-H are respectively Ykjmol^-1 and Zmol^-1 the value of Bond Energy N=N is

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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. Heat of formation of

2

moles of

{N H}_{3} (g)

- 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:

Q. Using the bond enthalpy data given below, calculate the enthalpy of formation of acetone (g)
Bond energy

C - H = 413.4 k J {m o l}^{- 1}

;
Bond energy

C - C = 347.0 k J {m o l}^{- 1}

;
Bond energy

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

;
Bond energy

O = O = 495.0 k J {m o l}^{- 1}

;
Bond energy

H - H = 435.8 k J {m o l}^{- 1}

;

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

Given the bond energies $N \equiv N$ , $H - H$ and $N - H$ bonds are $945, 436$ and $391 k J m o l e^{- 1}$ respectively, the enthalpy of the following reaction $N_{2} (g) + 3 H_{2} (g) \to 2 N H_{3} (g)$ is

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]

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