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Heat of Formation

The enthalpy ...

Question

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]

- 41.82 k J m o l^{- 1}

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+ 83.64 k J m o l^{- 1}

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

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- 833 k J m o l^{- 1}

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Solution

The correct option is A $- 41.82 k J m o l^{- 1}$
$3 H_{2}$ + $N_{2} \to 2 N H_{3}$

$Δ_{f} H (N H_{3}) = Bond energy of reactant - Bond energy of product$

$= (\frac{1}{2} B E o f N \equiv N b o n d + \frac{3}{2} B E o f H - H b o n d) - (3 B E o f N - H b o n d)$

$= (\frac{1}{2} \times 945.36 + \frac{3}{2} \times 435) - (3 \times 389)$

$= 472.68 + 652.5 - 1167$

$= - 41.82 k J / m o l$

Hence, the correct answer is option $A$ .

Suggest Corrections

Similar questions

: N \equiv N

946

m o l^{- 1}

H - H

435

m o l^{- 1}

N - N

159

m o l^{- 1}

N - H

389

m o l^{- 1}

m o l^{- 1}

N_{2} O

△ H^{\circ} (N_{2} O) = 82 k J m o l^{- 1}

N^{Θ} = N^{⨁} = O

(N \equiv N) b o n d = 950 k J m o l^{- 1}

(N = N) b o n d = 420 k J

^{- 1}

O = O) b o n d = 500 k J m o l^{- 1}

(O = N) b o n d = 610 k J m o l^{- 1}

N a C l (s)

Δ_{l a t t i c e} H = + 788 k J m o l^{- 1}

Δ_{s o l} H = + 4 k J m o l^{- 1}

N_{2} O (i n k J m o l^{- 1})

Δ H_{f}^{0}

N_{2} O

82 k J m o l^{- 1}

N \equiv N 946 k J m o l^{- 1}

N = N 418 k J m o l^{- 1}

O = O 498 k J m o l^{- 1}

N = O 607 k J m o l^{- 1}

K C l

K = 89 k J m o l^{- 1}

C l_{2} = 244 k J m o l^{- 1}

K = 425 k J m o l^{- 1}

C l = - 355 k J m o l^{- 1}

K C l = - 438 k J m o l^{- 1}

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