342

You visited us 342 times! Enjoying our articles? Unlock Full Access!

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}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

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

No worries! We‘ve got your back. Try BYJU‘S free classes today!

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

No worries! We‘ve got your back. Try BYJU‘S free classes today!

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

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

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}

Join BYJU'S Learning Program