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Heat of Formation

Bond dissocia...

Question

Bond dissociation enthalpies of $H_{2} (g) a n d N_{2} (g)$ are $436.0 k J m o l^{- 1}$ and $941.3 k J m o l^{- 1}$ respectively. Enthalpy of formation of $N H_{3}$ (g) is $- 46.0 k J m o l^{- 1}$ . What is the enthalpy of atomization of $N H_{3}$ .

A

490.2 k J / m o l

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B

- 490.2 k J / m o l

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C

390.2 k J / m o l

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D

- 390.2 k J / m o l

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Solution

The correct option is C $390.2 k J / m o l$
We are given,
B.E. of $H_{2} = 436.0 k J / m o l$
B.E. of $N_{2} = 941.3 k J / m o l$

Reaction taking place is:
$\frac{1}{2} N_{2} + \frac{3}{2} H_{2} \to N H_{3}$

$Δ H^{o} (N H_{3}) = B . E . o f r e a c t a n t s - B . E . o f p r o d u c t s$
$= \frac{1}{2} (B . E . o f N_{2} b o n d) + \frac{3}{2} (B . E . o f H_{2} b o n d) - 3 (B . E . o f N - H b o n d$
putting all the values,
$- 46 = \frac{1}{2} (941.3) + \frac{3}{2} (436.0) - 3 (B . E . o f N - H b o n d$

solving this, we get;
$\Rightarrow B . E . o f N - H b o n d = 390.2 k J / m o l$

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Heat of Formation

Standard XII Chemistry

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