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Enthalpy

Heat of atomi...

Question

Heat of atomisation of $N H_{3}$ and $N_{2} H_{4}$ are $x k c a l m o l^{- 1}$ and $y k c a l m o l^{- 1}$ respectively. What will be average bond energy of $N - N$ ?

A

\frac{- 3 y - 4 x}{3} k c a l m o l^{- 1}

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B

\frac{3 y - 4 x}{3} k c a l m o l^{- 1}

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C

\frac{3 y + 4 x}{3} k c a l m o l^{- 1}

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D

None of the above

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Solution

The correct option is B $\frac{3 y - 4 x}{3} k c a l m o l^{- 1}$
Given reactions are:
$N H_{3} \to N (g) + 3 H (g) Δ H_{a t o m} = x k c a l m o l^{- 1} = Δ H_{1} = 3 \times B E_{N - H} . . . . . .1$
$N_{2} H_{4} \to 2 N (g) + 4 H (g) Δ H_{a t o m} = y k c a l m o l^{- 1} = Δ H_{2} = 4 \times B E_{N - H} + B E_{N - N} . . . . . .2$
from equation 1 we have
$B E_{N - H} = \frac{x}{3}$
put $B E_{N - H}$ in equation 2 to get the value of $B E_{N - N}$
$Δ_{a t o m} H = y k c a l m o l^{- 1} = Δ H_{2} = 4 \times \frac{x}{3} + B E_{N - N}$
$B E_{N - N} = y - \frac{4 x}{3}$
$B E_{N - N} = \frac{3 y - 4 x}{3} k c a l m o l^{- 1}$
hence, the average bond energy of $N - N$ is $\frac{3 y - 4 x}{3} k c a l m o l^{- 1}$

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