For the reaction:
N₂H₄(g) → N₂H₂(g) + H₂(g) ; $109 k J / m o l$
Given bond energies:
$E_{N - N} = 163 k J / m o l$
$E_{N - H} = 391 k J / m o l$
$E_{H - H} = 436 k J / m o l$
Then the bond dissociation energy of N=N is

Question

For the reaction:

N₂H₄(g) → N₂H₂(g) + H₂(g) ; $109 k J / m o l$

Given bond energies:

$E_{N - N} = 163 k J / m o l$

$E_{N - H} = 391 k J / m o l$

$E_{H - H} = 436 k J / m o l$

Then the bond dissociation energy of N=N is

TestHub · Accepted Answer

109 = ( ϵ N − N ​ + 4 ϵ N − H ​ ) − ( ϵ N = N ​ + 2 ϵ N − H ​ ) − ϵ H − H ​ 109 109 ϵ N = N ​ ϵ N = N ​ ​ = ϵ N − N ​ − ϵ N = N ​ + 2 ϵ N − H ​ − ϵ H − H ​ = 163 − ϵ N = N ​ + 2 × 391 − 436 = 54 + 782 − 436 = 836 − 436 = 400 kJ ​

Chemistry - Thermochemistry Question with Solution | TestHub

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