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If the given reaction is carried out with 1 mol N2 and excess of H2 at a constant pressure of 40 bar then volume contraction of 2 L is observed. [Given : N2 (g) + 3H2 (g)  2NH3 H° = – 90 kJ] Now if 2 mol N2 and 8 mol H2 are taken then calculate (a) PV work done (b) U° for given amount
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Asked by kartikey.zopcart | 30 Oct, 2021, 04:17: PM
answered-by-expert Expert Answer
Based on the values of B.E. given, ΔfH0 of N2H4(g)  is :
Given BE of : 
N−N is 159 kJ mol−1,
H-H is 436 kJ mol−1,
 N≡N is 941 kJ mol−1,
 N−H is 398 kJ mol−1
 
N2 (g) + 2H2 (g) → 2N2H2
 
ΔfH0 of N2H4(g)  = 941 + (2× 436) - (4×398 +159)
 
                        = 1813 - 1751
 
                        = 62 kg mol-1
Answered by Ramandeep | 30 Oct, 2021, 07:07: PM

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