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Question

Consider the following data:

ΔfH0 (N2H4,l) = 50 kJ/mol

ΔfH0 (NH3,g) = 46k J/mol

B.E.(NH) = 393 kJ/mol and B.E.(HH) = 436 kJ/mol, also ΔvapH(N2H4,l) = 18 kJ/mol

The N- N bond energy in N2H4 is


A

226 kJ/mol

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B

154 kJ/mol

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C

190 kJ/mol

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D

45.45 kJ/mol

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Solution

The correct option is C

190 kJ/mol


12N2 + 32H2 NH3

(g) (g) (g)

Let BF of N≡ N s x

46 = x2 + 32 × 436 3 × 393

x = 958
N2H4(I) N2(g) + 2H2(g)
ΔH = 50kJ/mol

ΔH = [ΔHvap(N2H4,l) + 4 × BE(N H) + BE(N N) + BE(N N)] [B.E of (N N)] + 2BE(H H) 50 = (18 + 4 × 393 + y) [958 + 2 × 436]

50 = (1590 + y) (1830)

BE of (N N) (or)

Y = 190 KJ/mol


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