The enthalpy change (ΔH) for the process N2H4(g)→2N(g)+4H(g) is 1724kJmol−1 if the bond energy of N-H bond in ammonia is 391kJmol−1 what is the bond energy of N-H bond is N2H4

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160kJmol−1

391kJmol−1

1173kJmol−1

320kJmol−1

answer is A.

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Detailed Solution

H−NH−NH−H (So, 4N−H bond present means their energy of N−N in N2H4 so the bond energy of N−N in N2H4=1724−1564=160KJ/mol.

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