Q.

The enthalpy change (ΔH) for the process N2H4(g)2N(g)+4H(g) is 1724kJmol1 if the bond energy of N-H bond in ammonia is 391kJmol1 what is the bond energy of  N-H bond is N2H4

see full answer

Start JEE / NEET / Foundation preparation at rupees 99/day !!

21% of IItians & 23% of AIIMS delhi doctors are from Sri Chaitanya institute !!
An Intiative by Sri Chaitanya

a

160kJmol1

b

320kJmol1

c

391kJmol1

d

1173kJmol1

answer is A.

(Unlock A.I Detailed Solution for FREE)

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

Detailed Solution

 HNHNHH (So, 4NH bond present 

means their energy of NN in N2H4 so the bond energy of NN in N2H4

=17241564=160KJ/mol.

Watch 3-min video & get full concept clarity
score_test_img

Get Expert Academic Guidance – Connect with a Counselor Today!

whats app icon