Answer:
Option D
Explanation:
For the calaculation of C≡C bond energy we must first calculate dissociation energy of C2H2 as
C2H2(g)→ 2C(g)+2H(g).....(i)
Using the given bond energoes and enthalpies
C2H2(g)→2C(g)+2H(g);
△H=−225kJ ........(ii)
2C(s)→2C(g); △H=1410kJ..........(iii)
H2(g)→2H(g); △H=330kJ......(iv)
adding equation (ii),(iii) and (iv) gives equation(i)
⇒ C2H2(g)→2C(g)+2H(g);
△H=1515kJ
⇒ 1515kJ=2× (C-H)BE+(C≡C)BE
=2×350+(C≡C)BE
⇒ (C≡C)BE=1515-700=815 kJ/mol