Given the reactions CH_4 (g) rightarrow C (g) + 4H (g) partial differential H =

Question

Given the reactions CH_4 (g) rightarrow C (g) + 4H (g) partial differential H = 1656 kJ CF_4 (g) rightarrow C (g) + 4F (g) partial differential H = 1756 kJ CF_2 CH_2 rightarrow 2C (g) + 2H (g) + 2F (g) partial differential H = 2317 kJ Calculate the C = C bond energy in kilojoules per mole. Use Lewis Dot structures to predict the bonding in O_2 ^+ and O_2 and draw orbital energy diagrams of the ground state and hybridized state and explain the bonding patterns using Valence Bond Theory (the octet rule applies).

Explanation / Answer

Ans 1.

The bond energy of C-H bond from reaction 1 will be = 1656 / 4 = 414 KJ/mol

The bond energy of C-F from reaction 2 will be = 1756 / 4= 439 KJ/mol

Now substituting the bond energies of 2 C-H bonds and 2 C-F bonds in the enthalpy of reaction 3 , we can get the bond energy of C=C bond ,

(2 x 414) + (2 x 439) + BE = 2317

BE = 611 KJ/mol

So the bond energy of C=C is 611 KJ/mol

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