Question

The enthalpy of bond dissociation of CH4
constant presseure is 400 kcal at 25 ° C
Then calculate bond dissociation enthalpy
constant volume-​

Answer 1

Answer:

Dissociation reaction of CH

4

is,

CH

4

(g)→C(g)+4H(g) => ΔH=400 Kcal mol

−1

=4×(bond energy of C-H bond)

=> bond energy of C-H bond = 100Kcal

Similarly, dissociation reaction of C

2

H

6

is,

C

2

H

6

(g)→2C(g)+6H(g) => ΔH=670Kcal mol

−1

=(C-C bond energy)+6×(C-H bond energy)

=> (C-C bond energy)=670−6×100=70Kcal