Question

For the reactions, I. $C{{H}_{4}}(g)\xrightarrow[{}]{{}}C(g)+4H(g) ; \Delta H={{x}_{1}}$
II. $ {{C}_{2}}{{H}_{6}}(g)\xrightarrow[{}]{{}}2C(g)+6H(g);\Delta H={{x}_{2}}$
from I and II, bond energy of $C-C$ bond is

• A. $ {{x}_{1}}-{{x}_{2}}$
• B. ${{x}_{2}}-{{x}_{1}} $
• C. ${{x}_{2}}+1.5{{x}_{1}} $
• D. $ {{x}_{2}}-1.5{{x}_{1}}$

Answer 1

Answer: (D)

(I) $C{{H}_{4}}(g)\xrightarrow[{}]{{}}C(g)+4H(g)$
$\Delta H=x\text{ }for\text{ }four\text{ }C-H\text{ }bonds.$
$\therefore $ BE of $(C-H)$ bond $=\frac{{{x}_{1}}}{4}$
(II) ${{C}_{6}}{{H}_{6}}(g)\xrightarrow[{}]{{}}2C(g)+6H(g)$
$\Delta H={{x}_{2}}={{(BE)}_{C-C}}+6{{(BE)}_{C-H}}$
${{x}_{2}}=B{{E}_{C-C}}+6\times \frac{{{x}_{1}}}{4}$
${{(BE)}_{C-C}}={{x}_{2}}-1.5{{x}_{1}}$