| 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) \[{{x}_{2}}-1.5{{x}_{1}}\]
Correct Answer: D
Solution :
(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) \[{{C}_{6}}{{H}_{6}}(g)\xrightarrow[{}]{{}}2C(g)+6H(g)\] \[\Delta H={{x}_{2}}={{(BE)}_{C-C}}+6{{(BE)}_{C-H}}\] \[{{x}_{2}}=B{{E}_{C-C}}+6\times \frac{{{x}_{1}}}{4}\] \[{{(BE)}_{C-C}}={{x}_{2}}-1.5{{x}_{1}}\]
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