A) \[50\]Calories
B) \[80\] calories
C) \[40\] Calories
D) \[120\] calories
Correct Answer: B
Solution :
| given, enthalpy of \[C{{H}_{4}}=320cal\] |
| Enthalpy of \[{{C}_{2}}{{H}_{6}}=560\,cal\] |
| \[C{{H}_{4}}(g)\xrightarrow{{}}C(g)+4H(g)+4H(g);\Delta H=320cal\]\[4\Delta {{\operatorname{H}}_{\operatorname{C}-H}}=320\,cal\] |
| \[\therefore \] \[\Delta {{H}_{c-H}}=\frac{320}{4}=80cal\] |
| \[{{C}_{2}}{{H}_{6}}(g)\xrightarrow{{}}2C(g)+6H(g);\]\[\Delta \operatorname{H}=560\operatorname{cal}\] |
| \[\therefore \]\[6\Delta {{H}_{C-H}}+\Delta {{H}_{C-C}}=560\,cal\] |
| \[\therefore \]\[\Delta {{H}_{C-C}}=560-6\Delta {{H}_{C-H}}\] |
| \[=560-6\times 80\] |
| \[=560-480=80\,calories\] |
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