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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