A) 80 calories
B) 40 calories
C) 60 calories
D) 120 calories
Correct Answer: B
Solution :
\[C{{H}_{4}}\to C+4H,\,\,\Delta H=320\] \[{{E}_{C-H}}=90\,cal\] \[{{C}_{2}}{{H}_{6}}\to 2C+6H,\,\,\,\Delta H=360\] \[\therefore 360={{E}_{C-C}}+6{{E}_{C-H}}\] \[\therefore {{E}_{C-C}}=360-320=40\,\,cal\].You need to login to perform this action.
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