Combustion enthalpies of methane and ethane are `-210k. "cal mol"^(-1) and -368 k. "cal mol"^(-1)`. Combustion enthalpy of Decane is ......... .

Options

A. `-158 "k. cal"`
B. `-1632 "k. cal"`
C. `-1700 "k.cal"`
D. given mettre is not complete

Answer

B

Explanation

Combustion enthalpy of ethane `C_(2) H_(6) = - 368 "k cal mol"^(-1)`
Combustion enthalpy of methane `CH_(4) = -210 "k cal mol"^(-1)`
When, `CH_(4) to C_(2) H_(6)` is produced, one `- CH_(2)` is increased.
Increase in combustion enthalpy by increasing `- CH_(2) , - 210- (-368) = + 158 "k cal per" CH_(2)`
There is difference of nine `CH_(2)` between `C_(10) H_(22) and CH_(4)`.
Therefore, change in combustion enthalpy will be increase of nine `CH_(2)`
`9 xx 158 = 1422` K cal per nine `CH_(2)`
`therefore` Combustion energy of decane = sum of enthalpies of methane and nine `- CH_(2)`
`= 210+ 1422`
`= 1632 "K cal mol"^(-1)`
`rArr` Combustion enthalpy of Decane
`Delta H_("combustion") ("Decane")`
`= - 1632 "K cal mol"^(-1)`