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Combustion enthalpies of methane and ethane are -210k. "cal mol"^(-1) and -368 k. "cal mol"^(-1). Combustion enthalpy of Decane is ......... . |
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Answer» `-158 "k. cal"` 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)` |
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