The enthalpy of formation of `CH_(4), C_(2)H_(6)` and `C_(4)H_(10)` are -74.8, -84.7 and -126.1 kJ `mol^(-1)` respectively. Arrange them in the order of their efficiency as fuel per gram (enthalpy of formation of `CO_(2)(g)` and `H_(2)O(l)` are - 393.5 and `-285.8 kJ "mol"^(-1)` respectively).

Answer

`CH_(4) gt C_(2)H_(6) gt C_(4)H_(10)`

Explanation

The fuel efficiency can be calculated from the amount of heat evolved for every gram of fuel consumed.This can be done by first calculating the heat of combustion of these hydrocarbans.
(i) `CH_(4)(g) + 2O_(2)(g) to CO_(2)(g) + 2H_(2)O(l)`
`Delta_(c)H^(@) = [Delta_(f)H^(@)(CO_(2)) + 2Delta_(f)H^(@)(H_(2)O)] - [Delta_(f)H^(@)(CH_(4)) + 2 Delta_(f)H^(@) (O_(2))]`
` = [-393.5 + 2(-285.8)] - [(-74.8) + 0]`
` = -890.3 kJ "mol"^(-1)`
Heat evolved per gram = `(890.3)/(16) = 55.64 kJ g^(-1)`
(ii)`C_(2)H_(6)(g) + 7/2O_(2)(g) to 2CO_(2)(g) + 3H_(2)O(l)`
`Delta_(c)H^(@) = [2Delta_(f)H^(@)(CO_(2)) + 3Delta_(f)H^(@)(H_(2)O)] - [DeltaH^(@)(C_(2)H_(6)) + 7/2 Delta_(f)H^(@)(O_(2))]`
` = [2 xx (-393.5) + 3(-285.8)] - [-847 + 0]`
` = -1559.7 kJ "mol"^(-1)`
Heat evolved per gram = `(1569.7)/(30) = 51.99 kJ "g^(-1)`.
(iii)`C_(4)H_(10)(g) + 13/2O_(2)(g) to 4CO_(2)(g) + 5H_(2)O(l)`
`Delta_(c)JH^(@) = [4Delta_(f)H^(@)(CO_(2)) + 5Delta_(f)H^(@)(H_(2)O)] - [Delta_(f)H^(@)(C_(4)H_(10) + 13/2Delta_(f)H^(@)(O_(2))]`
` = [4(-393.5) + 5(-285.8)] -[(-126.1) + 0]`
` = -2876.9 kJ "mol"^(-1)`
Heat evolved per gram `= (2876.9)/(58) = 49.6 kJ "g^(-1)`
The decreasing order of fuel efficiency: `CH_(4) lt C_(2)H_(6) lt C_(4)H_(10)`.