The standard molar heats of formation of ethane, carbon dioxide, and liquid water ate `-21.1, -94.1`, and `-68.3kcal`, respectively. Calculate the standard molar heat of combustion of ethane.

Explanation

The required chemicla equation for combustion of ethane is
`2C_(2)H_(6)(g) +7O_(2)(g) = 4CO_(2)(g) +6H_(2)O(l), DeltaH^(Theta) = ?`
The equation involves `2mol` of `C_(2)H_(6)`, heta of combustion of thene will be `=(DeltaH^(Theta))/(2)`
`DeltaH^(Theta) = Delta_(f)H^(Theta) ("products") - Delta_(f)H^(Theta) ("reactants")`
`[4 xx Delta_(f)H_((CO_(2)))^(Theta)+ 6Delta_(f)H_((H_(2)O))^(Theta)]`
`-[2Delta_(f)H_((C_(2)H_(6)))^(Theta) + 7Delta_(f)H_((O_(2)))^(Theta)]`
`= [4 xx (-94.1) + 6 xx (-68.3)] -[2xx (-2.11) +7 xx 0]`
`=- 376.4 - 409.8 + 42.2`
`= 744.0 kcal`
`(DeltaH^(Theta))/(2)=` Heat of combustion of ethane
` = (744.0)/(2) = - 372.0 kcal`