The enthalpy change of formation of `CO_(2)(g)` is -393 kJ `mol^(-1)` and that of `H_(2)O(l)` is -286 kJ `mol^(-1)`. The enthalpy of combustion of one mole of ethanol `(C_(2)H_(5)OH)` is -1360. the enthalpy change for the formation of one mole of ethanol form its constituent elements is

Options

A. `-681kJ`
B. `-284kJ`
C. `+965kJ`
D. `1360kJ`

Answer

B

Explanation

(b) Write the proper chemical equations,
`C("graphite")+O_(2)(g)toCO_(2)(g)`
`DeltaH=-393kJ` . . (i)
`H_(2)(g)+(1)/(2)O_(2)(g)toH_(2)O(l)`
`DeltaH=-286kJ` . . (ii)
`C_(2)H_(5)OH(l)+3O_(2)(g)to2CO_(2)(g)+3H_(2)O(l),DeltaH=-1360kJ` . . (iii)
from `(2xxI+3xxII)`-(III) we get
`2C+3H_(2)+(1)/(2)O_(2) to C_(2)H_(5)OH`
`therefore 2(-393kJ)+3(-286kJ)-(-1360kJ)=-284kJ`