Calculate the amount of heat evolved during the complete combustion of `100 ml` of liquid benzene from the following data. Predict your answer as `(Delta H)/(100)` ( in KJ/mol).
(i) `18 gm` of graphite on complete combustion evolve 585 KJ heat
(ii) 15540 KJ heat is required to dissociate all the molecules of 1 litre water into `H_(2)` and `O_(2)`.
(iii) The heat of formation of liquid benzene is 48 kJ/mol
(iv) Density of `C_(6)H_(6)(l)=0.87 gm//ml`

Answer

36 KJ

Explanation

`(i)" "C(s) + O_(2)(g) rarr CO_(2)(g) , DeltaH_(f)^(0) = -390KJ//mol, " " DeltaH_(f)^(0) = (585)/(18)xx12`
`(ii) " " H_(2)(g)+(1)/(2)O_(2)(g) rarr H_(2)(l) , DeltaH_(f)^(0) =-280 KJ//mol ," " Delta H_(f)^(0) = (15540)/(55.5)`
`(iii)" " C_(6)H_(6)(l) + (15)/(2)O_(2)(g) rarr 6CO_(2)(g) + 3H_(2)O(l) ," " DeltaH_(f)^(0) = -48`
`therefore" "DeltaH^(0) = [6(-390) + 3(-280)] - 48 = -3228 KJ//mol`
`" "` Mass of benzene is `= 0.87 xx 100 = 87g`
`therefore" "` Heat evolved from `87` gm benzene `=3600KJ`
`" "` Hence , `(DeltaH)/(100)=36KJ.`