At `25^(@)C`, buring `0.2 "mole" H_(2)` with `0.1` mole `O_(2)` to produce `H_(2)O(l)` in a bomb calorimeter (constant volume) raises the temperature of the apperaturs `0.88^(@)C`. When `0.01mol` toulene is burned in this calorimeter, the temperature is raised by `0.615^(@)C`. Calculate `DeltaH^(Theta)` combustion of toluene. `Delta_(f)H^(Theta) H_(2)O(l) =- 286 kJ mol^(-1)`.

Explanation

`Delta_(f)H(H_(2)O) =- 286 kJ "mole"^(-1)`
`{:(H_(2)(g)+,(1)/(2)O_(2)(g)rarr,H_(2)O(l)),(0.2,0.1,0.2):}`
`Deltan_(g) = 0 -(3)/(2) =- (3)/(2)`
`DeltaU = DeltaH - Deltan_(g)RT`
`=- 286 -(-(3)/(2)) xx 8.314 xx 10^(-3) xx 298`
`=- 282.3 kJ mol^(-1)`
`=- 282.3 xx 0.2` (for 0.2 mole)
Heat capacity of calorimetre
`DeltaH = (ms) Deltat`
`(ms) = (DeltaH)/(DeltaT) = (282.3 xx 0.2)/(0.88) = 64.159`
`DeltaH` for `0.01ml` of Toluene
`DeltaU` (atomic constant volume) `= ms xx Deltat`
`= - 64.159 xx 0.615 kJ//0.01 mol`
`=- 39.458 kJ// 0.01 mol`
`=- 39.458 xx 100 kJ mol^(-1)`
`=- 3945.8 kJ mol^(-1)`
`=- 3945.8 kJ mol^(-1)`
`C_(7)H_(8)(l) +9O_(2)(g) rarr 7CO_(2)(g) +4H_(2)O(l)`
`Deltan_(g) = 7 - 9 = -2`
`DeltaH = DeltaU +Deltan_(g)RT`
`=- 3945.8 -2 xx 8.314 xx 10^(-3) xx 298`
`=- 3945.8 -5 =- 3950.8 kJ`mole