InterviewSolution
Saved Bookmarks
InterviewSolution
| 1. |
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)`. |
|
Answer» `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 |
|