Calculate the heat of glucose and its calorific value from following data :
(i)`C_"(graphite)"+O_(2(g)) to CO_(2(g)) , DeltaH`=-395 KJ
(ii)`H_(2(g)) +1//2O_2 to H_2O_((l)) , DeltaH`=-269.4 KJ
(iii)`6C+6H_(2(g)) +3O_(2(g)) to C_6H_12O_(6(s)) , DeltaH`=-1169.8 KJ

Explanation

Calorific value of glucose `DeltaH_C`= ?
`underset"Glucose"(C_6H_12O_6(s))+6O_2 to 6CO_(2(g)) + 6H_2O_((l)) " " DeltaH_C`= ?
Data given are ,
(i)`C_"(graphite)" + O_(2(g)) to CO_(2(g)) " " DeltaH`=-395 KJ ….(1)
(ii)`H_(2(g)) +1//2 O_2 to H_2O_((l)) " " DeltaH`=-269.4 KJ ...(2)
(iii)`C+6H_(2(g)) +3O_(2(g)) to C_6H_12O_(6(s)) " " DeltaH`=-1169.8 KJ ...(3)
Equation (1) x 6 `6C + 6O_2 to 6CO_2 " " DeltaH`=-2370 KJ
Equation (2) x 6 `6H_2 + 3O_2 to 6H_2O " " DeltaH`=-1616.4 KJ
Equation (3) is reversed ,
`C_6H_12O_6 to 6C + 6H_2 +3O_2 " " DeltaH`=+ 1169.8 KJ
Add all equations,
`C_6H_12O_6 +6O_2 to 6CO_2 + 6H_2O`
`DeltaH`=-2370 -1616.4 +1169.8
`DeltaH`=-3986.4 +1169.8
`DeltaH` =-2816.6 KJ
Calorific value of `C_6H_12 O_6=-2816.6 "KJ mol"^(-1)`