Nitroglycerine (MW =227.1) denotes according to the following equation :
`2C_(3)H_(5)(NO_(3))_(3)I to 3N_(2)(g)+1//2O_(2)(g)+6CO_(2)(g)+5H_(2)O(g)`
The standard molar enthalpies of formation, `DeltaH_(f)^(@)` for the compounds are given bellow:
`DeltaH_(f)^(@)[C_(3)H_(5)(NO_(3))_(3)]= -364 kJ//mol`
`DeltaH_(f)^(@)[CO_(2)(g)]= -395.5 kJ//mol`
`DeltaH_(f)^(@)[H_(2)O(g)]= -241.8 kJ//mol`
`DeltaH_(f)^(@)[N_(2)(g)]= 0 kJ//mol`
`DeltaH_(f)^(@)[O_(2)(g)]= 0 kJ//mol`
The enthalpy change when 10g of nitroglycerine is detonated is

Options

A. `-100.5 kJ`
B. `-62.5 kJ`
C. `-80.3 kJ`
D. `-74.9 kJ`

Answer

B

Explanation

`{:(2C_(3)H_(5)(NO_(3))_(3)(l) ,+ 3N_(2)(g),+1//2O_(2)(g),+6CO_(2)(g),+5H_(2)O(g)),(," " darr," " darr," " darr," " darr ),(,DeltaH_(f)^(@)=O,DeltaH_(f)^(@)=O,DeltaH_(f)^(@)=O,DeltaH_(f)^(@)=O):}`
`DeltaH_("reaction")^(@)=3xx0+(1)/(2)xx0+6xx-393.5+5xx-241.8-2xx-364`
= -2842kJ ` to` for 2 mole of nitroglycerine for 1 mole or for 227.1g ` = -(2842)/(2)`
For 1 g `= -(2842)/(2xx227.1)xx10= - 62.5 kJ`