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Nitroglycerine (MW =227.1) detonates 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 isA. `-100.5 kJ`B. `-62.5 kJ`C. `-80.3 kJ`D. `-74.9 kJ` |
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Answer» Correct Answer - B `{:(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` |
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