Saved Bookmarks
| 1. |
Reaction of gasous fluorine `(F_2)` with compound X yields a single product Y, whose mass precent composition is 61.7% F and 38.3% Cl.Calculate `Delta_fH^@` (in kJ/mol) for the synthesis of Y using following information `2ClF(g)+O_2(g)toCl_2O(g)+OF_2(g) DeltaH^@=205.6 kJ` `2ClF_3(g)+2O_2(g)toCl_2O(g)+3OF_2(g) DeltaH^@=533.0 kJ` `Delta_f -I^@(OF_2, g)`=24.7 kJ/mol |
|
Answer» Correct Answer - `DeltaH_f^0 =139` `DeltaH_f^@=139` no. of moles `overset(Cl)(38.7/35.5=1.09)oversetF(61.7/19=3.42)` Simplest relative ratio 1:3 So, product Y is `CiF_3` & compound X is CIF For the synthesis of Y reaction is `F_2(g)+CIF(g)toCIF_3(g)` `1//2Cl_2O(g)+3//2OF_2(g)toCIF_3(g)+O_2(g) " " DeltaH^@=-533.0//2 kJ implies -266.5 kJ` `CIF(g)+1//2O_2(g)to1//2Cl_2O(g)+1//2OF_2" " DeltaH^@=205.6//2 kJ implies 102.8 kJ` `(F_2(g)+1/2O_2(g)toF_2O(g) " " DeltaH^@=24.7 kJ)/(CIF(g)+F_2(g)toCIF_3(g) " "Delta_fH^@=-139kJ//mol ` |
|