question_answer

Consider the reaction : \[4N{{O}_{2(g)}}+{{O}_{2(g)}}\to 2{{N}_{2}}{{O}_{5}}_{(g)}'\]\[{{\Delta }_{r}}H=-111\,kJ.\] If \[{{N}_{2}}{{O}_{5(s)}}\]is formed instead of \[{{N}_{2}}{{O}_{5(g)}}\]in the above reaction, the \[{{\Delta }_{r}}H\]value will be: (given, \[\Delta H\]of sublimation for \[{{N}_{2}}{{O}_{5}}\]is \[54kJ\,mo{{l}^{-1}}\])     AIEEE  Solved  Paper (Held On 11 May  2011)

A)  +54kJ           

B)  +219kJ         

C)  -219kJ         

D) -165kJ

Correct Answer: D

Solution :

                                                \[-111-54=\Delta H'\] \[\Delta H'=-165KJ\]