NH_(3)+ 3CI_(2) hArr NCI_(3) + 3HCI, - Delta H_(1) N_(2) + 3H_(2) hArr 2NH_(3) , - Delta H_(2) H_(2) + CI_(2) hArrr 2HCI, Delta H_(3) Calculate the enthalpy of formation of NCI_(3)

NH_(3)+ 3CI_(2) hArr NCI_(3) + 3HCI, - Delta H_(1) N_(2) + 3H_(2) hAr

1.	NH_(3)+ 3CI_(2) hArr NCI_(3) + 3HCI, - Delta H_(1) N_(2) + 3H_(2) hArr 2NH_(3) , - Delta H_(2) H_(2) + CI_(2) hArrr 2HCI, Delta H_(3) Calculate the enthalpy of formation of NCI_(3)
Answer» `Delta H_(f) = - Delta H_(1) + (Delta H_(2) )/( 2) - (3)/(2) Delta H_(3)` `Delta H_(f) = - Delta H_(2) + (Delta H_(2) )/( 2) - (3)/(2) Delta H_(3)` `Delta H_(f) = - Delta H_(2) + (-Delta H_(2) )/( 2) - (3)/(2) Delta H_(3)` none of above Answer :D