The ionization enthalpy of Na+ formation from Nag is 495.8 kJ mol−1 , while the electron gain enthalpy of Br is -325.0 kJ mol-1. Given the lattice enthalpy of NaBr is −728.4 kJ mol−1 . The energy for the formation of NaBr ionic solid is −_______×10−1 kJ mol−1 .
see full answer
Your Exam Success, Personally Taken Care Of
1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.
An Intiative by Sri Chaitanya
answer is -5576.
(Unlock A.I Detailed Solution for FREE)
Best Courses for You
JEE
NEET
Foundation JEE
Foundation NEET
CBSE
Detailed Solution
=495.8+(-32.5)+(-728.4)=-557.6=-5576×10-1KJ/mol.Note : The above calculation is not for ΔHformation but for ΔHReactionBut on the basis of given data it is the best ans