Question
Tue January 25, 2011 By: Renu Suresh
I went through the question's answer in the "NCERT Solutions" but I didn't understand it.
How do we know how to do the calculations?

DOUBT: NCERT Exercise Q.no.15

Expert Reply
Tue January 25, 2011
ΔH = 715+(2x242)-30.5-(135.5)
ΔH =1304 kjmol-1
bond enthalpy =1304/4 =326 kjmol-1
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