Answer to Question #92415 in Physical Chemistry for Harish

Question

Find ∆H in terms of kj for the following reaction
M(g)+2X(g)--->M²+(g)+2X^-(g)
(IE)1 of M(g)=705Kj/mol
(IE)2 of M(g)=1151KJ/mol
(EA)1 of X(g)= -328KJ/mol
Answer 1200KJ

Answer 1

∆H=((IE)_1(M) +(IE)_2(M)+2(EA)_1(X))-(2·∆H_(X)+∆H_(M))=705+1151+2·(-328)-2·0+0=1200 KJ