Find ΔH in terms of KJ for the following reaction from the given dataM(g) + 2X(g) → M2+(g) + 2X–(g)(IE)1 of M(g) = 705 KJ mol–1(IE)2 of M(g) = 1151 KJ mol–1(EA)1 of X(g) = –328 KJmol–1

M(g)+2X(g)→M+2(g)+2X-(g)...∆HI....                M(g)→M+(g)      ∆H1= +705KJ.......(1)                    M+(g)→M+2(g)    ∆H2= +1151KJ.......(2)II.....              X(g)→X-(g)         ∆H= -328KJ                     2X(g)→2X-(g)      ∆H= -656KJ.......(3)∆H=∆H1+∆H2+∆H3     = [+705+1151+(-656)]KJ    =+1200KJ