Question

From the data below, calculate ∆Hºhydr(Z2+).

∆Hºlattice(ZBr2)=1987 kJ/mol

∆Hºsoln(ZBr2)=14 kJ/mol

∆Hºhydr(Br–)=–336 kJ/mol

Answer 1

Z^2+(g) + 2Br^-(g) -----> ZBr2(S)   DH0 lattice = -1987 Kj/mol

ZBr2(S)+aq ----> Z^2+(aq) + 2Br^-(aq) DH0 sol = 14 kj/mol
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Z^2+(g) + 2Br^-(g) + aq ----> Z^2+(aq) + 2Br^-(aq) DH0 = -1973 kj

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Z^2+(g) + aq ---> Z^2+(aq) DH0 hyd = ?

2Br^-(g) + aq ---> 2Br^-(aq) DH0 hyd = -336*2 = -672 kj

DH0 = DH0 hydr(Z2+) + DH0 hydr(2Br-)

-1973 = x -672

x = DH0 hydr(Z2+) = -1301 kj/mol