Q7.1(a) pg 235 - At 2257 K and 1.00 atm total pressure , water is 1.77 % dissociated at equilibrium by the way of the reaction 2H_{2}O -> 2H_{2} + O _{2} . Calculate K, molar G , del.(delta)G at this temperature.

This is my way of working it .

molar G = -2 X molar H_{2}O = -2 X -228.57 = 457.14 J/mol = 457.14 X 10^{3} kJ/mol

Since is equilibrium (dissociation ) , therefore del.G = 0

del.G = molar G + RT ln K

del.G =0, -molar G = RT ln K

457.14 X 10^{3} = 8.314X 2257 ln K

K = 2.63 X 10^{-11}

Which is wrong,

the answer given in book is K =2.85 X 10^{-6}

I don't know what is wrong with my calculation