Question

Heavy water, symbolized D2O (D = 2H) finds use as a neutron moderator in nuclear reactors. In a mixture with ordinary water, exchange of isotopes occurs according to the following equation: H2O+D2O⇌2HDO Kc = 3.86 at 298 K. When 1.50 mol of H2O is combined with 1.50 mol of D2O, what are the equilibrium amount of D2O (in moles) at 298 K? Assume the density of the mixture is constant at 1.05 g/cm3.

Answer 1

         H2O + D2O⇌2HDO

I         1.5      1.5         0

C        -x    -x             2x

E       1.5-x    1.5-x       2x

     Kc = [HDO]²/[H2O][D2O]

   3.86      = (2x)²/(1.5-x)(1.5-x)

    3.86     = {2x/(1.5-x)}²

     1.964   = 2x/1.5-x

   1.964(1.5-x) = 2x

     x = 0.743

   [H2O] = 1.5-x = 1.5-0.743 = 0.757Mole

   [D2O]   = 1.5-x    = 1.5-0.743 = 0.757Mole

[HDO] = x          = 0.743Mole