Heavy water, symbolized D2O 1D = 2H2 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 ∆ 2 HDO Kc = 3.86 at 298 K
When 1.00 mol of H2O is combined with 1.00 mol of D2O, what are the equilibrium amounts of H2O, D2O, and HDO (in moles) at 298 K? Assume the density of the mixture is constant at 1.05 gcm3.

Question

Heavy water, symbolized D2O 1D = 2H2 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 ∆ 2 HDO Kc = 3.86 at 298 K 
When 1.00 mol of H2O is combined with 1.00 mol of D2O, what are the equilibrium amounts of H2O, D2O, and HDO (in moles) at 298 K? Assume the density of the mixture is constant at 1.05 g>cm3.

Accepted Answer

Hello. In this problem, we're told that generated by chloral methane is used as an NMR solvent in the presence of decorum methane. The following exchange of hydrogen isotopes occurs. We're told the equilibrium constants 2.65 at 298 Kelvin. We are asked what are the equilibrium concentration and moles of are reacting and products? And when two moles of the deteriorated decor methane is mixed with two moles of decor methane at two or 90 kelvin. And we're told that we can assume the mixtures density is fixed at 1.33 g per cubic centimeter. Let's begin by writing the balanced reaction equation. So we have deteriorated deco methane reacting with the coral methane to give us our product where we had a hygiene isotope exchange occur. And then we want to write down what we have initially so initially have two moles of both of our reactant since and no moles of our product. The change then based on the coefficient in the balanced reaction equations is minus X. For reactant which are being consumed and plus two X. For our product. The equilibrium then we're combining the initial and the change. So this is two minus X. And two minus X. And then two X. Our equilibrium constant expression. We have our product concentration and that's squared based on the cold fish and the balanced reaction equation. And we have the concentration of our reactant The one power since they both have a coefficient of one plugging in the values from the table. We then get two X squared all over 2.0 minus x times 2.0 minus x. And we're told that that is equal to 2.65. Simplifying this. We get then two X squared over 2.0 minus x squared is equal to 2.65. So we're gonna take the square root of both sides. So we get two x Over 2 - is equal to the square root of 2.65, Which is equal 1.6-788, then have two X. Is equal to 1.62788 times two minus X, Which works out to 3. -1.62788 x. Moving our excess to one side, we now have two X plus 1.62788 X is equal to 3.25576. On the left hand side, we now have 3.6 to 788 X equals 3.25576, solving for X. X is now equal to 3.25576 by two by 3.6 to 788, Which works out to 0.89743. So then to determine our moles of our reactant, an equilibrium, we look back at our ice table And the equilibrium amount of both of our reactant is found by taking 2 -1. So we will have to -0.89743, Which works out to 1.10 moles for both our reactant and the moles of our product. As indicated, an ice table is equal to two x. So we have then two times 0.89743, Which is equal to 1.79 moles. So our moles of Product, we're sorry reacting to the equilibrium. So the duty iterated methane and the methane is equal to 1.10 moles. And the moles of our product Is equal to 1.79 moles. Keep in mind that we can plug just molds into our equilibrium constant expression because we would have concentration squared in the numerator, divided by concentration squared in the denominator. Sorry, units of volume would cancel. And so we can safely ignore volume. In this case. Thanks for watching. Hope this helps

Verified Solution

Key Concepts

Chemical Equilibrium

Isotope Exchange Reactions

Stoichiometry