Like heavy water (${\mathbf{\text{D}}}_{\mathbf{\text{2}}}\mathbf{\text{O}}$), so-called “semi-heavy water” (HDO) undergoes H/D exchange. The scenes below depict an initial mixture of HDO and ${\mathbf{\text{H}}}_{\mathbf{\text{2}}}$reaching equilibrium.

a) Write the balanced equation for the reaction. (b) Is the value of K greater or less than? (c) If each molecule depicted represents0.10M, calculate K.

a) We can see five $H_2$molecules and five HDO molecules on the left image. We can see three $H_2$molecules, two HD molecules, three HDO molecules, and two ${\text{H}}_{\text{2}}\text{O}$molecules on the right. The following equation may be used to describe that reaction:

b) We have 10 reactant molecules and product molecules. Because not all reactants have changed into products, we have both R and P species in equilibrium. In situations like these, we can't be certain of the K value, but we can estimate it to be between ${\text{1×10}}^{\text{3}}$ and ${\text{1×10}}^{\text{-3}}$. If just reactant species were present, the K value would be less than ${\text{1×10}}^{\text{3}}$. Also, if just product species were present, we would argue that K is greater than ${\text{10}}^{\text{3}}$.

Therefore, the value is greater than one or else ${\text{1×10}}^{\text{3}}{\text{<k<1×10}}^{\text{-3}}$

c) The value of K is obtained as: