Q22.82CP - Chemistry: Molecular Nature Of Matter And Change

Question

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 represents 0.10 M, calculate K.

Step-by-Step Solution

Answer

a) The balanced equation for the reaction is: ${\text{5HDO + 5H}}_{\text{2}}\to {\text{3HD + 3HDO + 2H}}_{\text{2}}\text{O}$.

b) The value of K is:$1\times {10}^{-3}<k<1\times {10}^3$ .

c) The value of K is obtained as: ${\text{K = 1.194×10}}^{\text{ - 3}}$.

Step 1: Define Elements

An element is a pure material made up entirely of atoms with the same number of protons in their nuclei, as defined by chemistry. Chemical elements, unlike chemical compounds, cannot be broken down into smaller substances by chemical reactions.

Step 2: Write the balanced equation

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:

${\text{5HDO + 5H}}_{\text{2}}\to {\text{3HD + 3HDO + 2H}}_{\text{2}}\text{O}$

Therefore, the balanced equation is: ${\text{5HDO + 5H}}_{\text{2}}\to {\text{3HD + 3HDO + 2H}}_{\text{2}}\text{O}$.

Step 3: Is the value of K greater or less than one?

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 ${\text{1×10}}^{\text{3}}{\text{<k<1×10}}^{\text{-3}}$

Step 4: Calculate the value of K

c) The value of K is obtained as:

$$\begin{gathered} {\text{[HDO]}}_{\text{left }}\text{ = 0.5M} \\ {\left[{\text{H}}_{\text{2}}\right]}_{\text{left }}\text{ = 0.5M} \\ {\left[{\text{H}}_{\text{2}}\right]}_{\text{right }}\text{ = 0.3M} \\ {\text{[HDO]}}_{\text{right }}\text{ = 0.3M} \\ \text{[HD] = 0.2M} \\ \left[{\text{H}}_{\text{2}}\text{O}\right]\text{ = 0.2M} \\ \text{K = }\frac{{\left[{\text{H}}_{\text{2}}\right]}^{\text{2}}{\text{[HD]}}^{\text{2}}{\text{[HDO]}}^{\text{3}}{\left[{\text{H}}_{\text{2}}\text{O}\right]}^{\text{2}}}{{\text{[HDO]}}^{\text{5}}{\left[{\text{H}}_{\text{2}}\right]}^{\text{5}}} \\ \text{K = }\frac{{\text{(0.3M)}}^{\text{3}}{\text{(0.2M)}}^{\text{2}}{\text{(0.3M)}}^{\text{3}}{\text{(0.2M)}}^{\text{2}}}{{\text{(0.5M)}}^{\text{5}}{\text{(0.5)}}^{\text{5}}} \\ {\text{K = 1.194×10}}^{-3} \end{gathered}$$

Therefore, the value of K is:${\text{K = 1.194×10}}^{-3}$ .